Vector Addition Terms: Resultant Vs. Equilibrant

Fill in the blanks: The resultant vector and equilibrant vector have (the) _________ magnitudes and point in the _________ direction. Solution Video

1. Same, Same
2. Different, Same
3. Same, Opposite
4. Different, Opposite
5. None of the above

The resultant vector and equilibrant vector have the same magnitudes and point in the opposite direction.

(The resultant vector is the net sum of more than one vector, and the equilibrant vector is the opposite vector that would cancel another vector out, equal in magnitude and in the opposite direction.)

Solving Tension Forces and the Resultant Vector using Sine Law

An 8kg mass is suspended in equilibrium from two massless cables. Cable 1 is attached to a vertical wall at an angle of 10˚ to the horizontal while Cable 2 is attached to the ceiling at an angle of 50˚ to the horizontal. Calculate the tension in each cable. Solution

Hint Clear Info

T₁ = T₂ =

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N

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Velocity Vector Addition Tip-to-Tail

A canoe is being paddled 5km/h upstream against a current of 1km/h. A 2km/h crosswind is blowing against the canoe at an angle of 30˚ East from the path of the canoe. Calculate the resultant velocity of the canoe relative to the ground. Solution Video

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km/h

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Linear Combination using Substitution or Elimination of Equations

Show that the vector $\vec{x} = \langle 5, 19 \rangle$ can be written as a linear combination of the set {(5, 3), (-2, 2)} Solution

Hint Clear Info

(5, 3) + (-2, 2) = (5, 19)

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Lines in 3-Space: Vector, Parametric, and Symmetric Equations

Calculate the distance between the two points V(3, 2, -4) and W(11, -7, 8) in 3-space. Solution

d =

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units

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Linear Combination in R³

Which of the following is a unit vector? Solution

1. $\left(\dfrac{1}{\sqrt{3}}, \dfrac{1}{\sqrt{3}}, \dfrac{1}{\sqrt{3}} \right)$
2. $\left( -1, -1, -1 \right)$
3. $\left(\dfrac{1}{3}, \dfrac{1}{3}, \dfrac{1}{3} \right)$
4. $\left(1, 1, 1 \right)$

Remember that unit vectors have magnitude equal to 1. $\begin{align} |\vec{i}| + |\vec{j}| + |\vec{k}| & = 1 \\ \\ \end{align}$ The only one whose magnitude equals 1 is: $\begin{align} & = \sqrt{\left(\dfrac{1}{\sqrt{3}}\right)^2 + \left(\dfrac{1}{\sqrt{3}}\right)^2 + \left(\dfrac{1}{\sqrt{3}}\right)^2} \\ \\ & = \sqrt{\dfrac{1}{3} + \dfrac{1}{3} + \dfrac{1}{3}} \\ \\ & = \sqrt{\dfrac{3}{3}} \\ \\ & = 1 \\ \\ \end{align}$ The rest of the choices will not equal 1.

Write the 3-dimensional vector $\langle 4, 2, -1 \rangle$ as a linear combination of the unit vectors: $\vec{i}$, $\vec{j}$, and $\vec{k}$ Solution

Simplifying with $\vec{i}$, $\vec{j}$, and $\vec{k}$

Given the following information, $-3\vec{m} + 2\vec{n}$ equals $5\vec{i} - 6\vec{j} - 16\vec{k}$. Solution $\begin{align} \vec{m} & = -\vec{i} + 4\vec{j} - 6\vec{k} \\ \vec{n} & = \vec{i} + 3\vec{j} - \vec{k} \end{align}$

1. True
2. False

Use substitution... $= 5\vec{i} - 6\vec{j} + 16\vec{k}$

Linear Combination in R³

Solve for $x$, y, and z in the following equation. Solution $3(4, 2y, 3z) - 2(2x, y, -7) = (2x, -5, 8)$

Hint Clear Info

x = y = ━━ z = ━━

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Linear Combination in R³

Find the value of $x$ so that the vector $\vec{c} = \langle 4, -x, 7 \rangle$ can be written as a linear combination of the vectors $\vec{a} = \langle 2, 4, 3 \rangle$ and $\vec{b} = \langle -1, 2, 1 \rangle$. Solution

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━━

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The Dot Product

The result of a dot product is a(n): Solution

1. Vector
2. Scalar
3. Absolute value
4. Positive value always
5. Force

The dot product is also known as the scalar product.

Two vectors are parallel if the ________ of the vectors equals one, and perpendicular if the ________ of the vectors equals zero. Solution

1. Sum
2. Difference
3. Dot product
4. Cross product
5. Parameter

Two vectors are parallel if the dot product of the vectors equals one, 1.

Two vectors are perpendicular if the dot product of the vectors equals zero, 0.

Calculate the dot product of the unit vector k, $\ \vec{k}\bullet\vec{k}$ Solution

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Which of the following dot product properties is incorrect? Solution

1. $\vec{a} •\vec{b} = \vec{b} •\vec{a}$
2. $\vec{a}•\vec{a} = \Big|\vec{a}\Big|^2$
3. $\vec{a}•(\vec{b} + \vec{c}) = \vec{a}•\vec{b} + \vec{a}•\vec{c}$
4. $n(\vec{a} •\vec{b}) = (n)(\vec{a}) •\vec{b} = (n)(\vec{a}) •(n)(\vec{b})$
5. None are incorrect

All answer choices are true, except for... $n(\vec{a} •\vec{b}) = (n)(\vec{a}) •\vec{b} = (n)(\vec{a}) •(n)(\vec{b})$ Which should be... $n(\vec{a} •\vec{b}) = (n)(\vec{a}) •\vec{b} = \vec{a} •(n)(\vec{b})$

The Dot Product of Vectors

Calculate the dot product of the following

In 2-dimensions: $\vec{a}$ = (1, 2) and $\vec{b}$ = (-3, 4) Solution

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In 3-dimensions: $\vec{a}$ = (1, -2, 3) and $\vec{b}$ = (4, 5, -6) Solution

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Determine the angle θ between the three-dimensional vectors from part 'b' above. Solution

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˚

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Perpendicular Vectors in 3-Space using Dot Product

Which vector is perpendicular to both of the following vectors? Solution $\begin{align} \vec{a} & = \langle 4, 2, -1 \rangle \\ \vec{b} & = \langle 5, 6, 3 \rangle \\ \end{align}$

1. $\!\langle 20, 12, -3 \rangle$
2. $\!\langle 9, 8, 2 \rangle$
3. $\!\langle -\frac{12}{17}, 1, -\frac{14}{17} \rangle$
4. $\!\langle \frac{7}{12}, -3, -\frac{11}{4} \rangle$
5. $\!\langle \frac{4}{5}, \frac{1}{3}, -\frac{1}{3} \rangle$

Scalar Projection Equation

Given the following vectors, the scalar projection of $\vec{a}$ on $\vec{b}$ equals Solution $\begin{align} \vec{a} & = \langle 5, 3, -1 \rangle \\ \vec{b} & = \langle 1, 2, 6 \rangle \\ \end{align}$

1. $\! \dfrac{(5)(1)\ -\ (3)(2)\ -\ (-1)(6)}{\sqrt{(5)^2\ -\ (3)^2\ -\ (-1)^2}}$
2. $\! \dfrac{(5)(1)\ +\ (3)(2)\ +\ (-1)(6)}{\sqrt{(5)^2\ +\ (3)^2\ +\ (-1)^2}}$
3. $\! \dfrac{(5)(1)\ +\ (3)(2)\ +\ (-1)(6)}{\sqrt{(1)^2\ -\ (2)^2\ -\ (6)^2}}$
4. $\! \dfrac{(5)(1)\ +\ (3)(2)\ +\ (-1)(6)}{\sqrt{(1)^2\ +\ (2)^2\ +\ (6)^2}}$
5. $\! \dfrac{(5)(1)\ +\ (3)(2)\ +\ (-1)(6)}{(1)^2\ +\ (2)^2\ +\ (6)^2}$

$\begin{align} & = \dfrac{(a_1)(b_1)\ +\ (a_2)(b_2)\ +\ (a_3)(b_3)}{\sqrt{(b_1)^2\ +\ (b_2)^2\ +\ (b_3)^2}} \\ \\ & = \dfrac{(5)(1)\ +\ (3)(2)\ +\ (-1)(6)}{\sqrt{(1)^2\ +\ (2)^2\ +\ (6)^2}} \\ \\ & = \dfrac{5}{\sqrt{41}} \bbox[Bisque, 3pt]{ × \dfrac{\sqrt{41}}{\sqrt{41}} } \\ \\ & = \dfrac{5\sqrt{41}}{41} \\ \\ \end{align}$ Notice only the magnitude on the bottom changes in a scalar projection since a•b = b•a... $\vec{a} \ on \ \vec{b} = \dfrac{\vec{a}\ \cdot\ \vec{b}}{|\vec{b}|}$

$\vec{b} \ on \ \vec{a} = \dfrac{\vec{a}\ \cdot\ \vec{b}}{|\vec{a}|}$

Vector Projections

The formula for a vector projection of a on b is written as: Solution

1. $\! \dfrac{\vec{a}\cdot \vec{b}}{|\vec{a}|} × \dfrac{\vec{a}}{|\vec{b}|}$
2. $\! \dfrac{\vec{a}\cdot \vec{b}}{|\vec{b}|} × \dfrac{\vec{a}}{|\vec{b}|}$
3. $\! \dfrac{\vec{a}\cdot \vec{b}}{|\vec{a}|} × \dfrac{\vec{b}}{|\vec{b}|}$
4. $\! \dfrac{\vec{a}\cdot \vec{b}}{|\vec{b}|} × \dfrac{\vec{b}}{|\vec{b}|}$
5. $\! \dfrac{\vec{a}\cdot \vec{b}}{|\vec{b}|} × \dfrac{\vec{a}}{|\vec{a}|}$

Notice that the vector projection is the scalar projection that gets... $\dfrac{\vec{a}\cdot \vec{b}}{|\vec{b}|}$ ...multiplied by the unit vector (which gives the direction) of the second vector in the statement that gets projected onto, i.e. "a on b", in the form: $\dfrac{\vec{b}}{|\vec{b}|}$

Determine the Direction Vector using Vector Projection

Determine the direction vector given the following vectors of v on w: Solution $\begin{align} \vec{v} & = \langle 0, 1, 2 \rangle \\ \vec{w} & = \langle 3, 4, 5 \rangle \\ \end{align}$

1. $\! \dfrac{7}{25}·\langle 3, 4, 5 \rangle$
2. $\! \dfrac{14}{5}·\langle 3, 4, 5 \rangle$
3. $\! \dfrac{25}{7}·\langle 3, 4, 5 \rangle$
4. $\! \dfrac{14}{5}·\langle 0, 1, 2 \rangle$
5. $\! \dfrac{7}{25}·\langle 0, 1, 2 \rangle$

Simplified 2 × 2 Review of Matrices for Cross Products

The following matrix can be simplified to the determinant: $\ (a)(d) - (b)(c)$ Solution $\begin{vmatrix} a & b \\ c & d \end{vmatrix}$

1. True
2. False

True: $\begin{vmatrix} a & b \\ c & d \end{vmatrix}$

$= (a)(d) - (b)(c)$

Calculate the cross product $\vec{m} × \vec{n}$ of the vectors: $\vec{m} = \langle 2, -5 \rangle, \vec{n} = \langle 3, 2 \rangle$ Solution

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2 × 2 matrix: $\begin{vmatrix} 2 & -5 \\ 3 & 2 \end{vmatrix}$ Calculate the determinant: $\begin{align} & = (a)(d) - (b)(c) \\ \\ & = (2)(2) - (-5)(3) \\ \\ & = 4 + 15 \\ \\ & = 19 \\ \\ \end{align}$

Calculate the magnitude $\Big|\vec{m} × \vec{n}\Big|$ of the vectors: $\vec{m} = \langle 2, -5 \rangle, \vec{n} = \langle 3, 2 \rangle$ Solution

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The magnitude of 2 × 2 vectors can only be calculated with the formula: $\begin{align} & \Big|\vec{m}\Big| \Big|\vec{n}\Big| \sin θ \\ \\ \end{align}$ Where the magnitudes are: $\begin{align} \Big|\vec{m}\Big| & = \sqrt{2^2 + (-5)^2} \\ \\ \Big|\vec{m}\Big| & = \sqrt{29} \\ \\ \Big|\vec{n}\Big| & = \sqrt{3^2 + 2^2} \\ \\ \Big|\vec{n}\Big| & = \sqrt{13} \\ \\ \end{align}$ The angle is calculated with the cosine of the dot product of m and n, divided by the product of the magnitudes of m and n: $\begin{align} \cos θ & = \dfrac{\vec{m}•\vec{n}}{\Big|\vec{m}\Big|\Big|\vec{n}\Big|} \\ \\ \cos θ & = \dfrac{(2)(3) + (-5)(2)}{\sqrt{29}\sqrt{13}} \\ \\ \cos θ & = \dfrac{-4}{\sqrt{377}} \\ \\ θ & = \cos^{-1}\left( \dfrac{-4}{\sqrt{377}} \right) \\ \\ θ & = 101.9˚ \\ \\ \end{align}$ Then calculate this thing finally... $\begin{align} \Big|\vec{m} × \vec{n}\Big| & = \Big|\vec{m}\Big| \Big|\vec{n}\Big| \sin θ \\ \\ & = \sqrt{29} \sqrt{13} \sin 101.9 \\ \\ & = \sqrt{377} \sin 101.9 \\ \\ & = 19 \\ \\ \end{align}$

Match the vectors into their correct property. Solution

$\vec{v_1} = \langle 2, -6 \rangle$
$\vec{v_2} = \langle 1, -3 \rangle$

$\vec{v_1} = \langle -3, 2 \rangle$
$\vec{v_2} = \langle 3, 4 \rangle$

Parallel

Perpendicular

Vectors are perpendicular in 3-space unless the cross product equals zero. Calculate the cross products using the 2 × 2 matrix (not shown).. $\begin{vmatrix} a & b \\ c & d \end{vmatrix}$

$= (a)(d) - (b)(c)$

Cross Product Theory

The equation stated below Solution $\vec{a}\ ×\ \big(\vec{e}\ +\ \vec{c}\big)\ =\ \vec{a}\ ×\ \vec{e}\ +\ \vec{a}\ ×\ \vec{c}$

1. Uses scalar multiplication
2. Uses the commutative property
3. Uses the distributive property
4. Uses the associative propert
5. Is false

This is the distributive property.

Theory of Scalar and Vector Properties of Cross Product, Dot Product, etc.

Which of the following statements is incorrect? Solution

1. Adding vectors produces a vector
2. Cross product of a scalar and a vector produces a vector
3. Cross product produces a vector
4. Dot product of a vector and a scalar produces a scalar
5. Dot product of a vector and a vector produces a scalar
6. Dot product produces a scalar
7. Product of a scalar and a vector produces a vector

"Cross product of a scalar and a vector produces a vector"; there is no such thing.

Any dot product produces a scalar.

Adding vectors produces a vector because they all have directions.

Product of a scalar and a vector produces a vector because the scalar is a multiplication factor of the vector, which maintains its direction.

Cross Product Theory: Direction and the Order of the Cross

Given the vectors below, the cross products are equal: $\vec{v_1}\ ×\ \vec{v_2}\ =\ \vec{v_2}\ ×\ \vec{v_1}$ Solution $\begin{align} \vec{v_1} &= \big\langle 1, 6, 7 \big\rangle \\ \vec{v_2} &= \big\langle 2, 7, -1 \big\rangle \\ \end{align}$

1. True
2. False

First using the right-hand-rule will show the cross products point in different directions. To prove numerically:

Cross product formula in the form $\big\langle a_1, a_2, a_3 \big\rangle$ and $\big\langle b_1, b_2, b_3 \big\rangle$: $\vec{a} × \vec{b} = \big\langle a_2b_3 - a_3b_2,\ a_3b_1 - a_1b_3,\ a_1b_2 - a_2b_1 \big\rangle$ Check to see each cross product: $\begin{align} \vec{v_1}\ ×\ \vec{v_2} &= (1, 6, 7) \ ×\ (2, 7, -1) \\ &= (6)(-1) - (7)(7), \ (7)(2) - (1)(-1), \ (1)(7) - (6)(2) \\ &= \big\langle -55, 15, -5 \big\rangle \end{align}$ And, $\begin{align} \vec{v_2}\ ×\ \vec{v_1} &= (2, 7, -1) \ ×\ (1, 6, 7) \\ &= (7)(7) - (-1)(6), \ (-1)(1) - (2)(7), \ (2)(6) - (7)(1) \\ &= \big\langle 55, -15, 5 \big\rangle \end{align}$ Clearly they are not equal: $\vec{v_1}\ ×\ \vec{v_2}\ ≠\ \vec{v_2}\ ×\ \vec{v_1}$ This is because they point in opposite directions. Use the right-hand-rule to determine the direction the cross product points.

Cross Product of Two Vectors

The cross product of the following two vectors can be calculated as... Solution $\begin{align} \vec{a} & = \langle a_1, a_2, a_3 \rangle \\ \vec{b} & = \langle b_1, b_2, b_3 \rangle \\ \end{align}$

1. $\vec{a} × \vec{b} = \left \langle a_1b_2 - a_2b_1,\ a_2b_3 - a_3b_2,\ a_3b_1 - a_1b_3 \right \rangle$
2. $\vec{a} × \vec{b} = \left \langle a_2b_3 - a_3b_2,\ a_3b_1 - a_1b_3,\ a_1b_2 - a_2b_1 \right \rangle$
3. $\vec{a} × \vec{b} = \left \langle a_1b_1 - a_2b_2,\ a_2b_2 - a_3b_3,\ a_3b_3 - a_1b_1 \right \rangle$
4. $\vec{a} × \vec{b} = \left \langle a_1b_2 + a_2b_1,\ a_2b_3 + a_3b_2,\ a_3b_1 + a_1b_3 \right \rangle$
5. $\vec{a} × \vec{b} = \left \langle a_2b_3 + a_3b_2,\ a_3b_1 + a_1b_3,\ a_1b_2 + a_2b_1 \right \rangle$

This equation is hard to remember: $\vec{a} × \vec{b} = \big \langle a_2b_3 - a_3b_2,\ a_3b_1 - a_1b_3,\ a_1b_2 - a_2b_1 \big \rangle$

Cross Product of Two Vectors

It is generally considered conventient to calculate the cross product using matrix notation...

Using matrices to calculate the cross product of two vectors results in which of the following computations? Solution $\begin{vmatrix} \vec{i} & \vec{j} & \vec{k} \ \\ a_1 & a_2 & a_3 \ \\ b_1 & b_2 & b_3 \ \end{vmatrix} \ $ $\begin{matrix} \vec{i} & \vec{j} \\ a_1 & a_2 \\ b_1 & b_2 \end{matrix}$

1. $\vec{a} × \vec{b} = \left \langle (a_1b_2 - a_2b_1)\vec{i},\ (a_2b_3 - a_3b_2)\vec{j},\ (a_3b_1 - a_1b_3)\vec{k} \right \rangle$
2. $\vec{a} × \vec{b} = \left \langle (a_2b_3 - a_3b_2)\vec{i},\ (a_3b_1 - a_1b_3)\vec{j},\ (a_1b_2 - a_2b_1)\vec{k} \right \rangle$
3. $\vec{a} × \vec{b} = \left \langle (a_1b_1 - a_2b_2)\vec{i},\ (a_2b_2 - a_3b_3)\vec{j},\ (a_3b_3 - a_1b_1)\vec{k} \right \rangle$
4. $\vec{a} × \vec{b} = \left \langle (a_1b_2 + a_2b_1)\vec{i},\ (a_2b_3 + a_3b_2)\vec{j},\ (a_3b_1 + a_1b_3)\vec{k} \right \rangle$
5. $\vec{a} × \vec{b} = \left \langle (a_2b_3 + a_3b_2)\vec{i},\ (a_3b_1 + a_1b_3)\vec{j},\ (a_1b_2 + a_2b_1)\vec{k} \right \rangle$

Cross product using matrix: $\begin{vmatrix} \vec{i} & \vec{j} & \vec{k} \ \\ a_1 & a_2 & a_3 \ \\ b_1 & b_2 & b_3 \ \end{vmatrix} \ $ $\begin{matrix} \vec{i} & \vec{j} \\ a_1 & a_2 \\ b_1 & b_2 \end{matrix}$ $\vec{a} × \vec{b} = \left \langle [a_2b_3 - a_3b_2]\vec{i},\ [a_3b_1 - a_1b_3]\vec{j},\ [a_1b_2 - a_2b_1]\vec{k} \right \rangle$ Or you could use the form: $\begin{vmatrix} i_1 \ & \ j_1 \ & \ k_1 \ \\ i_2 \ & \ j_2 \ & \ k_2 \ \end{vmatrix} \ $ $\begin{matrix} i_1 \ & \ j_1 \\ i_2 \ & \ j_2 \end{matrix}$ $\vec{a} × \vec{b} = \Big\langle [j_1k_2 - k_1j_2],\ [k_1i_2 - i_1k_2],\ [i_1j_2 - j_1i_2] \Big\rangle$

Determine a vector that is orthogonal to the following vectors by calculating the cross product determinant of the matrix. Solution $\begin{vmatrix} \vec{i} & \vec{j} & \vec{k} \\ 1 \ & 2 \ & 3 \\ 4 \ & 5 \ & 6 \end{vmatrix}$

Hint Clear Info

$\vec{i} +$ $\vec{j} -$ $\vec{k}$

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$\begin{vmatrix} \vec{i} & \vec{j} & \vec{k} \ \\ a_1 & a_2 & a_3 \ \\ b_1 & b_2 & b_3 \ \end{vmatrix} \ $ $\begin{matrix} \vec{i} & \vec{j} \\ a_1 & a_2 \\ b_1 & b_2 \end{matrix}$ $\begin{vmatrix} \vec{i} & \vec{j} & \vec{k} \\ 1 \ & 2 \ & 3 \\ 4 \ & 5 \ & 6 \end{vmatrix} \ \ $ $\begin{matrix} \vec{i} & \vec{j} \\ 1 \ & 2 \\ 4 & 5 \end{matrix}$ $\begin{align} \vec{a} × \vec{b} &= \left \langle (a_2b_3 - a_3b_2)\vec{i},\ (a_3b_1 - a_1b_3)\vec{j},\ (a_1b_2 - a_2b_1)\vec{k} \right \rangle \\ &= \left \langle \big[(2)(6) - (3)(5)\big]\vec{i},\ \big[(3)(4) - (1)(6)\big]\vec{j},\ \big[(1)(5) - (2)(4)\big]\vec{k} \right \rangle \\ &= \left \langle (12 - 15)\vec{i},\ (12 - 6)\vec{j},\ (5 - 8)\vec{k} \right \rangle \\ &= \left \langle -3\vec{i},\ 6\vec{j},\ -3\vec{k} \right \rangle \\ \end{align}$

Combining the Cross Product and Dot Product

Lets see what happens when there are different combinations of dot products and cross products...

What's calculated first, cross product or dot product? Solution $\langle 2, 5, -3 \rangle • \langle 1, -2, 4 \rangle × \langle -7, 3, 2\rangle$

1. Cross product
2. Dot product

Do cross product first, no matter which way it is written: $\langle 2, 5, -3 \rangle • \langle 1, -2, 4 \rangle × \langle -7, 3, 2\rangle$ Or $\langle 2, 5, -3 \rangle × \langle 1, -2, 4 \rangle • \langle -7, 3, 2\rangle$

Calculate $\langle 2, 5, -3 \rangle • \langle 1, -2, 4 \rangle × \langle -7, 3, 2\rangle$ Solution

Hint Clear Info

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Do cross product first. $\begin{vmatrix} \vec{i} & \vec{j} & \vec{k} \\ 1 & -2 & 4 \\ -7 & 3 & 2 \end{vmatrix} \ \ $ $\begin{matrix} \vec{i} & \vec{j} \\ 1 & -2 \\ -7 & 3 \end{matrix}$ $\begin{align} & = (-2)(2) - (4)(3), (4)(-7) - (1)(2), (1)(3) - (-2)(-7) \\ \\ & = -4 - 12, -28 - 2, 3 - 14 \\ \\ & = {\color{WildStrawberry}\langle -16, -30, -11 \rangle} \\ \\ \end{align}$ Then do the dot product $\begin{align} & \langle 2, 5, -3 \rangle • \langle 1, -2, 4 \rangle × \langle -7, 3, 2\rangle \\ \\ & = \langle 2, 5, -3 \rangle • {\color{WildStrawberry}\langle -16, -30, -11 \rangle} \\ \\ & = (2)(-16) + (5)(-30) + (-3)(-11) \\ \\ & = -32 - 150 + 33 \\ \\ & = -149 \\ \\ \end{align}$

Calculate $\langle 1, -2, 3 \rangle × \Big[ \langle -2, 2, 2 \rangle × \langle -3, -2, 5\rangle \Big ]$ Solution

Hint Clear Info

⟨ , , ⟩

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Need to calculate the cross product in brackets first $\Big[ \langle -2, 2, 2 \rangle × \langle -3, -2, 5\rangle \Big ]$ ... $\begin{vmatrix} \vec{i} & \vec{j} & \vec{k} \ \\ a_1 & a_2 & a_3 \ \\ b_1 & b_2 & b_3 \ \end{vmatrix} \ $ $\begin{matrix} \vec{i} & \vec{j} \\ a_1 & a_2 \\ b_1 & b_2 \end{matrix}$ $\begin{vmatrix} \vec{i} & \vec{j} & \vec{k} \\ -2 & 2 & 2 \\ -3 & -2 & 5 \end{vmatrix} \ \ $ $\begin{matrix} \vec{i} & \vec{j} \\ -2 & 2 \\ -3 & -2 \end{matrix}$ $\begin{align} &= \left\langle (a_2b_3 - a_3b_2),\ (a_3b_1 - a_1b_3),\ (a_1b_2 - a_2b_1) \right\rangle \\ \\ &= \left \langle [(2)(5) - (2)(-2)],\ [(2)(-3) - (-2)(5)],\ [(-2)(-2) - (2)(-3)] \right \rangle \\ \\ &= \left \langle 10 + 4,\ -6 + 10,\ 4 + 6 \right \rangle \\ \\ &= \left \langle 14,\ 4,\ 10 \right \rangle \\ \\ \end{align}$ So we can simplify to, $\begin{align} & \langle 1, -2, 3 \rangle × \Big[ \langle -2, 2, 2 \rangle × \langle -3, -2, 5\rangle \Big ] \\ \\ &= \langle 1, -2, 3 \rangle × \langle 14, 4, 10 \rangle \\ \\ \end{align}$ Finally calculate the cross product $\langle 1, -2, 3 \rangle × \langle 14, 4, 2 \rangle$ $\begin{vmatrix} \vec{i} & \vec{j} & \vec{k} \\ 1 & -2 & 3 \\ 14 & 4 & 10 \end{vmatrix} \ \ $ $\begin{matrix} \vec{i} & \vec{j} \\ 1 & -2 \\ 14 & 4 \end{matrix}$ $\begin{align} &= \left\langle (a_2b_3 - a_3b_2),\ (a_3b_1 - a_1b_3),\ (a_1b_2 - a_2b_1) \right\rangle \\ \\ &= \left \langle [(-2)(10) - (3)(4)],\ [(3)(14) - (1)(10)],\ [(1)(4) - (-2)(14)] \right \rangle \\ \\ &= \left \langle [-20 - 12],\ [42 - 10],\ [4 + 28] \right \rangle \\ \\ &= \left \langle -32,\ 32,\ 32 \right \rangle \\ \\ \end{align}$

Combining the Cross Product and Dot Product

The following equation is used to calculate the volume of a parallelepiped. Solution $\Big| \ \vec{a}\ ·\ \vec{b}\ ×\ \vec{c}\ \Big|$

1. True
2. False

Do the cross product $\vec{b}\ ×\ \vec{c}$ first, followed by the dot product of that cross product you just did. This gives you the area of a parallelogram (the base of the parallelepiped in this case). $\Big| \ \vec{a}\ ·\ \vec{b}\ ×\ \vec{c}\ \Big|$ Then this area of the bast is dotted with the 'height' of the parallelepiped to give the volume... use the absolute value of this.

Given the vectors:   $\vec{a} = \langle 2, 3, -1 \rangle$, $\vec{b} = \langle 3, -2, 0 \rangle$, and $\vec{c} = \langle 1, -3, 4 \rangle$, calculate the volume of the parallelepiped. Solution

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units³

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Determine if the following vectors are coplanar, or non-coplanar. (Show your work in your own notes) Solution $\begin{align} \vec{a} & = \langle 1, 2, 3 \rangle \\ \vec{b} & = \langle -1, -2, 1 \rangle \\ \vec{c} & = \langle 3, 5, 2\rangle \\ \end{align}$

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Vectors are coplanar if the volume = 0. (Do cross product first) $\begin{align} Volume & = \Big| \ \vec{a}\ ·\ \vec{b}\ ×\ \vec{c}\ \Big| \\ \\ & = \Big| \ \vec{a}\ ·\ (-2)(2) - (1)(5), (1)(3) - (-1)(2), (-1)(5) - (-2)(3) \\ \\ & = \Big| \langle 1, 2, 3 \rangle ·\ \langle -9, 5, 1 \rangle \Big| \\ \\ & = \Big| (1)(-9) + (2)(5) + (3)(1) \Big| \\ \\ & = 4 \\ \\ \end{align}$ Therefore the vectors are non-coplanar.

Application of Cross Product Using Torque

A person applies a 10N force (F) to push open a door. If this person pushes on the door with a radius (r) of 0.45m from the hinges applying a torque (T) of 3.45J, determine the angle in degrees, of the person's force with the door. Solution $\vec{T} = \big|\,\vec{r}\,\big|\,\big|\,\vec{F}\,\big|\,sinθ$

θ =

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˚

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$θ$ is the angle between the force (F) and the object that is pushed. $\begin{align} \vec{T} & = \big|\,\vec{r}\,\big|\,\big|\,\vec{F}\,\big|\,sinθ \\ \\ 3.45 & = |0.45||10| \sinθ \\ \\ \sinθ & = 0.766 \\ \\ θ & = \sin^{-1}(0.766) \\ \\ θ & = 50˚ \\ \\ \end{align}$

Parametric Equations and Direction Vectors

Determine the direction vector given the following parametric equations: $x = 2 + 5m,\quad y = -10 - 3m,\quad z = 4 + \frac{7}{4}m$ Solution

1. 2, -10, 4
2. -2, 10, -4
3. -5, 3, 7/4
4. 7, -13, 23/4
5. 5, -3, 7/4

The coefficients on 'm'... 5, -3, 7/4

Lines in 2-Space: Vector and Parametric Equations

A line passes through points A(5, 2) and B(1, 10). Determine a direction vector $\vec{m}$, for this line. (Use proper form with parentheses) Solution

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Lines in 2-Space: Vector and Parametric Equations

A line passing through the point N(1, 5) has a direction vector $\vec{m}$ = (4, -5). Determine the vector equation. (Use proper form with parentheses) Solution

$\vec{r} =$

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(   ,   )   +   t (   ,   )

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Lines in 2-Space: Vector and Parametric Equations

Determine (and in your notes, sketch) two points on the line represented by the vector equation $\vec{r}$ = (1, 4) + t(-2, -2)... for -5 ≤ t ≤ 5. (Use proper form with parentheses) [2] Solution

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Lines in 2-Space: Determine the Angle Between Direction Vectors

Determine the angle $\theta$ between the given direction vectors $\vec{m}$ = (8, 3) and $\vec{r}$ = (-3, 5) Solution

θ =

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Lines in 2-Space: Vector and Parametric Equations

Given the vector equation $\vec{r}$ = (2, 3) + t(3, -4)

Determine if point C(14, -13) is on the line. (Show your work in your notes) Solution

1. Yes it is on the line
2. No it's not on the line

Determine the coordinate where the line crosses the y-axis. Solution

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$\bigg( \quad , \dfrac{\quad}{\quad} \bigg)$

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Lines in 2-Space: Cartesian Equations

Two lines are parallel if their normals are:

1. Inverse
2. Perpendicular
3. Reciprocals
4. Scalar Multiples
5. Dot Products

The cartesian equation of a line in 2-space is written in the form:

1. Ax + By + C = 0
2. y = mx + b
3. By = Ax + C
4. Ax² + By + C = 0
5. Ax² + Bx + C = 0

The 'normal' is the same as the: Solution

1. Positive reciprocal
2. Perpendicular line
3. Regular
4. Slope
5. X-intercept

Perpendicular line

Lines in 2-Space: Cartesian Equations

Given the equations of two lines, 8 + 4y = 2x and 9 - 4y + $\color{Magenta}{k}\,$x = 0, for what value of $\color{Magenta}{k}$ are the lines perpendicular? Solution

k =

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Lines in 3-Space: Vector, Parametric, and Symmetric Equations

Two vector equations represent the same line when they have only:

1. Direction vectors that are scalar multiples of eachother
2. The same slope
3. A point of intersection
4. Direction vectors that are scalar multiples of eachother and points of intersection
5. Two different vector equations can never represent the same line

Lines in 3-Space: Vector, Parametric, and Symmetric Equations

The ________ of two direction vectors gives you a third direction vector that is perpendicular to the first two.

1. Sum
2. Difference
3. Dot product
4. Cross product
5. Parameter

Lines in 3-Space: Vector, Parametric, and Symmetric Equations

Find the value of k given the two perpendicular direction vectors: $\vec{G}$ = (4, -1, 6) and $\vec{H}$ = (3, -12k, 2k) Solution

k =

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━━

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Lines and Planes

A line can intersect a plane: (check all that apply) Solution

1. 0 times
2. 1 time
3. 2 times
4. 3 times
5. Infinite times

A line can intersect a plane an infinite number of times when the line is on the plane.

Lines and Planes: No Point of Intersection

A line that does not intersect a plane, is proven when: Solution

1. The line is orthogonal (perpendicular) to the normal of the plane, and not on the plane.
2. The line is parallel to the plane
3. When the dot product of the line and the plane is equal to zero
4. There are many points of intersection
5. None of the above

When the vector line and the normal of the plane are perpendicular, then the vector line is parallel to the plane.

Lines and Planes: Line on the Plane

A line is verified to be on a plane when: Solution

1. There is only one possible value for the parameter (scalar multiple)
2. When the dot product of the line and the plane is equal to zero
3. The line is parallel to the plane
4. There are infinite values for the parameter (scalar multiple), e.g. 0m = 0
5. None of the above

There are infinite values for the scalar multiple when the line is 'on' the plane.

Lines and Planes: Finding a Point of Intersection

Determine the point of intersection between the vector $\vec{v}$ = (1, -2, 3) + m(2, 3, 1) and the plane defined below. (Show exact values and reduce fully) Solution 2x - y + 3z + 4 = 0

Hint Clear Info

$\bigg( \dfrac{\quad}{\quad}, \dfrac{\quad}{\quad}, \dfrac{\quad}{\quad} \bigg)$

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Coplanar Vectors in 2-Space

In 2-space, any two noncollinear (not lying on the same straight line) vectors are coplanar. Solution

1. True
2. False

In 2-space, the vectors are coplanar.

Lines and Planes: Finding the Point of Intersection

Determine the point(s) of intersection of the following lines: $\vec{r_1}$ = (0, 1, 2) + m(1, -1, 2) and $\vec{r_2}$ = (-3, 4, -4) + n(0, 1, 2) Solution

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(   ,   ,   )

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The Intersection of a Vector Line and a Plane

Determine the point of intersection between the plane z = 5, and the vector line: x = -1 + s, y = 3 + 2s, and z = 6 + 7s. (Reduce fully) Solution

Hint Clear Info

$\bigg( \dfrac{\quad}{\quad}, \dfrac{\quad}{\quad}, \quad \bigg)$

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Set the two 'z' equations equal, given that z = 5 and z = 6 + 7s... solve for 's', $s = - \dfrac{1}{7}$ Substitute 's' value into the parametric equations to solve for 'x', 'y' and 'z'.
...
The point of intersection between the vector and the plane = (x, y, z) $= \left( \dfrac{-8}{7}, \dfrac{19}{7}, 5 \right)$

Intersection of Lines

Skew lines exist in R² and R³. Solution

1. True
2. False

Skew lines are lines that are not parallel and do not intersect. This is not possible in 2D, it is only possible in 3D.

The Intersection of Planes

It is possible for two or more planes to intersect at... (More than one answer: check all that apply) Solution

1. 1 point
2. 2 points
3. 3 points
4. 4 points
5. Infinite points

If the planes overlap, they would intersect at an infinite number of points. Three planes intersect at one point.

Checking a Solution to a System of Two Equations with Three Unknowns

Determine if the point, (2, 7, 1) is a solution to the system, hence if this is the point of intersection of the following planes: Solution $\begin{align} x - 2y + 2z & = -10 \\ \\ 5x + 2y - z & = 12 \end{align}$

1. Yes it is a point of intersection (POI) of the given planes
2. No it is not a point of intersection (POI) of the given planes
3. It cannot be determined with the given information

Sub (2, 7, -2) into both equations and check if the solutions are the same. If they are not the same, then the given point is not a point of intersection, i.e. not a solution to the system. $\begin{align} 2 - 2(7) + 2(1) & = -10 \\ \\ \\ \\ 5(2) + 2(7) - (1) & = 23 \end{align}$ Therefore the point, (2, 7, 1) is not a point of intersection (POI) of the given planes.

Solving a System with Three Equations and Three Unknowns

Solve the system, hence determine the point of intersection. (Use exact values and reduce fully) Solution $\begin{align} 2x + 3y - z & = 2 \\ \\ 3x + y + 2z & = 5 \\ \\ x - 2y + z & = 4 \end{align}$

Hint Clear Info

$\bigg( \dfrac{\quad}{\quad}, \dfrac{\quad}{\quad}, \dfrac{\quad}{\quad} \bigg)$

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Solve using a combination of substitution and elimination, with 3 equations and 3 unknowns... (not shown)

$\bigg( \dfrac{33}{14}, \dfrac{-15}{14}, \dfrac{-1}{2} \bigg)$

The Intersection of Planes: Parallel Planes

Solve the system, if possible: Solution ① $\ 4x + 3y - z = 4$
② $\ 8x + 6y - 2z = 7$

Planes with the same normal (or scalar multiple of normal) are parallel so they do not intersect and there is no solution to the system. If the last number on the right side of the equal sign was also a scalar multiple, then the two planes would be considered 'coincident', meaning there are an infinite number of solutions because the planes overlap...

The following planes are... Solution ① Ax + By + Cz + D = 0
② Ax + By + Cz + E = 0

1. Coplanar
2. Coincident
3. Parallel only
4. It cannot be determined

The planes are not 'coincident' (overlapping) because the last values ('D', 'E') differ. Since the planes have the same normals (A, B, C) but different constants at the end, then they are only parallel.

[Note that "coplanar" is typically only used when describing whether or not 'points' lie on the same plane.]

The Intersection of Planes: Concept

Which of the following statements about plane intersection is false? Solution

1. Two planes that do not intersect are parallel and non-coincident.
2. Two planes can intersect along a line.
3. Two planes can intersect at a point.
4. Two coincident planes have infinite intersection points and solutions.

While planes can intersect along a continuous line, they cannot intersect at a single point in space.

The Intersection of Two Cartesian (Scalar) Planes; Equation of a Line of Intersection

Solve the system, if possible: Solution ① $\quad 3x - 2y + z = 3$
② $\quad 7x - 6y + 3z = -2$

Notice that the normal of the two equations are not equal, meaning the planes are not parallel. Normals are found from the coefficients of the equation when written only in standard (Cartesian) form: $\begin{align} & (3, -2, 1) \\ \\ & (7, -6, 3) \end{align}$ Also note that if the planes are parallel and the constant on the right side of the Cartesian equation is the same in both, then the planes are coincident with infinite number of solutions or intersection points. Also note that two planes intersect along a line, and not at a single point.

To solve, we solve for one variable and introduce a parameter. Right away we can solve for x... $\begin{align} 3x - 2y + z & = 3 \\ \\ z & = 3 - 3x + 2y \end{align}$ Sub ① in ② $\begin{align} 7x - 6y + 3z & = -2 \\ \\ 7x - 6y + 3(3 - 3x + 2y) & = -2 \\ \\ 7x - \cancel{6y} + 9 - 9x + \cancel{6y} & = -2 \\ \\ x & = \dfrac{11}{2} \end{align}$ Substituting 'x' into one of the equations, we get a solution to this system of planes in the form of an equation of a line of intersection: $\begin{align} 3x - 2y + z & = 3 \\ \\ 3\left(\dfrac{11}{2}\right) - 2y + z & = 3 \\ \\ \dfrac{33}{2} - 2y + z & = 3 \\ \\ z & = 2y - \dfrac{27}{2} \\ \\ \end{align}$ Other forms are accepted: $y = \dfrac{1}{4}(2z + 27)$

The Intersection of Planes: Equation of the Line of Intersection of Planes Passing Through a Point

Determine the equation of a line that passes through the point N(1, 2, 3) and is parallel to the equation of the line of intersection of the following planes that are in Cartesian form: Solution $\begin{align} 2x + y - z & = 4 \\ \\ 3x + 2y & = 2 \end{align}$

The line of intersection is perpendicular to the normals of both planes: (2, 1, -1) and (3, 2, 0). To find the direction vector of the line that is perpendicular to both normals - the line of intersection - we can take the cross product as follows:

$\begin{align} \vec{m} &= (2, 1, -1) \ × \ (3, 2, 0) \\ \\ &= \left \langle a_2b_3 - a_3b_2,\ a_3b_1 - a_1b_3,\ a_1b_2 - a_2b_1 \right \rangle \\ \\ &= \left \Big\langle (1)(0) - (-1)(2), \ (-1)(3) - (2)(0), \ (2)(2) - (1)(3) \right \Big\rangle \\ \\ &= \left \langle 3, -5, 1 \right \rangle \end{align}$

A line parallel to the line of intersection will have the same direction vector, $\vec{m}$. The parametric equations can be found by combining the point N(1, 2, 3) and the direction vector $\vec{m} = \left \langle 3, -5, 1 \right \rangle$.

x = 1 + 3s
y = 2 - 5s
z = 3 + 1s

The line equation can also be written in vector form: $\vec{v} = (1, 2, 3) + t(3, -5, 1)$

Vector Equations of Planes

Given the following vector equations of planes:

Determine the normals to each plane equation. Solution

Get the normal from the cross product of the direction vectors. $\begin{align} \vec{N_1} & = \langle -3, -2, -2\rangle \times \langle 0, 0, -2\rangle \\ \\ & = \langle 4, -6, 0\rangle \\ \\ \vec{N_2} & = \langle -3, -2, -2\rangle \times \langle -6, -4, -8\rangle \\ \\ & = \langle 8, -12, 0\rangle \\ \\ & = 2\langle 4, -6, 0\rangle \end{align}$

Determine whether the planes are coplanar. Solution

First check if the planes are at least parallel. Planes are parallel when their normals are collinear. [If the planes are not parallel then they are definitely not coplanar] Get the normal from the cross product of the direction vectors. $\begin{align} \vec{N_1} & = \langle -3, -2, -2\rangle \times \langle 0, 0, -2\rangle \\ \\ & = \langle 4, -6, 0\rangle \\ \\ \vec{N_2} & = \langle -3, -2, -2\rangle \times \langle -6, -4, -8\rangle \\ \\ & = \langle 8, -12, 0\rangle \\ \\ & = 2\langle 4, -6, 0\rangle \end{align}$ Since the normals are scalar multiples of eachother, then we can conclude that the two planes are at least parallel. Finally, check if the planes are coplanar by substituting the position vector from one plane, into the other plane equation, to see if (L.S. = R.S.). $\begin{align} (1, 2, 3) & = (-5, -2, -5) + s\langle -3, -2, -2\rangle + t\langle -6, -4, -8\rangle \\ \\ 1 & = -5 -3s - 6t \\ \\ 6 & = -3s - 6t \\ \\ s & = -2t - 2 \end{align}$ Get one more equation... And then substitute 's' into it... $\begin{align} 2 & = -2 - 2s - 4t \\ \\ 4 & = -2s - 4t \\ \\ 4 & = -2(-2t - 2) - 4t \\ \\ 4 & = 4t + 4 - 4t \\ \\ 4 & = 4 \quad \text{(L.S. = R.S.)} \end{align}$ Therefore the planes are coplanar.

The Distance from a Point to a Line in 2-Dimensions

Calculate the distance between the point N(1, 5) and a line with the equation 3x + y - 4 = 0. (Reduce your exact answer fully) Solution

d =

Hint Clear Info

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Write the equation in Cartesian form: Ax + By + C = 0

Use the point (X, Y) $\begin{align} d & = \dfrac{\big|(A)(x) \ + \ (B)(y) \ + \ (C)\big|}{\sqrt{A^2 \ + \ B^2}} \\ \\ & = \dfrac{\big|(3)(1) \ + \ (1)(5) \ + \ (-4)\big|}{\sqrt{(3)^2 \ + \ (1)^2}} \\ \\ & = \dfrac{2}{3} \\ \\ \end{align}$

Find the distance between a point M(2, 3) and a line with vector equation $\vec{r}$ = (1, -8) + s(4, 6). (Reduce your exact answer fully) Solution

d =

Hint Clear Info

$\dfrac{\quad\sqrt{\begin{matrix} \\ \quad \end{matrix}}}{\quad}$

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Convert the vector into Cartesian form: x = 1 + 4s and y = -8 + 6s

Substitute for 's'

6x - 4y - 33 = 0

Use the point (X, Y) $\begin{align} d & = \dfrac{\big|(A)(x) \ + \ (B)(y) \ + \ (C)\big|}{\sqrt{A^2 \ + \ B^2}} \\ \\ & = \dfrac{\big|(6)(2) \ + \ (-4)(3) \ + \ (-33)\big|}{\sqrt{(6)^2 \ + \ (-4)^2}} \\ \\ & = \dfrac{33}{2\sqrt{13}} \\ \\ & = \dfrac{33}{2\sqrt{13}} \times \dfrac{\sqrt{13}}{\sqrt{13}} \\ \\ & = \dfrac{33\sqrt{13}}{2(13)} \\ \\ & = \dfrac{33\sqrt{13}}{26} \\ \\ \end{align}$

Calculating the Distance Between Two Parallel Lines in 2-Dimensions

Find the shortest distance between two parallel lines. Round your answer to the nearest decimal place. Solution $\begin{align} 3x - y - 7 & = 0 \\ \\ -6x + 2y - 13 & = 0 \end{align}$

d =

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First make any point out of one of the line equations. It is easiest to choose a y-intercept or x-intercept.

E.g. sub x = 0 into the top equation: 3(0) - y - 7 = 0

y = - 7

Point: (0, -7)

Then sub this point and the other line equation into the distance formula.... $\begin{align} d & = \dfrac{\big|(A)(x) \ + \ (B)(y) \ + \ (C)\big|}{\sqrt{A^2 \ + \ B^2}} \\ \\ d & = \dfrac{\big|(-6)(0) \ + \ (2)(-7) \ + \ (-13)\big|}{\sqrt{(-6)^2 \ + \ (2)^2}} \\ \\ & = \dfrac{27}{\sqrt{40}} \\ \\ & = 4.269 \\ \\ & = 4.3 \\ \\ \end{align}$

The Distance Between a Point and a Line in 3-Dimensions

Calculate the shortest distance from the point N(2, 8, 5) and a line with equation $\vec{r}$ = (2, 5, -1) + s(1, -5, 3). (Reduce your exact answer fully) Solution

d =

Hint Clear Info

$\dfrac{\quad\sqrt{\begin{matrix} \\ \quad \ \end{matrix}}}{\quad}$

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First find $\vec{NP}$ = P - N = [ 2 - 2, 5 - 8, -1 - 5 ] = (0, -3, -6)

Use the direction vector given in the equation $\vec{m}$ = (1, -5, 3)

The new distance equation uses the given point 'N' and the point on the vector 'P'. Notice that the numerator involves the cross product between the direction vector and the line 'NP'. $\begin{align} d & = \dfrac{\big| \vec{m} \ × \ \vec{NP} \big|}{\vec{m}} \\ \\ & = \dfrac{\big| (1,\ -5,\ 3) \ × \ (0,\ -3,\ -6) \big|}{\sqrt{ (1)^2\ +\ (-5)^2\ +\ (3)^2}} \\ \\ & = \dfrac{\big| (39,\ 6,\ -3) \big|}{\sqrt{37}} \\ \\ & = \dfrac{\big| \sqrt{ (39)^2\ +\ (6)^2\ +\ (-3)^2} \big|}{\sqrt{37}} \\ \\ & = \dfrac{3\sqrt{174}}{\sqrt{37}} \times \dfrac{\sqrt{37}}{\sqrt{37}} \\ \\ & = \dfrac{3\sqrt{6438}}{37} \\ \\ \end{align}$

The Distance Between a Point and a Plane

Find the shortest distance from the point P(6, 7, 8) to the plane: 2x + 3y - 4z + 5 = 0 Solution

d =

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Sub the point P(6, 7, 8) and equation 2x + 3y - 4z + 5 = 0 into the new distance equation for planes:

x = 6
y = 7
z = 8
A = 2
B = 3
C = -4
D = 5 $\begin{align} d & = \dfrac{\big|(A)(x) \ + \ (B)(y) \ + \ (C)(z) + (D) \big|}{\sqrt{A^2 \ + \ B^2 \ + \ C^2}} \\ \\ & = \dfrac{\big|(2)(6) \ + \ (3)(7) \ + \ (-4)(8) + (5) \big|}{\sqrt{(2)^2 \ + \ (3)^2 \ + \ (-4)^2}} \\ \\ & = \dfrac{6}{\sqrt{29}} \\ \\ & = \dfrac{6\sqrt{29}}{29} \\ \\ & = 1.1142 \\ \\ \end{align}$

The Distance Between Two Parallel Planes

Find the distance between the planes Solution $\begin{align} x - 3y + 4z + 2 &= 0 \\ \\ x - 3y + 4z - 5 &= 0 \end{align}$

d =

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First we must find a point (x, y, z) on one of the planes, and use this with the other plane equation in the distance formula.

To find a point on the plane, it is easiest to find an x, y, or z intercept.

(E.g.) to find the x-intercept, plug y=0 and z=0 into one of the equations.

x - 3y + 4z + 2 = 0     →     x - 3(0) + 4(0) + 2 = 0     →     x = -2

Now we have the point P(-2, 0, 0) and the other equation x - 3y + 4z - 5 = 0

x = -2, y = 0, z = 0, A = 1, B = -3, C = 4, D = -5. $\begin{align} d & = \dfrac{\big|(A)(x) \ + \ (B)(y) \ + \ (C)(z) + (D) \big|}{\sqrt{A^2 \ + \ B^2 \ + \ C^2}} \\ \\ & = \dfrac{\big|(1)(-2) \ + \ (-3)(0) \ + \ (4)(0) + (-5) \big|}{\sqrt{(1)^2 \ + \ (-3)^2 \ + \ (4)^2}} \\ \\ & = \dfrac{7}{\sqrt{26}} \\ \\ & = \dfrac{7\sqrt{26}}{26} \\ \\ & = 1.373 \\ \\ \end{align}$

The Distance Between Lines that are Not Parallel and Do Not Intersect, in 3-Dimensions

Find the shortest distance between the skew lines. Solution $\begin{align} \vec{r_1} &= (2, 3, 4) + s(1, 6, 7) \\ \\ \vec{r_2} &= (1, 5, -3) + t(2, 7, -1) \\ \end{align}$

d =

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The shortest distance between the two skew lines is the magnitude of a third vector connecting the two lines. This third vector is perpendicular to both skew lines. $P_1 = (2, 3, 4)\ $ and $\ \vec{m_1} = \langle 1, 6, 7 \rangle$
$P_2 = (1, 5, -3)\ $ and $\ \vec{m_2} = \langle 2, 7, -1 \rangle$ Direction (position) vector of positions P₁, P₂: $\begin{align} \vec{P_1 P_2} & = \langle 1 - 2, 5 - 3, -3 - 4 \rangle \\ \\ & = \langle -1, 2, -7 \rangle \end{align}$ Cross product of $\big|\vec{m_2}\big|\ ×\ \big|\vec{m_1}\big|$ or $\vec{m_2} × \vec{m_1}$ or $\vec{m_1} × \vec{m_2}$, either way you will get the same answer in the end... $\begin{align} \big|\vec{m_2}\big|\ ×\ \big|\vec{m_1}\big|\ &= \langle 2, 7, -1 \rangle \ ×\ \langle 1, 6, 7 \rangle \\ \\ &= \Big\langle (7)(7) - (-1)(6), \ (-1)(1) - (2)(7), \ (2)(6) - (7)(1) \Big\rangle \\ \\ &= \langle 55, -15, 5 \rangle \end{align}$ Scalar projection of the position vector, $\vec{P_1 P_2}$, onto the cross product, $\vec{m_2} × \vec{m_1}$ or $\vec{m_1} × \vec{m_2}$... $\begin{align} magnitude & = \dfrac{\Big| \vec{P_1 P_2} \ \cdot \ \langle \vec{m_2} × \vec{m_1} \rangle \Big|}{\Big| \langle \vec{m_2} × \vec{m_1} \rangle \Big|} \\ \\ & = \dfrac{\Big| (-1)(-55) + (2)(15) + (-7)(-5) \Big|}{\Big| 5\sqrt{131} \Big|} \\ \\ & = \dfrac{120}{5\sqrt{131}} \\ \\ & = 24\sqrt{131} \\ \\ & = 274.7 \end{align}$

Limit Properties

The limit, $\lim\limits_{ x\rightarrow 3 }{ \sqrt{x - 3} }$ exists. Solution

1. True
2. False

If the limit (dependent, y-value) was not the same on either side of 'n', $\lim\limits_{ x\rightarrow { n }^{ + } }{ f\left( x \right) } \ne \lim\limits_{ x\rightarrow { n }^{ - } }{ f\left( x \right) }$ then the limit, $\lim\limits_{ x \rightarrow \, n }{ f \left( x \right) }$ does not exist.

While the function $f(x) = \sqrt{x - 3}$ is defined at x = 3, and $\lim\limits_{ x\rightarrow 3^{+} }{ \sqrt{x - 3} }$ exists, the limit from the negative side, $\lim\limits_{ x\rightarrow 3^{-} }{ \sqrt{x - 3} }$ does not exist (D.N.E.). Therefore, since the limit on either side is not the same (DNE & 0), then $\lim\limits_{ x\rightarrow 3 }{ \sqrt{x - 3} }$ DNE.

Limits

Solve the limits.

Solution $\lim \limits_{ x\rightarrow -3 }{ \, \dfrac{\dfrac{1}{3} + \dfrac{1}{x}}{9 - x^2} }$

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$\begin{align} & = \underset{ x \rightarrow -3 }{lim} \dfrac{\dfrac{x}{3x} + \dfrac{3}{3x}}{(3 + x)(3 - x)} \\ \\ & = \underset{ x \rightarrow -3 }{lim} \dfrac{\dfrac{x + 3}{3x}}{(3 + x)(3 - x)} \\ \\ & = \underset{ x \rightarrow -3 }{lim} \dfrac{\cancel{x + 3}}{3x} \times \dfrac{1}{(\cancel{3 + x})(3 - x)} \\ \\ & = \underset{ x \rightarrow -3 }{lim} \dfrac{1}{(3x)(3 - x)} \\ \\ & = \dfrac{1}{(3)(-3)(3 - (-3))} \\ \\ & = -\dfrac{1}{54} \end{align}$

Solution $\lim \limits_{ x\rightarrow 1 }{ \, \dfrac{ \sqrt{x} - 1 }{x - 1} }$

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$\begin{align} & = \underset{ x \rightarrow 1 }{\lim} \ \dfrac{ x^{\frac{1}{2}} - 1 }{x - 1} \\ \\ & = \underset{ x \rightarrow 1 }{\lim} \ \dfrac{ x^{\frac{1}{2}} - 1 }{\left( x^{\frac{1}{2}} \right)^2 - 1^2} \\ \\ & = \underset{ x \rightarrow 1 }{\lim} \ \dfrac{ x^{\frac{1}{2}} - 1 }{\left( x^{\frac{1}{2}} - 1 \right)\left( x^{\frac{1}{2}} + 1 \right)} \\ \\ & = \underset{ x \rightarrow 1 }{\lim} \ \dfrac{\cancel{ x^{\frac{1}{2}} - 1 }}{\left( \cancel{x^{\frac{1}{2}} - 1} \right)\left( x^{\frac{1}{2}} + 1 \right)} \\ \\ & = \dfrac{1}{(1)^{\frac{1}{2}} + 1} \\ \\ & = \dfrac{1}{2} \end{align}$

Solution $\lim \limits_{ x\rightarrow ∞ }{ \, \dfrac{3x^2 - 2x}{5x^3 + 2x^2} }$

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$\begin{align} & = \lim \limits_{ x\rightarrow ∞ }{ \, \dfrac{3x^2 - 2x}{5x^3 + 2x^2} } \div \dfrac{x^3}{x^3} \\ \\ & = \lim \limits_{ x\rightarrow ∞ }{ \, \dfrac{\dfrac{3x^2 - 2x}{x^3}}{\dfrac{5x^3 + 2x^2}{x^3}} } \\ \\ & = \lim \limits_{ x\rightarrow ∞ }{ \, \dfrac{\dfrac{3x^2}{x^3} - \dfrac{2x}{x^3}}{\dfrac{5x^3}{x^3} + \dfrac{2x^2}{x^3}} } \\ \\ & = \lim \limits_{ x\rightarrow ∞ }{ \, \dfrac{\dfrac{3}{x} - \dfrac{2}{x^2}}{5 + \dfrac{2}{x}} } \\ \\ & = \dfrac{\cancelto{0}{\dfrac{3}{∞}} - \cancelto{0}{\dfrac{2}{∞}}}{5 + \cancelto{0}{\dfrac{2}{∞}}} \\ \\ & = \dfrac{0}{5} \\ \\ & = 0 \end{align}$

Solution $\lim \limits_{ x\rightarrow ∞ }{ \, \dfrac{ \sqrt{x^2 + 8} }{2x + 4} }$

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$\begin{align} & = \lim \limits_{ x\rightarrow ∞ }{ \, \dfrac{ \sqrt{x^2 + 8} }{2x + 4} } \div \dfrac{x}{x} \\ \\ & = \lim \limits_{ x\rightarrow ∞ }{ \, \dfrac{ \dfrac{\sqrt{x^2 + 8}}{x} }{\dfrac{2x + 4}{x}} } \\ \\ & = \lim \limits_{ x\rightarrow ∞ }{ \, \dfrac{ \dfrac{\sqrt{x^2 + 8}}{\sqrt{x^2}} }{\dfrac{2x}{x} + \dfrac{4}{x}} } \\ \\ & = \lim \limits_{ x\rightarrow ∞ }{ \, \dfrac{\sqrt{\dfrac{x^2 + 8}{x^2}}}{2 + \dfrac{4}{x}} } \\ \\ & = \lim \limits_{ x\rightarrow ∞ }{ \, \dfrac{\sqrt{\dfrac{x^2}{x^2} + \dfrac{8}{x^2}}}{2 + \dfrac{4}{x}} } \\ \\ & = \lim \limits_{ x\rightarrow ∞ }{ \, \dfrac{\sqrt{1 + \dfrac{8}{x^2}}}{2 + \dfrac{4}{x}} } \\ \\ & = \dfrac{\sqrt{1 + \cancelto{0}{\dfrac{8}{∞}}}}{2 + \cancelto{0}{\dfrac{4}{∞}}} \\ \\ & = \dfrac{1}{2} \end{align}$

Solution $\lim \limits_{ x\rightarrow ∞ }{ \, \left[ \dfrac{6x + 1}{2x - 3} \ + \ \dfrac{x + \sqrt{2}}{x + 2} \right] }$

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Divide everything by a factor of 'x'.. $\begin{align} & = \underset{ x \rightarrow ∞ }{\lim} \ \left[ \dfrac{\dfrac{6x}{x} + \dfrac{1}{x}}{\dfrac{2x}{x} - \dfrac{3}{x}} + \dfrac{\dfrac{x}{x} + \dfrac{\sqrt{2}}{x}}{\dfrac{x}{x} + \dfrac{2}{x}} \right] \\ \\ & = \underset{ x \rightarrow ∞ }{\lim} \ \left[ \dfrac{6 + \dfrac{1}{x}}{2 - \dfrac{3}{x}} + \dfrac{1 + \dfrac{\sqrt{2}}{x}}{1 + \dfrac{2}{x}} \right] \\ \\ & = \dfrac{6 + \cancelto{0}{\dfrac{1}{∞}}}{2 - \cancelto{0}{\dfrac{3}{∞}}} + \dfrac{1 + \cancelto{0}{\dfrac{\sqrt{2}}{∞}}}{1 + \cancelto{0}{\dfrac{2}{∞}}} \\ \\ & = \dfrac{6}{2} + \dfrac{1}{1} \\ \\ & = 4 \end{align}$

Solution $\lim \limits_{ x\rightarrow ∞ }{ \left[2x - \sqrt{4x^2 + 4x - 3}\right] }$

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First, use the difference of squares procedure, essentially by multiplying by 1, $\begin{align} & = \underset{ x \rightarrow ∞ }{lim} \ 2x - \sqrt{4x^2 + 4x - 3} \times \dfrac{2x + \sqrt{4x^2 + 4x - 3}}{2x + \sqrt{4x^2 + 4x - 3}} \\ \\ & = \underset{ x \rightarrow ∞ }{lim} \ \dfrac{4x^2 - (4x^2 + 4x - 3)}{2x + \sqrt{4x^2 + 4x - 3}} \end{align}$ [Tricky] Common factor the x² under the root to take it out of the root, and at the same time, put 'x's in the denominator $\begin{align} & = \underset{ x \rightarrow ∞ }{lim} \ \dfrac{4x^2 - (4x^2 + 4x - 3)}{2x + \sqrt{x^2\left(4 + \dfrac{4}{x} - \dfrac{3}{x^2}\right)}} \\ \\ & = \underset{ x \rightarrow ∞ }{lim} \ \dfrac{-4x + 3}{2x + x\sqrt{4 + \dfrac{4}{x} - \dfrac{3}{x^2}}} \end{align}$ Then, Divide the numerator and denominator by x, $\begin{align} & = \underset{ x \rightarrow ∞ }{lim} \ \dfrac{\left(\dfrac{-4x + 3}{x}\right)}{\left(\dfrac{2x + x\sqrt{4 + \dfrac{4}{x} - \dfrac{3}{x^2}}}{x}\right)} \\ \\ & = \underset{ x \rightarrow ∞ }{lim} \ \dfrac{\dfrac{-4x}{x} + \dfrac{3}{x}}{\dfrac{2\cancel{x}}{\cancel{x}} + \dfrac{\cancel{x}}{\cancel{x}}\sqrt{4 + \dfrac{4}{x} - \dfrac{3}{x^2}}} \\ \\ & = \underset{ x \rightarrow ∞ }{lim} \ \dfrac{-4 + \dfrac{3}{x}}{2 + 1\sqrt{4 + \dfrac{4}{x} - \dfrac{3}{x^2}}} \end{align}$ Simplify, $\begin{align} & = \dfrac{-4 + \cancelto{0}{\dfrac{3}{∞}}}{2 + 1\sqrt{4 + \cancelto{0}{\dfrac{4}{∞}} - \cancelto{0}{\dfrac{3}{∞^2}}}} \\ \\ & = \dfrac{-4}{2 + \sqrt{4}} \\ \\ & = -1 \end{align}$

Limits using Change of Variable

To denationalize the denominator in the function below, we can multiply both the numerator and denominator by the conjugate, as shown. Solution $\begin{align} y & = \dfrac{1}{\sqrt[3]{x} - 1} \\ \\ & = \dfrac{1}{\sqrt[3]{x} - 1} \times \dfrac{\sqrt[3]{x} + 1}{\sqrt[3]{x} + 1} \\ \\ & = \dfrac{\sqrt[3]{x} + 1}{(\sqrt[3]{x})^2 - 1} \\ \\ & =\ ... \end{align}$

1. True
2. False

Multiplying by the conjugate is typically only useful to derationalize the denominator for square roots, like: $\begin{align} y & = \dfrac{1}{\sqrt[2]{x} - 1} \\ \\ & = \dfrac{1}{\sqrt[2]{x} - 1} \times \dfrac{\sqrt[2]{x} + 1}{\sqrt[2]{x} + 1} \\ \\ & = \dfrac{\sqrt[2]{x} + 1}{(\sqrt[2]{x})^2 - 1} \\ \\ & = \dfrac{\sqrt[2]{x} + 1}{x - 1} \end{align}$ Therefore, a different technique, called 'Change of Variable' must be used...

Determine the following. Solution $\lim \limits_{ x\rightarrow 0 }{ \dfrac{x}{\sqrt[3]{x + 8} - 2} }$

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Let 't' = $\sqrt[3]{x + 8}$, and write 'x' in terms of 't' $\begin{align} t & = \sqrt[3]{x + 8} \\ \\ t^3 & = (\sqrt[3]{x + 8})^3 \\ \\ t^3 & = x + 8 \\ \\ x & = t^3 - 8 \end{align}$ Find, as x -> 0, t -> ? $\begin{align} t & = \sqrt[3]{(x) + 8} \\ \\ t & = 2 \end{align}$ Now, substitute into the limit, and factor the difference of cubes: $\begin{align} & \lim \limits_{ x\rightarrow 0 }{ \dfrac{(x)}{(\sqrt[3]{x + 8}) - 2} } \\ \\ & = \lim \limits_{ t\rightarrow 2 }{ \dfrac{(t^3 - 8)}{(t) - 2} } \\ \\ & = \lim \limits_{ t\rightarrow 2 }{ \dfrac{(t - 2)\left[(t)^2 + (t)(2) + (2)^2\right]}{t - 2} } \\ \\ & = \lim \limits_{ t\rightarrow 2 }{ \dfrac{\cancel{(t - 2)}\left[(t)^2 + (t)(2) + (2)^2\right]}{\cancel{(t - 2)}} } \\ \\ & = \lim \limits_{ t\rightarrow 2 }{ \ t^2 + 2t + 4 } \\ \\ & = (2)^2 + 2(2) + 4 \\ \\ & = 12 \end{align}$

Instantaneous and Average Rates of Change (Without Difference Quotient)

Given the function ƒ(x) = x⁴ - x³, determine:

the average rate of change on the interval -3 ≤ x ≤ 1 Solution

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the instantaneous rate of change at x = -5 Solution

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Instantaneous Rate of Change at the Vertex of a Quadratic Function (Without Difference Quotient)

For the function ƒ(x) = 3(x - 4)² + 1

Estimate the instantaneous rate of change at the vertex, without using difference quotient. Solution

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Hint: Slope at the vertex point.

Determine the average rate of change from x = 2 to x = 8. Solution

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$\begin{align} AROC & = \dfrac{f(x_2) - f(x_1)}{x_2 - x_1} \\ \\ & = \dfrac{[3(x_2 - 4)^2 + 1] - [3(x_1 - 4)^2 + 1]}{x_2 - x_1} \\ \\ & = \dfrac{[3(8 - 4)^2 + 1] - [3(2 - 4)^2 + 1]}{8 - 2} \\ \\ & = 6 \end{align}$

Average Rate of Change for a Trig Function

For the function ƒ(x) = 8sin(x), calculate the exact value of the average rate of change (AROC) on the interval [0, 90] Solution $\text{AROC} = \dfrac{f(x_2) - f(x_1)}{x_2 - x_1}$

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Estimating Instantaneous Rate of Change for a Trig Function

For the function ƒ(x) = 6sin(30x) + 10, estimate the instantaneous rate of change (IROC), at x = 2 Solution $\text{IROC} = \dfrac{f(x + h) - f(x - h)}{(x + h) - (x - h)}$

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Instantaneous Rate of Change (Limit of Difference Quotient)

Given the function ƒ(x) = 2x - 4x², find the slope of the tangent using the difference quotient, at x = -2. Solution $m = f ’ (x) = \lim\limits_{ h \rightarrow 0 } {\dfrac{f(x + h) - f(x)}{h}}$

m =

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Instantaneous Rate of Change (Limit of Difference Quotient)

A baseball is launched into the air and its height can be modelled by the function h(t) = -5t² + 15t + 1, where h(t) is height in metres, and t is time in seconds. Determine the exact instantaneous rate of change in the height of the baseball at 2s. Solution

s =

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m/s

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Use the difference quotient, at t = 2, as h --> 0 $\begin{align} & = \lim\limits_{ h \rightarrow 0 } {\dfrac{f(t + h) - f(t)}{h}} \\ \\ & = \lim\limits_{ h \rightarrow 0 } {\dfrac{\left[-5(t + h)^2 + 15(t + h) + 1\right] - \left[-5(t)^2 + 15(t) + 1\right]}{h}} \\ \\ & = \lim\limits_{ h \rightarrow 0 } {\dfrac{\left[-5(2 + h)^2 + 15(2 + h) + 1\right] - \left[-5(2)^2 + 15(2) + 1\right]}{h}} \\ \\ & = \lim\limits_{ h \rightarrow 0 } {\dfrac{\left[-5(4 + 4h + h^2) + 30 + 15h + 1\right] - \left[11\right]}{h}} \\ \\ & = \lim\limits_{ h \rightarrow 0 } {\dfrac{\left[-20 - 20h - 5h^2 + 30 + 15h + 1\right] - 11}{h}} \\ \\ & = \lim\limits_{ h \rightarrow 0 } {\dfrac{-5h^2 - 5h}{h}} \\ \\ & = \lim\limits_{ h \rightarrow 0 } {\dfrac{-\cancel{h}(5h + 5)}{\cancel{h}}} \\ \\ & = \lim\limits_{ h \rightarrow 0 } {-(5\cancelto{0}{h} + 5)} \\ \\ & = -5\ m/s \end{align}$

Equation of a Line using Instantaneous Rate of Change (Difference Quotient)

Determine the equation of the line, in slope-y-intercept form, that passes through the point P(1, 2) and is parallel to the line tangent to the function below, at the point P(7, 5). (Solve using the limit of the difference quotient.) Solution $f(x) = \sqrt{3x + 4}$

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$\dfrac{}{\quad\quad}x \ + \ \dfrac{}{\quad\quad}$

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First, determine the slope at x = 7, with the limit of the difference quotient, $\begin{align} & = \lim\limits_{ h \rightarrow 0 } {\dfrac{f(x + h) - f(x)}{h}} \\ \\ & = \lim\limits_{ h \rightarrow 0 } {\dfrac{\sqrt{3(x + h) + 4} \ - \ \sqrt{3x + 4}}{h}} \\ \\ & = \lim\limits_{ h \rightarrow 0 } {\dfrac{\sqrt{3(7 + h) + 4} \ - \ \sqrt{3(7) + 4}}{h}} \\ \\ & = \lim\limits_{ h \rightarrow 0 } {\dfrac{\sqrt{25 \ + \ 3h} \ - \ \sqrt{25}}{h}} \ \times \ \ \dfrac{\sqrt{25 \ + \ 3h} \ + \ \sqrt{25}}{\sqrt{25 \ + \ 3h} \ + \ \sqrt{25}} \\ \\ & = \lim\limits_{ h \rightarrow 0 } {\, \dfrac{(25 \ + \ 3h) \ - \ (25)}{h\,(\sqrt{25 \ + \ 3h} \ + \ \sqrt{25})}} \\ \\ & = \lim\limits_{ h \rightarrow 0 } {\, \dfrac{3\cancel{h}}{\cancel{h}\,(\sqrt{25 \ + \ 3h} \ + \ \sqrt{25})}} \\ \\ & = \lim\limits_{ h \rightarrow 0 } {\, \dfrac{3}{\sqrt{25 \ + \ 3\cancelto{0}{h}} \ + \ \sqrt{25}}} \\ \\ & = \dfrac{3}{\sqrt{25} \ + \ \sqrt{25}} \\ \\ & = \dfrac{3}{10} \end{align}$ Then just use this slope with the given point to determine the linear equation, using the slope 'm', from the limit, and the point P(1, 2) $\begin{align} y & = m(x) + b \\ \\ 2 & = \dfrac{3}{10}(1) + b \\ \\ b & = \dfrac{20}{10} - \dfrac{3}{10} \\ \\ b & = \dfrac{17}{10} \end{align}$ Therefore the equation is, $y = \dfrac{3}{10}x + \dfrac{17}{10}$

Instantaneous Rate of Change Word Problems

The growth of a cellular culture is modeled by the function below, where 't' is time in days, and g(x) is mass in mg. Determine the rate of change of the mass with respect to time, on the 32nd day. Solution $g(t) = \sqrt{\dfrac{1}{4}t + 1}$

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mg/day

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Determine the rate of change by taking the limit, at t = 32, $\begin{align} & = \lim\limits_{ h \rightarrow 0 } {\dfrac{g(t + h) - g(t)}{h}} \\ \\ & = \lim\limits_{ h \rightarrow 0 } {\dfrac{\sqrt{\dfrac{1}{4}(t + h) + 1} - \sqrt{\dfrac{1}{4}t + 1}}{h}} \\ \\ & = \lim\limits_{ h \rightarrow 0 } {\dfrac{\sqrt{\dfrac{1}{4}(32 + h) + 1} - \sqrt{\dfrac{1}{4}(32) + 1}}{h}} \\ \\ & = \lim\limits_{ h \rightarrow 0 } {\dfrac{\sqrt{8 + \dfrac{1}{4}h + 1} - \sqrt{8 + 1}}{h}} \\ \\ & = \lim\limits_{ h \rightarrow 0 } {\dfrac{\sqrt{\dfrac{1}{4}h + 9} - 3}{h}} \\ \\ & = \lim\limits_{ h \rightarrow 0 } {\dfrac{\sqrt{\dfrac{1}{4}h + 9} - 3}{h}} \ \times \ \dfrac{\sqrt{\dfrac{1}{4}h + 9} + 3}{\sqrt{\dfrac{1}{4}h + 9} + 3} \\ \\ & = \lim\limits_{ h \rightarrow 0 } {\dfrac{\left(\dfrac{1}{4}h + 9\right) - 9}{h\left(\sqrt{\dfrac{1}{4}h + 9} + 3\right)}} \\ \\ & = \lim\limits_{ h \rightarrow 0 } {\dfrac{\dfrac{1}{4}\cancel{h}}{\cancel{h}\left(\sqrt{\dfrac{1}{4}h + 9} \ + 3\right)}} \\ \\ & = \dfrac{1}{4\left(\sqrt{\cancelto{0}{\dfrac{1}{4}(0)} + 9} \ + 3\right)} \\ \\ & = \dfrac{1}{4\left(3 + 3\right)} \\ \\ & = \dfrac{1}{24}\ \tfrac{mg}{day} \\ \\ & ≈ 0.0417\ \tfrac{mg}{day} \end{align}$

For a cylinder that has a height 'h' double its radius 'r', determine the rate of change of volume with respect to radius, when the radius is 10 cm. [Diagram not to scale] Solution

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cm³/cm

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For volume of a cylinder $\begin{align} Volume & = \pi r^2 h \\ \\ V & = \pi r^2 (2r) \\ \\ V & = 2\pi r^3 \end{align}$ Determine the instantaneous rate of change using the difference quotient, $\begin{align} & = \lim\limits_{ h \rightarrow 0 } {\dfrac{V(r + h) - V(r)}{h}} \\ \\ & = \lim\limits_{ h \rightarrow 0 } {\dfrac{2\pi (r + h)^3 - 2\pi (r)^3}{h}} \\ \\ & = \lim\limits_{ h \rightarrow 0 } {\dfrac{2\pi (10 + h)^3 - 2\pi (10)^3}{h}} \\ \\ & = \lim\limits_{ h \rightarrow 0 } {\dfrac{2\pi [1000 + 200h + 10h^2 + 100h + 200h^2 + h^3] - 2\pi (1000)}{h}} \\ \\ & = \lim\limits_{ h \rightarrow 0 } {\dfrac{2\pi [h^3 + 210h^2 + 300h]}{h}} \\ \\ & = \lim\limits_{ h \rightarrow 0 } {\dfrac{2\pi \cancel{h}[h^2 + 210h + 300]}{\cancel{h}}} \\ \\ & = \lim\limits_{ h \rightarrow 0 } {\, 2\pi[h^2 + 210h + 300]} \\ \\ & = 2\pi[(0)^2 + 210(0) + 300] \\ \\ & = 600\pi \\ \\ & = 1,884\ \dfrac{cm^3\ volume}{cm\ radius} \end{align}$

Slope of a Tangent to a Curve

Determine the slope of the tangent to $y = 2x^{\frac{1}{2}} - 2x^2 + 1$, at x = 1. Solution

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Differentiate using power rule, $\begin{align} y’ & = (2)\left(\dfrac{1}{2}\right) x^{\frac{1}{2} - 1} - (2)(2)x^{2 - 1} \\ \\ & = (\cancel{2})\left(\dfrac{1}{\cancel{2}}\right) x^{\frac{-1}{2}} - 4x \\ \\ & = x^{\frac{-1}{2}} - 4x \end{align}$ At x = 1, $\begin{align} & = (1)^{\frac{-1}{2}} - 4(1) \\ \\ & = \dfrac{1}{\sqrt{1}} - 4 \\ \\ & = -3 \end{align}$

Chain Rule.. f ∘ g(x) = ƒ(g(x)) = ƒ'[g(x)] · g'(x)

Differentiate. Solution Video $y = (5x^2 + 2)^2$

1. 50x
2. 20x
3. 10x² + 4
4. 50x³ + 20x
5. 100x³ + 40x

For some function, using the Chain Rule like, $\begin{align} F(x) & = [f(x)]^2 \\ \\ F’(x) & = 2[f(x)]^{2-1}·f’(x) \end{align}$ So, $\begin{align} y &= (5x^2 + 2)^2 \\ \\ \dfrac{dy}{dx} &= 2(5x^2 + 2)^{2 - 1}\cdot(10x) \\ \\ &= (5x^2 + 2)\cdot(20x) \\ \\ &= 100x^3 + 40x \\ \\ \end{align}$

Chain Rule

Differentiate, use a calculator do evaluate your values.

Given, $g(x) = \left[ 4x^3 - (3x^2 - 12)^{-3} \right]^{-\frac{5}{2}}$ determine, g'(1). Solution

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Using chain rule, multiple times, $\begin{align} g’(x) & = \dfrac{-5}{2} \left[ 4x^3 - (3x^2 - 12)^{-3} \right]^{\frac{-5}{2} - 1} \left[ 3(4)x^2 - (-3)(3x^2 - 12)^{-3-1}(2(3)x^{2-1}) \right] \\ \\ & = \dfrac{-5}{2} \left[ 4x^3 - (3x^2 - 12)^{-3} \right]^{\frac{-7}{2}} \left[ 12x^2 + 3(3x^2 - 12)^{-4}(6x) \right] \\ \\ & = \dfrac{-5(6x)}{2} \left[ 4x^3 - (3x^2 - 12)^{-3} \right]^{\frac{-7}{2}} \left[ 2x + 3(3x^2 - 12)^{-4} \right] \\ \\ & = -15x \left[ 4x^3 - (3x^2 - 12)^{-3} \right]^{\frac{-7}{2}} \left[ 2x + 3(3x^2 - 12)^{-4} \right] \end{align}$ So, $\begin{align} g’(1) & = -15(1) \left[ 4(1)^3 - (3(1)^2 - 12)^{-3} \right]^{\frac{-7}{2}} \left[ 2(1) + 3(3(1)^2 - 12)^{-4} \right] \\ \\ & = -0.23415 \end{align}$

Given, $f(x) = (3x + x^{-4})^{\tfrac{1}{3}}$ determine, f'(1). Solution

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Leibniz Notation

Using Leibniz notation, differentiate for 'y' in terms of 'x' given the following. Solution $\begin{align} y & = 4u^2 + 6u - 3 \\ \\ u & = 3x^2 + 2 \\ \\ x & = -1 \end{align}$

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Leibniz: $\begin{align} \dfrac{d\,y}{d\,x} & = \dfrac{d\,y}{d\,u} \times \dfrac{d\,u}{d\,x} \\ \\ & = \left[ 8u + 6 \right] \times \left[ 6x \right] \end{align}$ Substitute 'u' $\begin{align} & = \left[ 8u + 6 \right] \times \left[ 6x \right] \\ \\ & = \left[ 8(3x^2 + 2) + 6 \right] \times \left[ 6x \right] \end{align}$ Substitute 'x' $\begin{align} & = \left[ 8(3x^2 + 2) + 6 \right] \times \left[ 6x \right] \\ \\ & = \left[ 24x^2 + 22 \right] \times \left[ 6x \right] \\ \\ & = \left[ 24(-1)^2 + 22 \right] \times \left[ 6(-1) \right] \\ \\ & = -276 \end{align}$

Using Leibniz notation, differentiate for 'y' in terms of 'x' given the following. Solution $\begin{align} y & = \dfrac{2u^2 + 1}{u - 5} \\ \\ u & = \dfrac{1}{(2x^2 - 3)^2} \\ \\ x & = 1 \end{align}$

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Rearrange 'u', in order to use chain rule for $\dfrac{d\,u}{d\,x}$ $\begin{align} u & = \dfrac{1}{(2x^2 - 3)^2} \\ \\ & = (2x^2 - 3)^{-2} \end{align}$ Use quotient rule when differentiating 'dy' by 'du' $\begin{align} \dfrac{d\,y}{d\,x} & = \dfrac{d\,y}{d\,u} \times \dfrac{d\,u}{d\,x} \\ \\ & = \left[ \dfrac{a’b - ab’}{b^2} \right] \times \left[ -2(2x^2 - 3)^{-3}(4x) \right] \\ \\ & = \left[ \dfrac{(4u)(u - 5) - (2u^2 + 1)(1)}{(u - 5)^2} \right] \times \left[ (-8x)(2x^2 - 3)^{-3} \right] \\ \\ & = \left[ \dfrac{2u^2 - 20u - 1}{(u - 5)^2} \right] \times \left[ (-8x)(2x^2 - 3)^{-3} \right] \end{align}$ Substitute 'x' into 'u' $\begin{align} u & = \dfrac{1}{(2x^2 - 3)^2} \\ \\ & = \dfrac{1}{(2(1)^2 - 3)^2} \\ \\ & = 1 \end{align}$ Substitute 'x' and 'u' into the original equation $\begin{align} & \left[ \dfrac{2u^2 - 20u - 1}{(u - 5)^2} \right] \times \left[ (-8x)(2x^2 - 3)^{-3} \right] \\ \\ & = \left[ \dfrac{2(1)^2 - 20(1) - 1}{((1) - 5)^2} \right] \times \left[ (-8(1))(2(1)^2 - 3)^{-3} \right] \\ \\ & = \dfrac{-19}{2} \\ \\ & = -9.5 \end{align}$

Using Leibniz notation, determine $\dfrac{dy}{dx}$ at x = 3, given the following information. Solution $y = 2u^3 - 3u^2 \quad\quad u = 5v^2 \quad\quad v = 2(x - 1)$

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First get 'u' in terms of 'x' $\begin{align} u & = 5v^2 \\ \\ u & = 5\left(2(x - 1)\right)^2 \\ \\ u & = 20\left(x - 1\right)^2 \end{align}$ Then, using Leibniz notation, $\begin{align} \dfrac{dy}{dx} & = \dfrac{du}{dx} \times \dfrac{dy}{du} \\ \\ & = \left(2(20)(x - 1)^{2-1}(1)\right) \times \left((2)(3)(u)^2 - (3)(2)(u) \right) \\ \\ & = 40(x - 1)\left(6(u)^2 - 6(u)\right) \end{align}$ Then substitute 'u', then 'x', $\begin{align} & = 40(x - 1)\left(6\left(20\left(x - 1\right)^2\right)^2 - 6\left(20\left(x - 1\right)^2\right)\right) \\ \\ & = 40(3 - 1)\left(6\left(20\left(3 - 1\right)^2\right)^2 - 6\left(20\left(3 - 1\right)^2\right)\right) \\ \\ & = 3,033,600 \end{align}$

Quotient Rule with Other Rules

Differentiate.

$f(x) = \dfrac{a^3 b^5}{c^2}$, where a = 2, b = 3, and c = 4. Solution

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This mainly combines the product rule and quotient rule (and power rule). $\begin{align} f(x) & = \dfrac{u}{v} \\ \\ f’(x) & = \dfrac{u’·v - u·v’}{v^2} \\ \\ f’(x) & = \dfrac{[h’·k + h·k’]·v - u·v’}{v^2} \end{align}$ Where $\begin{align} u & = a^3b^5 \\ \\ v & = c^2 \\ \\ h & = a^3 \\ \\ k & = b^5 \end{align}$ Differentiate, $\begin{align} f’(x) & = \dfrac{[h’·k + h·k’]·v - u·v’}{v^2} \\ \\ & = \dfrac{[(3a^2)(b^5) + (a^3)(5b^4)]·(c^2) - (a^3b^5)(2c)}{(c^2)^2} \\ \\ & = \dfrac{3a^2b^5c^2 + 5a^3b^4c^2 - 2a^3b^5c}{c^4} \\ \\ & = \dfrac{a^2b^4c(3cb + 5ac - 2ab)}{c^4} \\ \\ & = \dfrac{a^2b^4(3cb + 5ac - 2ab)}{c^3} \end{align}$ Where a = 2, b = 3, and c = 4... $\begin{align} f’(x) & = \dfrac{a^2b^4(3cb + 5ac - 2ab)}{c^3} \\ \\ & = \dfrac{(2)^2(3)^4\Big(3(4)(3) + 5(2)(4) - 2(2)(3)\Big)}{(4)^3} \\ \\ & = 324 \end{align}$

$g(x) = \dfrac{-1}{7} \sqrt{\dfrac{2x - 5}{3 - 4x}}$ at x = 1. Solution

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$\dfrac{\sqrt{\begin{matrix} \\ \quad \end{matrix}}}{\quad}$

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Start out with the chain rule, $\begin{align} & \dfrac{-1}{7} \sqrt{\dfrac{2x - 5}{3 - 4x}} \\ \\ & = \dfrac{-1}{7} \left(\dfrac{2x - 5}{3 - 4x}\right)^{\frac{-1}{2}} \\ \\ & = \dfrac{-1}{7} \cdot \dfrac{1}{2} \cdot \left(\dfrac{2x - 5}{3 - 4x}\right)^{\frac{-1}{2}} \left[ \dfrac{d}{dx}\,\left(\dfrac{2x - 5}{3 - 4x}\right) \right] \end{align}$ Then apply the quotient rule to the interior, let $\begin{align} a & = 2x - 5 \\ \\ b & = 3 - 4x \end{align}$ So, $\begin{align} & = \dfrac{-1}{7} \cdot \dfrac{1}{2} \cdot \left(\dfrac{2x - 5}{3 - 4x}\right)^{\frac{-1}{2}} \left[ \dfrac{a’·b - a·b’}{b^2} \right] \\ \\ & = \dfrac{-1}{7} \cdot \dfrac{1}{2} \cdot \left(\dfrac{2x - 5}{3 - 4x}\right)^{\frac{-1}{2}} \left[ \dfrac{2(3 - 4x) - (2x - 5)(-4)}{(3 - 4x)^2} \right] \\ \\ & = \dfrac{2[3 - 4x + 2(2x - 5)]}{2(-7)\sqrt{\dfrac{2x - 5}{3 - 4x}} \cdot (3 - 4x)^2} \\ \\ & = \dfrac{-7}{-7\dfrac{(2x - 5)^{\frac{1}{2}}}{(3 - 4x)^{\frac{1}{2}}} \cdot (3 - 4x)^2} \\ \\ & = \dfrac{1}{(2x - 5)^{\frac{1}{2}} \cdot (3 - 4x)^{2-\frac{1}{2} } } \\ \\ & = \dfrac{1}{(2x - 5)^{\frac{1}{2}} \cdot (3 - 4x)^{\frac{3}{2}} } \end{align}$ At x = 1, $\begin{align} & = \dfrac{1}{[2(1) - 5]^{\frac{1}{2}} \cdot [3 - 4(1)]^{\frac{3}{2}} } \\ \\ & = \dfrac{1}{[-3]^{\frac{1}{2}} \cdot [-1]^{\frac{3}{2}} } \\ \\ & = \dfrac{1}{\sqrt{-3} \cdot \sqrt{-1}\sqrt{-1}\sqrt{-1} } \\ \\ & = \dfrac{1}{\sqrt{-3} \cdot (1)\sqrt{-1} } \\ \\ & = \dfrac{1}{\sqrt{-3 \times -1} } \\ \\ & = \dfrac{1}{\sqrt{3} } \\ \\ & = \dfrac{1}{\sqrt{3}} \times \dfrac{\sqrt{3}}{\sqrt{3}} \\ \\ & = \dfrac{\sqrt{3}}{3} \\ \\ \end{align}$

$\dfrac{(2x + 3)^{\frac{1}{2}}}{(x - 4)^{\frac{1}{4}}}$, at x = 20. Solution

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Use quotient rule, in the form, $\begin{align} f(x) & = \dfrac{a}{b} \\ \\ f’(x) & = \dfrac{a’·b - a·b’}{b^2} \\ \\ & = \ ... \end{align}$ Differentiate, $\begin{align} & = \dfrac{\left[\dfrac{1}{2}(2x+3)^{\frac{-1}{2}}(2)\right](x-4)^{\frac{1}{4}} - (2x + 3)^{\frac{1}{2}}\left[\dfrac{1}{4}(x-4)^{\frac{-3}{4}}(1)\right]}{\left[(x-4)^{\frac{1}{4}}\right]^2} \end{align}$ Simplify, [simplification steps may vary] $\begin{align} & = \dfrac{(2x+3)^{\frac{-1}{2}}(x-4)^{\frac{1}{4}} - \dfrac{1}{4}(2x + 3)^{\frac{1}{2}} (x-4)^{\frac{-3}{4}}}{(x-4)^{\frac{1}{2}}} \\ \\ & = (2x+3)^{\frac{-1}{2}}(x-4)^{\frac{1}{4} - \frac{1}{2}} - \dfrac{1}{4}(2x + 3)^{\frac{1}{2}} (x-4)^{\frac{-3}{4} - \frac{1}{2}} \\ \\ & = (2x+3)^{\frac{-1}{2}}(x-4)^{\frac{-1}{4}} - \dfrac{1}{4}(2x + 3)^{\frac{1}{2}} (x-4)^{\frac{-5}{4}} \\ \\ & = (2x+3)^{\frac{-1}{2}}(x-4)^{\frac{-1}{4}} \left[ 1 - \dfrac{1}{4}(2x+3)(x-4)^{-1} \right] \\ \\ & = (2x+3)^{\frac{-1}{2}}(x-4)^{\frac{-1}{4}} \left[ 1 - \dfrac{2x + 3}{4(x-4)}\right] \\ \\ & = (2x+3)^{\frac{-1}{2}}(x-4)^{\frac{-1}{4}} \left[ \dfrac{4(x-4)}{4(x-4)} - \dfrac{2x + 3}{4(x-4)}\right] \\ \\ & = (2x+3)^{\frac{-1}{2}}(x-4)^{\frac{-1}{4}} \left[ \dfrac{2x -19}{4(x-4)}\right] \\ \\ & = \dfrac{1}{4}(2x+3)^{\frac{-1}{2}}(x-4)^{\frac{-1}{4} - 1} \left[ 2x -19 \right] \\ \\ & = \dfrac{1}{4}(2x+3)^{\frac{-1}{2}}(x-4)^{\frac{-5}{4}} \left( 2x -19 \right) \\ \\ & = \ ... \end{align}$ At, x = 20 $\begin{align} & = \dfrac{1}{4}(2(20)+3)^{\frac{-1}{2}}((20)-4)^{\frac{-5}{4}} \left( 2(20) -19 \right) \\ \\ & = 0.02502 \end{align}$

Differentiation with Compositions and Combinations of Functions

Determine $c ∘ b’ ∘ a(\sqrt{5})$ given the following. Solution $\begin{align} a(x) & = \dfrac{8}{2x^2 - 4} \\ \\ b(x) & = 2\sqrt{3x + 5} \\ \\ c(x) & = 2x^2 + 1 \end{align}$

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Understand the notation, $\begin{align} & c ∘ b’ ∘ a(\sqrt{5}) \\ \\ & = c\bigg[b’\Big(a(\sqrt{5})\Big)\bigg] \end{align}$ First determine, $\begin{align} a(\sqrt{5}) & = \dfrac{8}{2(\sqrt{5})^2 - 4} \\ \\ & = \dfrac{4}{3} \end{align}$ Differentiate b(x), $\begin{align} b(x) & = 2\sqrt{3x + 5} \\ \\ b(x) & = 2(3x + 5)^{\frac{1}{2}} \\ \\ b’(x) & = 2\left(\dfrac{1}{2}\right)(3x + 5)^{-\frac{1}{2}}(3) \\ \\ b’(x) & = \dfrac{3}{\sqrt{3x + 5}} \end{align}$ So, $\begin{align} b’\Big(a(\sqrt{5})\Big) & = \dfrac{3}{\sqrt{3\left(\dfrac{4}{3}\right) + 5}} \\ \\ & = \dfrac{3}{\sqrt{9}} \\ \\ & = 1 \end{align}$ Then, $\begin{align} c\bigg[b’\Big(a(\sqrt{5})\Big)\bigg] & = 2(1)^2 + 1 \\ \\ & = 3 \end{align}$

Determine the value of 'x' given the following. Solution $\begin{align} a(x) & = 6x^2 \\ \\ b(x) & = 2x^2 - 4 \\ \\ c(x) & = 3\sqrt{2(x - 1)} \\ \\ a’\bigg[b\Big(c’(x)\Big)\bigg] & = -60 \end{align}$

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First, $\begin{align} c’(x) & = 3\left(\dfrac{1}{2}\right)(2x - 2)^{\frac{-1}{2}}(2) \\ \\ & = \dfrac{3}{\sqrt{2x - 2}} \end{align}$ So, $\begin{align} b\Big(c’(x)\Big) & = 2\left(c’(x)\right)^2 - 4 \\ \\ & = 2\left(\dfrac{3}{\sqrt{2x - 2}}\right)^2 - 4 \\ \\ & = 2\left(\dfrac{9}{2x - 2}\right) - 4 \\ \\ & = \dfrac{9}{x - 1} - 4 \\ \\ & = \dfrac{9 - 4(x - 1)}{x - 1} \\ \\ & = \dfrac{-4x + 13}{x - 1} \\ \\ \end{align}$ And since, $\begin{align} a’(x) & = 12(x) \end{align}$ Then, $\begin{align} a’\bigg[b\Big(c’(x)\Big)\bigg] & = 12\left(b\Big(c’(x)\Big)\right) \\ \\ -60 & = 12\left(\dfrac{-4x + 13}{x - 1}\right) \\ \\ -5 & = \dfrac{-4x + 13}{x - 1} \\ \\ -5(x - 1) & = -4x + 13 \\ \\ -5x + 5 & = -4x + 13 \\ \\ -x & = 8 \\ \\ x & = -8 \end{align}$

Differentiation Word Problems

A medication is metabolized in the body and the concentration, in mg/L, in the bloodstream is given by c(t), where 't' is in hours. For t > 0.5 hr, determine f'(8), and interpret it in your own notes. Solution $c(t) = \dfrac{25}{2t^2 + 32t - 12}$

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$\tfrac{mg}{L \cdot hr}$

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Differentiate, $\begin{align} f ’ (t) & = \dfrac{a’b - ab’}{b^2} \\ \\ & = \dfrac{\cancelto{0}{(0)(2t^2 + 32t - 12)} - (25)(4t + 32)}{(2t^2 + 32t - 12)^2} \\ \\ & = \dfrac{-100t - 800}{(2t^2 + 32t - 12)^2} \end{align}$ Substitute, $\begin{align} f ’ (8) & = \dfrac{-100(8) - 800}{[2(8)^2 + 32(8) - 12]^2} \\ \\ & = -0.012\ \tfrac{mg}{L \cdot hr} \end{align}$

A hamburger store can sell 120 hamburgers per week at $4.00 per hamburger. For each $0.50 decrease in price, they can sell 20 more hamburgers. Determine the ‘marginal revenue’ function in terms of 'x'. Solution

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Let 'x be the number of changes, $\begin{align} R(x) & = (price)(\#\ sold) \\ \\ & = (4 - 0.5x)(120 + 20x) \\ \\ & = -10x^2 + 20x + 480 \end{align}$ Marginal revenue, $\begin{align} R(x) & = -10x^2 + 20x + 480 \\ \\ R’(x) & = -20x + 20 \end{align}$

Optimization: Using Similar Triangles

A straight section of railroad crosses two roads at points that are 800 m and 1200 m from an intersection. Find the dimensions of the largest rectangular lot that can be laid out in the triangle formed by the railroad and roads in terms of x. Solution

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m

You need two equations; one from similar triangles, the other from area. You need to differentiate for Area.

Let the dimensions of the rectangular lot be, “x” and “y”.

Using similar triangles, starting with a rectangle that fits with a height of 'y' and a width of 'x', $\begin{align} \frac {y}{ 800 } & = \frac { 1200-x }{1200} \\ \\ y & = \dfrac{2}{3}(1200 - x) \\ \\ y & = -\dfrac{2}{3}x + 800 \end{align}$ Calculate the area in terms of x: $\begin{align} Area & = (x)(y) \\ \\ A & = (x)\left( -\dfrac{2}{3}x + 800 \right) \\ \\ A & = -\dfrac{2}{3}x^2 + 800x \end{align}$ The max occurs when the derivative (A‘) equals zero, $\begin{align} A & = -\dfrac{2}{3}x^2 + 800x \\ \\ A‘ & = -\dfrac{4}{3}x + 800 \\ \\ 0 & = -\dfrac{4}{3}x + 800 \\ \\ x & = 600 \end{align}$ Find 'y' $\begin{align} y & = -\dfrac{2}{3}x + 800 \\ \\ y & = -\dfrac{2}{3}(600) + 800 \\ \\ y & = 400 \end{align}$ When x = 600, y = 400, the maximum dimensions are 600 m x 400 m

A tall ladder is propped on the ground and a 6 m wall, that is 4 m from an other wall with height 'h'. Determine the minimum length of the ladder that can reach the wall. Solution

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m

You need two equations; one from similar triangles, the other from Pythagorean theorem. You need to differentiate for Length.

Use special triangles to write a relation between 'h' and 'x' $\begin{align} \dfrac{h}{x} & = \dfrac{6}{x - 4} \\ \\ h & = \dfrac{6x}{x - 4} \end{align}$ Since we want to minimize 'L', write a relation between all variables involved. This uses the pythagorean theorem $\begin{align} L^2 & = x^2 + h^2 \\ \\ L & = \sqrt{x^2 + h^2} \end{align}$ Now substitute the equation for 'h' that you derived earlier, $\begin{align} L & = \sqrt{x^2 + h^2} \\ \\ L & = \sqrt{x^2 + \left( \dfrac{6x}{x - 4} \right)^2} \end{align}$ Now take the derivative, using chain rule for the square root, and quotient rule for the fraction under the root, $\begin{align} L & = \sqrt{x^2 + \left( \dfrac{6x}{x - 4} \right)^2} \\ \\ L & = \left(x^2 + \dfrac{36x^2}{(x - 4)^2} \right)^{\frac{1}{2}} \\ \\ L’ & = f’\big[g(x)\big] · g’(x) \\ \\ L’ & = f’\big[g(x)\big] · \bigg[ a(x)^{a - 1} + \dfrac{a’·b - a·b’}{b^2} \bigg] \\ \\ 0 & = \dfrac{1}{2} \left( x^2 + \dfrac{36x^2}{(x - 4)^2} \right)^{-\frac{1}{2}} \bigg[ 2x + \dfrac{(72x)(x - 4)^2 - (36x^2)(2(x - 4))}{\left( (x - 4)^2 \right)^2} \bigg] \\ \\ 0 & = \dfrac{ 2x + \dfrac{(72x)(x - 4)^2 - (72x^2)(x - 4)}{ (x - 4)^4 } }{2\sqrt{ x^2 + \dfrac{36x^2}{(x - 4)^2}}} \\ \\ 0 & = \dfrac{ 2x + \dfrac{72x[(x - 4) - (x)]}{ (x - 4)^3 } }{2\sqrt{ x^2 + \dfrac{36x^2}{(x - 4)^2}}} \\ \\ 0 & = \dfrac{ 2x + \dfrac{-288x}{ (x - 4)^3 } }{2\sqrt{ x^2 + \dfrac{36x^2}{(x - 4)^2}}} \\ \\ 0 & = (x)\dfrac{ 1 + \dfrac{-144}{ (x - 4)^3 } }{\sqrt{ x^2 + \dfrac{36x^2}{(x - 4)^2}}} \\ \\ 0 & = (x)\left( 1 + \dfrac{-144}{ (x - 4)^3 } \right) \end{align}$ Now determine the 'roots', if ab = 0, then (a) = 0, or (b) = 0

$\begin{align} x & = 0 \end{align}$ $\begin{align} 0 & = 1 + \dfrac{-144}{ (x - 4)^3 } \\ \\ -1(x - 4)^3 & = -144 \\ \\ (x - 4)^3 & = 144 \\ \\ \sqrt[3]{(x - 4)^3} & = \sqrt[3]{144} \\ \\ x - 4 & ≈ 5.241 \\ \\ x & ≈ 9.241 \\ \\ \end{align}$

[not shown here] Also it's best practice to do a check of the increasing [f'(x) > 0] & decreasing [f'(x) < 0] intervals, to make sure that 'L' is at a minimum.

Solve for the approximate value of 'L' ... $\begin{align} L & = \sqrt{x^2 + \left( \dfrac{6x}{x - 4} \right)^2} \\ \\ L & = \sqrt{(9.241)^2 + \left( \dfrac{6(9.241)}{(9.241) - 4} \right)^2} \\ \\ L & ≈ 14.05\ m \\ \\ \end{align}$

Optimization: Geometry

A farmer wants to build a corral with 3 parallel partitions to make 4 equal sized, rectangular pens to keep her livestock. She already has 650 m of fencing to use. Determine the total dimensions that will maximize the total area of the corral. Solution

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m

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The total amount of fencing used is, $\begin{align} 650\ m & = 5(y) + 2(x) \\ \\ y & = 130 - \dfrac{2}{5}x \end{align}$ To maximize the area, $\begin{align} Area & = (x)(y) \\ \\ A & = (x)\left(130 - \dfrac{2}{5}x\right) \\ \\ A & = -\dfrac{2}{5}x^2 + 130x \end{align}$ Differentiate to find the critical point, when A’ = 0 $\begin{align} A’ & = -(2)\dfrac{2}{5}x + 130 \\ \\ 0 & = -\dfrac{4}{5}x + 130 \\ \\ x & = 162.5\ m \end{align}$ Now, to get 'y', use your equation from before, [Testing to verify max/min not shown here. Could use intervals or slopes] $\begin{align} y & = 130 - \dfrac{2}{5}(162.5) \\ \\ & = 65\ m \end{align}$

A box with a square base, and an open top is to be made from 6,000 cm² of material. Determine the dimensions that will make the make the largest possible box, negating the width of the material. Solution

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x = h =

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cm

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Make an equation given the area of the base and 4 sides, $\begin{align} Area & = A_{base} + 4(A_{side}) \\ \\ 6,000\ cm^2 & = x^2 + 4(xh) \\ \\ 4xh & = -x^2 + 6,000 \\ \\ h & = -\dfrac{1}{4}x + \dfrac{1,500}{x} \end{align}$ To maximize the volume, write an equation for volume. You want everything 'in terms of' one variable, so substitute 'h' out... $\begin{align} Volume & = (x)(x)(h) \\ \\ V & = x^2\left(\dfrac{-1}{4}x + \dfrac{1,500}{x}\right) \\ \\ V & = \dfrac{-1}{4}x^3 + 1,500x \end{align}$ Differentiate to find the critical point, when V’ = 0 $\begin{align} V’ & = (3)\dfrac{-1}{4}x^2 + 1,500 \\ \\ 0 & = \dfrac{-3}{4}x^2 + 1,500 \\ \\ x^2 & = 2,000 \\ \\ x & = \pm \sqrt{2,000} \\ \\ x & = + \sqrt{2,000}\ cm \end{align}$ Since 'x' represents a length, it cannot be negative, so there is one answer. [Testing to verify max/min not shown here. Could use intervals or slopes] $\begin{align} h & = -\dfrac{1}{4}(\sqrt{2,000}) + \dfrac{1,500}{\sqrt{2,000}} \\ \\ h & = 22.36\ cm \end{align}$

A straight, 10 cm wire is cut into two lengths to make two hoops for bubble-blowing wands - one square, and one circle. To make smaller bubbles, determine the minimum total area of the circle and square combined. Solution

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cm²

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Cut the wire into two lengths, (x) and (10 - x). For this solution, use the piece (x) for the circle and the piece (10 - x) for the square,

$\begin{align} \text{Perimeter}_{\, circle} & = 2 \pi r \\ \\ x & = 2 \pi r \\ \\ r & = \dfrac{x}{2 \pi} \end{align}$ $\begin{align} \text{Perimeter}_{\, square} & = 4(side) \\ \\ 10 - x & = 4(side) \\ \\ side & = \dfrac{10 - x}{4} \\ \\ & = \dfrac{10}{4} - \dfrac{1}{4}x \\ \\ & = \dfrac{5}{2} - \dfrac{1}{4}x \end{align}$

To minimize the total area, $\begin{align} A_{Total} & = A_{Circle} + A_{Square} \\ \\ & = \pi r^2 + (side)^2 \\ \\ & = \pi \left(\dfrac{x}{2 \pi}\right)^2 + \left(\dfrac{5}{2} - \dfrac{1}{4}x\right)^2 \end{align}$ To minimize, differentiate Area using power and chain rules, and set equal to zero, $\begin{align} A’ & = (2)\left(\dfrac{1}{4 \pi}\right)x + 2\left(\dfrac{5}{2} - \dfrac{1}{4}x\right)\left(- \dfrac{1}{4}\right) \\ \\ 0 & = \dfrac{1}{2 \pi}x - \dfrac{1}{2}\left(\dfrac{5}{2} - \dfrac{1}{4}x\right) \\ \\ \dfrac{5}{4} & = x\left(\dfrac{1}{2\pi} + \dfrac{1}{8}\right) \\ \\ \dfrac{5}{4} & = x\left(\dfrac{4}{8\pi} + \dfrac{\pi}{8\pi}\right) \\ \\ \dfrac{5}{4} & = x\left(\dfrac{4 + \pi}{8\pi}\right) \\ \\ x & = \dfrac{5(8\pi)}{4(4 + \pi)} \\ \\ x & = \dfrac{10\pi}{4 + \pi} \end{align}$ Now determine the maximum area at this 'x' value, $\begin{align} A & = \pi \left(\dfrac{x}{2 \pi}\right)^2 + \left(\dfrac{5}{2} - \dfrac{1}{4}x\right)^2 \\ \\ & = \pi \left(\dfrac{\left(\frac{10\pi}{4 + \pi}\right)}{2 \pi}\right)^2 + \left(\dfrac{5}{2} - \dfrac{1}{4}\left(\frac{10\pi}{4 + \pi}\right)\right)^2 \\ \\ & = 3.5\ cm^2 \end{align}$

The bulk retail cost of grade 2, 0.035" titanium sheet metal is $0.04 per cm³. To make super strong (for fun) titanium soda cans with 113π cm³ volume, Determine the radius and height of one can that is most economical. Solution

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r = h =

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Calculate volume, to get an equation for 'h' to plug into the cost function later, $\begin{align} Volume & = (area\ of\ base)(height) \\ \\ 113\pi & = (\pi r^2)(h) \\ \\ h & = \dfrac{113}{r^2} \end{align}$ Determine an equation for cost, $\begin{align} Cost & = (price)(area) \\ \\ C & = (0.04)(2\pi r^2 + 2\pi r h) \\ \\ C & = 0.08 \pi r^2 + 0.08 \pi r h \end{align}$ Substitute 'h', $\begin{align} C & = 0.08 \pi r^2 + 0.08 \pi r h \\ \\ C & = 0.08 \pi r^2 + 0.08 \pi r \left( \dfrac{113}{r^2} \right) \\ \\ C & = 0.08 \pi r^2 + 9.04 \pi r^{-1} \end{align}$ Differentiate to minimize cost, C, [Testing to verify max/min not shown here. Could use intervals or slopes] $\begin{align} C’ & = 0.16 \pi r - 9.04 \pi r^{-2} \\ \\ 0 & = \dfrac{0.16 \pi r^3}{r^2} - \dfrac{9.04 \pi}{r^2} \\ \\ 0(r^2) & = 0.16 \pi r^3 - 9.04 \pi \\ \\ 0 & = 0.16 \pi r^3 - 9.04 \pi \\ \\ r^3 & = \dfrac{9.04 \pi}{0.16} \\ \\ r & = 5.62\ cm \end{align}$ Determine 'h' $\begin{align} h & = \dfrac{113}{r^2} \\ \\ h & = \dfrac{113}{(5.62\ cm)^2} \\ \\ h & = 3.58\ cm \end{align}$

Determine the maximum volume of a cone that can be placed (inscribed) inside a sphere with a diameter of 16 cm. Solution $V_{cone} = \dfrac{1}{3} \pi (r_{c})^2 h$

V =

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cm³

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See that the radius of the sphere accounts for some lengths as shown, You have the radius of the sphere, and you can use the pythagorean theorem in the triangle with sides. This means the radius 'r_c' of the cone, can be written in terms of the height 'h' $\begin{align} (8) ^2 & = r_c ^2 + (h-8)^2 \\ \\ r_c & = \sqrt{8^2 - (h-8) ^2} \\ \\ r_c & = \sqrt{64 - (h^2 - 16h + 64)} \\ \\ r_c & = \sqrt{-h^2 + 16h} \end{align}$ To determine the maximum volume of the cone, differentiate for volume... $\begin{align} V_{cone} & = \dfrac{1}{3} \pi (r_{c})^2 h \\ \\ V & = \dfrac{1}{3} \pi \left(\sqrt{-h^2 + 16h}\right)^2 h \\ \\ V & = \dfrac{1}{3} \pi \left(-h^3 + 16h^2 \right) \\ \\ V’ & = \dfrac{1}{3} \pi \left(-3h^2 + 32h \right) \\ \\ 0 & = -\dfrac{1}{3} \pi h \left(3h - 32 \right) \end{align}$ [Testing to verify max/min not shown here. Could use intervals or slopes]

$\begin{align} h & = 0 \end{align}$ $\begin{align} 3h - 32 & = 0 \\ \\ h & = \dfrac{32}{3} \end{align}$

Now with the height, you can determine the volume, using your equation from before, $\begin{align} V_{cone} & = \dfrac{1}{3} \pi \left(-h^3 + 16h^2 \right) \\ \\ V_{cone} & = \dfrac{1}{3} \pi \left(-\left(\dfrac{32}{3}\right)^3 + 16\left(\dfrac{32}{3}\right)^2 \right) \\ \\ & = 635.5\ cm^3 \end{align}$

Optimization: Distance, Speed, Time

A group of campers are trying to get to their campsite as fast as possible, across a narrow part of a lake, before it starts to rain. They can canoe at 5 hm/hr and briskly walk (portage) at 6 km/hr, and can only travel in a straight path by canoe or portage, but can land any point along the opposite side of the lake. If the lake is 2 km wide, and the distance from the campsite to the point directly across the lake from where they started is 10 km, determine the fastest time they can make it across. [Diagram not to scale] Solution

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hours

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Set up a sketch, to represent the distances canoed and walked, Use pythagorean theorem to make an expression for the distance canoed, $\begin{align} d_{canoe} & = \sqrt{2^2 + x^2} \\ \\ d_{walk} & = (10 - x) \end{align}$ Since, distance = (speed)(time)... $\begin{align} \text{Total Time} & = \dfrac{d_{canoe}}{s_{canoe}} \ + \ \dfrac{d_{walk}}{s_{walk}} \\ \\ T & = \dfrac{\sqrt{4 + x^2}}{5} + \dfrac{10 - x}{6} \\ \\ T & = \dfrac{1}{5}\sqrt{4 + x^2} + \dfrac{10}{6} - \dfrac{1}{6}x \end{align}$ To determine the minimum time, differentiate for T, and set equal to zero, $\begin{align} T’ & = \left(\dfrac{1}{5}\right)\left(\dfrac{1}{2}\right)\left(4 + x^2\right)^{-\frac{1}{2}}(2x) \ - \ \dfrac{1}{6} \\ \\ 0 & = \dfrac{x}{5\sqrt{4 + x^2}} - \dfrac{1}{6} \\ \\ \dfrac{1}{6} & = \dfrac{x}{5\sqrt{4 + x^2}} \\ \\ \sqrt{4 + x^2} & = \dfrac{6}{5}x \\ \\ \left(\sqrt{4 + x^2}\right)^2 & = \left(\dfrac{6}{5}x\right)^2 \\ \\ 4 + x^2 & = \dfrac{36}{25}x^2 \\ \\ x^2 & = \dfrac{100}{11} \\ \\ x & = \dfrac{10}{\sqrt{11}} \end{align}$ [Testing to verify max/min not shown here. Could use intervals or slopes] $\begin{align} T & = \dfrac{1}{5}\sqrt{4 + x^2} + \dfrac{10}{6} - \dfrac{1}{6}x \\ \\ T & = \dfrac{1}{5}\sqrt{4 + \left(\dfrac{10}{\sqrt{11}}\right)^2} + \dfrac{10}{6} - \dfrac{1}{6}\left(\dfrac{10}{\sqrt{11}}\right) \\ \\ T & = \dfrac{5}{3} + \dfrac{\sqrt{11}}{15} \\ \\ T & = 1.89\ hr \end{align}$

Car 'D' starts 9 km East of Car 'G' and starts moving North at 70 km/hr at the same time that Car 'G' starts moving East at 80 km/hr. Determine the minimum distance between the cars, if the cars travel at a constant speed. Solution

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km

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At some point in time 't' the cars will form a right angle triangle at point 'D' where the distance between the cars is the hypotenuse length, 'H' $\begin{align} H^2 & = (9 - x)^2 + (y)^2 \\ \\ H & = \sqrt{(9 - x)^2 + (y)^2} \end{align}$ First we must write 'x' and 'y' in terms of the same variable 't', so we use their constant speed and distance,

$\begin{align} distance_{\,G} & = (speed_{\,G})(time) \\ \\ x & = (80)(t) \end{align}$ $\begin{align} distance_{\,D} & = (speed_{\,D})(time) \\ \\ y & = (70)(t) \end{align}$

To minimize the distance between the cars 'H' we must write 'H' in terms of the same variable, 't', $\begin{align} H & = \sqrt{(9 - x)^2 + (y)^2} \\ \\ & = \sqrt{(9 - 80t)^2 + (70t)^2} \end{align}$ To minimize 'H' differentiate, $\begin{align} H’ & = \dfrac{1}{2} \left[ (9 - 80t)^2 + (70t)^2 \right] ^{\frac{-1}{2}} \left[ 2(9 - 80t)(-80) + 2(70t)(70) \right] \\ \\ H’ & = \dfrac{ -1440 + 22,600t }{ 2\sqrt{(9 - 80t)^2 + (70t)^2} } \end{align}$ And set equal to zero, and solve, $\begin{align} H’ & = 0 \\ \\ 0 & = \dfrac{ -1440 + 22,600t }{ 2\sqrt{(9 - 80t)^2 + (70t)^2} } \\ \\ 0 & = -1440 + 22,600t \\ \\ t & = \dfrac{36}{565}\ hr \\ \\ t & ≈ 0.0637\ hr \end{align}$ Use a sign chart to substitute values surrounding this point, to inspect the slopes, to ensure the point is a minimum.

	t < 0.0637	t = 0.0637	t > 0.0637
L'	(-)	0	(+)

Now that you have verified the point is a minimum, determine the hypotenuse length, 'H' $\begin{align} H & = \sqrt{(9 - 80t)^2 + (70t)^2} \\ \\ & = \sqrt{[9 - 80(0.0637)]^2 + [70(0.0637)]^2} \\ \\ & = 85.02\ km \end{align}$

Optimization: Graphs of Functions

Determine the area of the largest quadrilateral, that fits between the function, $y = x^3 + 12$, the x-axis, and the y-axis, for x < 0. Round your answer to the nearest decimal, and state as a positive value. Solution

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units²

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Let one vertex of the quadrilateral be a point on the curve, (-x, y)

First, determine an equation for area, in terms of 'x' $\begin{align} Area & = (base)(height) \\ \\ A & = (-x)(y) \\ \\ A & = (-x)(x^3 + 12) \\ \\ A & = -x^4 - 12x \end{align}$ Now to maximize area, use power rule, and set equal to zero, [Testing to verify max/min not shown here. Could use intervals or slopes] $\begin{align} A’ & = -4x^3 - 12 \\ \\ 0 & = -4x^3 - 12 \\ \\ 4x^3 & = - 12 \\ \\ x^3 & = - 3 \\ \\ x & = \sqrt[3]{-3} \end{align}$ Calculate the area, $\begin{align} A & = -(x)^4 - 12(x) \\ \\ & = -(\sqrt[3]{-3})^4 - 12(\sqrt[3]{-3}) \\ \\ & = -6.49 \\ \\ & = 6.5\ units^2 \end{align}$ The value for area is negative because it is in the domain x < 0, but we can state the area as a positive scalar magnitude.

Determine the coordinate on the graph of the function, $y = \sqrt{-2x + 4}$, with a minimum distance from the origin. Solution

Hint Clear Info

$\bigg( \quad , \sqrt{\begin{matrix} \\ \quad \end{matrix}} \bigg)$

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Determine the minimum distance between the point, (x, y) and the origin, (4, 0) using length of a line equation: $\begin{align} L & = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \\ \\ L & = \sqrt{(x - 0)^2 + (y - 0)^2} \\ \\ L & = \sqrt{x^2 + y^2} \end{align}$ We need to write the function in terms of only 'x', so substitute the original function for 'y', $\begin{align} L & = \sqrt{x^2 + \left(\sqrt{-2x + 4}\right)^2} \\ \\ L & = \sqrt{x^2 -2x + 4} \end{align}$ Now differentiate using chain, and determine the minimum, [Testing to verify max/min not shown here. Could use intervals or slopes] $\begin{align} L’ & = \dfrac{1}{2}\left( x^2 - 2x + 4 \right) (2x - 2) \\ \\ 0 & = \left( x^2 - 2x + 4 \right) (x - 1) \end{align}$ Using the discriminant (b² - 4ac), we can see that x² - 2x + 4 is undefined, so that leaves us with only, $\begin{align} x - 1 & = 0 \\ \\ x & = 1 \end{align}$ Determine the corresponding 'y' value in the coordinate, $\begin{align} y & = \sqrt{-2(x) + 4} \\ \\ y & = \sqrt{-2(1) + 4} \\ \\ y & = \sqrt{2} \\ \\ \end{align}$

Determine the minimum slope of the tangent to the curve. Solution $y = \dfrac{-1}{x^2 + 5}$

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The minimum slope is not zero, it is the most negative number

The first derivative is the function of the slope (of the tangent), $\begin{align} y’ & = \dfrac{(0)(x^2 + 5) - (-1)(2x)}{(x^2 + 5)^2} \\ \\ y’ & = \dfrac{2x}{(x^2 + 5)^2} \end{align}$ Take the second derivative to determine the maximum or minimum points of the slope (y’). $\begin{align} y’’ & = \dfrac{(2)(x^2 + 5)^2 - (2x)[2(x^2 + 5)(2x)]}{(x^2 + 5)^4} \\ \\ y’’ & = \dfrac{(x^2 + 5)\Big[2(x^2 + 5) - 8x^2\Big]}{(x^2 + 5)^4} \\ \\ y’’ & = \dfrac{-6x^2 + 10}{(x^2 + 5)^3} \end{align}$ The max or min points occur at, $\begin{align} 0 & = \dfrac{-6x^2 + 10}{(x^2 + 5)^3} \\ \\ 6x^2 & = 10 \\ \\ x & = \pm 1.3 \end{align}$ Since there are two possible values, use a table test whether slope (of y'') is increasing or decreasing...

Each part of function y''	x < -1.3	-1.3 < x < 1.3	x > 1.3
-6x2 + 10	(—)	(+)	(—)
(x2 + 5)3	(+)	(+)	(+)
Result:	(—)	(+)	(—)

We can conclude that a minimum occurs at the point (x = -1.3) where the slope changes from (-) to (+) $\begin{align} Slope = y’ & = \dfrac{2(-1.3)}{\Big((-1.3)^2 + 5\Big)^2} \\ \\ & = -0.058 \end{align}$

Determine the area of the triangle with the shortest hypotenuse that crosses the point (4, 1) and the x-axis and y-axis. Solution

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cm²

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Use similar triangles to make an equation for 'y' in terms of 'x' $\begin{align} \dfrac{y}{1} & = \dfrac{x}{x - 4} \\ \\ y & = \dfrac{x}{x - 4} \end{align}$ Use the pythagorean theorem to make an expression for the hypotenuse, 'c', $\begin{align} c^2 & = x^2 + y^2 \\ \\ c & = \sqrt{x^2 + y^2} \\ \\ c & = \sqrt{x^2 + \left(\dfrac{x}{x - 4}\right)^2} \end{align}$ Differentiate, this may be tricky to see because it uses chain rule, quotient rule, and power rule, $\begin{align} c’ & = \dfrac{1}{2}\left(x^2 \ + \ \left(\dfrac{x}{x-4}\right)^2\right)^{\frac{-1}{2}} \ \left[ 2x \ + \ 2 \ \left(\dfrac{x}{x-4}\right)^1\left(\dfrac{(1)(x-4) \ - \ (x)(1)}{(x-4)^2}\right) \right] \\ \\ c’ & = \dfrac{1}{2}\left(x^2 \ + \ \left(\dfrac{x}{x-4}\right)^2\right)^{\frac{-1}{2}} \ (2) \ \left[ x \ + \left( \dfrac{-4x}{(x -4)^3}\right) \right] \\ \\ c’ & = \dfrac{ x \ - \ \dfrac{4x}{(x -4)^3} }{\sqrt{x^2 \ + \ \left(\dfrac{x}{x-4}\right)^2}} \end{align}$ Determine the minimum, $\begin{align} 0 & = \dfrac{ x \ - \ \dfrac{4x}{(x -4)^3} }{\sqrt{x^2 \ + \ \left(\dfrac{x}{x-4}\right)^2}} \\ \\ x \ - \ \dfrac{4x}{(x -4)^3} & = 0 \\ \\ x\left( 1 - \dfrac{4}{(x -4)^3} \right) & = 0 \end{align}$ [Testing to verify max/min not shown here. Could use intervals or slopes]

$x = 0$ $\begin{align} 1 - \dfrac{4}{(x -4)^3} & = 0 \\ \\ (x -4)^3 & = 4 \\ \\ x & = \sqrt[3]{4} + 4 \\ \\ x & = 5.59\ cm \end{align}$

Solve for height, 'y' $\begin{align} y & = \dfrac{x}{x - 4} \\ \\ & = \dfrac{5.59}{5.59 - 4} \\ \\ & = 3.52\ cm \end{align}$ Finally, the minimum area is $\begin{align} Area & = \dfrac{1}{2}(base)(height) \\ \\ & = \dfrac{1}{2}(5.59)(3.52) \\ \\ & = 9.84\ cm^2 \end{align}$

Optimization: with Trigonometry

A certain IMAX® screen is 18 m tall and 2 m above the floor. Determine the optimal distance to sit from the screen, to maximize the viewing angle 'θ' of the screen. Solution

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m

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For the lower, inner triangle, $\begin{align} \tan \beta & = \dfrac{2}{d} \\ \\ \beta & = \tan^{-1} \left(\dfrac{2}{d}\right) \end{align}$ For the whole, entire triangle, $\begin{align} \tan \left( \theta + \beta \right) & = \dfrac{18 + 2}{d} \\ \\ \theta + \beta & = \tan^{-1} \left(\dfrac{20}{d}\right) \\ \\ \theta & = \tan^{-1} \left(\dfrac{20}{d}\right) - \beta \end{align}$ Combining these two equations, $\begin{align} \theta & = \tan^{-1} \left(\dfrac{20}{d}\right) - \beta \\ \\ \theta & = \tan^{-1} \left(\dfrac{20}{d}\right) - \tan^{-1} \left(\dfrac{2}{d}\right) \end{align}$ To maximize the viewing angle, 'θ', differentiate, and set equal to zero, $\begin{align} \theta & = \tan^{-1} \left(\dfrac{20}{d}\right) - \tan^{-1} \left(\dfrac{2}{d}\right) \\ \\ \theta \, ’ & = \dfrac{1}{1 + (20d^{-1})^2} (-20d^{-2}) - \dfrac{1}{1 + (2d^{-1})^2} (-2d^{-2}) \\ \\ 0 & = \dfrac{-20}{d^2(1 + 400d^{-2})} - \dfrac{-2}{d^2(1 + 4d^{-2})} \\ \\ 0 & = \dfrac{-20}{d^2 + 400} + \dfrac{2}{d^2 + 4} \end{align}$ Solve for the distance 'd', $\begin{align} \dfrac{20}{d^2 + 400} & = \dfrac{2}{d^2 + 4} \\ \\ 20(d^2 + 4) & = 2(d^2 + 400) \\ \\ 18d^2 & = 720 \\ \\ d^2 & = 40 \\ \\ d & = \pm \sqrt{40} \\ \\ d & ≈ 6.3\ m \end{align}$ [Max/min verification (first or second derivative) check not shown here].

Determine the angle 'θ' that gives the following isosceles triangle a maximum area, for 0 < θ < 180˚. Solution

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˚

Hint Unavailable

Make a sketch of a right angle triangle with the following properties, Determine expressions for 'b' and 'h' using trig,

$\begin{align} \cos \left(\dfrac{\theta}{2}\right) & = \dfrac{opposite}{hypotenuse} \\ \\ \sin \left(\dfrac{\theta}{2}\right) & = \left(\dfrac{\left(\frac{b}{2}\right)}{5}\right) \\ \\ b & = 10\sin \left(\dfrac{\theta}{2}\right) \end{align}$ $\begin{align} \cos \left(\dfrac{\theta}{2}\right) & = \dfrac{adjacent}{hypotenuse} \\ \\ \cos \left(\dfrac{\theta}{2}\right) & = \dfrac{h}{5} \\ \\ h & = 5\cos \left(\dfrac{\theta}{2}\right) \end{align}$

To maximize the area of the whole isosceles triangle, $\begin{align} Area_{\, triangle} & = \dfrac{1}{2}(base)(height) \\ \\ & = \dfrac{1}{2}\left[ 10\sin \left(\dfrac{\theta}{2}\right) \right] \left[ 5\cos \left(\dfrac{\theta}{2}\right) \right] \\ \\ & = 25\sin\left(\dfrac{\theta}{2}\right) \cos\left(\dfrac{\theta}{2}\right) \end{align}$ To differentiate 'A', let $u = 25\sin\left(\dfrac{\theta}{2}\right)$ and $v = \cos\left(\dfrac{\theta}{2}\right)$ $\begin{align} A ’ & = u’v + uv’ \\ \\ A ’ & = \left[ 25\cos\left(\dfrac{\theta}{2}\right)\left(\dfrac{1}{2}\right) \right]\left[ \cos\left(\dfrac{\theta}{2}\right) \right] + \left[ 25\sin\left(\dfrac{\theta}{2}\right) \right]\left[ -\sin\left(\dfrac{\theta}{2}\right)\left(\dfrac{1}{2}\right) \right] \\ \\ 0 & = \left[ \dfrac{25}{2} \right]\left[ \cos ^2 \left(\dfrac{\theta}{2}\right) - \sin ^2 \left(\dfrac{\theta}{2}\right)\right] \end{align}$ So, solve using double angle formula, and since it's the interior angle of a triangle, it should make sense that the domain is 0 < θ < 180˚ $\begin{align} \cos ^2 \left(\dfrac{\theta}{2}\right) - \sin ^2 \left(\dfrac{\theta}{2}\right) & = 0 \\ \\ \cos \left[ 2 \left(\dfrac{\theta}{2}\right) \right] & = 0 \\ \\ \cos(\theta) & = 0 \\ \\ \theta & = 90˚ \quad\quad\quad \text{This is a standard value} \end{align}$ [Verify that a maximum occurs with either a first, or second derivative test, not shown here.]

Positive and Negative Intervals

Given the graph of the function

Determine the positive and negative intervals for the diagram of the function above. Solution

1. Positive: -2 ≤ x ≤ 0 and Negative: 0 ≤ x ≤ 2
2. Positive: -1 ≤ x ≤ 1 and Negative: 1 ≤ x ≤ -1
3. Positive: 1 ≤ x ≤ -1 and Negative: -1 ≤ x ≤ 1
4. Positive: x ≤ -2, 0 ≤ x ≤ 2 and Negative: x ≥ 2, -2 ≤ x ≤ 0

Positive and negative intervals occur are based on where the function exists above or below the x-axis for positive or negative values of y.

Determine the intervals of increase and decrease for the diagram of the function above. Solution

1. Increase: -2 ≤ x ≤ 0 and Decrease: 0 ≤ x ≤ 2
2. Increase: -1 < x < 1 and Decrease: -1 > x > 1
3. Increase: 1 ≤ x ≤ -1 and Decrease: -1 ≤ x ≤ 1
4. Increase: x < -2, 0 < x < 2 and Decrease: x > 2, -2 < x < 0

Think of the slope of the tangent at any one point. The intervals of increase have a tangent with a positive slope, and the intervals of decrease have a tangent with a negative slope.

Determine the where the function is concave up for the diagram of the function above. Solution

1. x < -1, x > 1
2. -1 < x < 1
3. x < -2, 0 < x < 2
4. -2 < x < 0, x > 2

The function is concave up on the intervals: -2 < x < 0, and x > 2

Absolute and Local Maximum and Minimum

Determine the correct statement. Solution

1. The function has an absolute maximum at (3, 3) and an absolute minimum at (-1, -2)
2. The function has a local maximum at (3, 3) and a local minimum at (-1, -2)

The points are considered local when the function tends to infinity in the space.

Determine the correct statement. Solution

1. The function has an absolute maximum at (1, 3) and an absolute minimum at (7, -2)
2. The function has a local maximum at (1, 3) and a local minimum at (7, -2)

The points are considered absolute (or global) when the 'y' value is the most extreme over the entire function.

Characterizing Points of Inflection

You know there is a point of inflection (POI) at 'x' because it has opposite concavity on either side of 'x', and the same sign (positive/negative) of slope on either side of 'x'. Given the following characteristics, classify the POI. Solution f'(x) ≠ 0, and f'(x) defined

1. Horizontal POI
2. Vertical POI
3. Oblique POI
4. It cannot be determined

For each:

• Horizontal POI: f'(x) = 0
• Vertical POI: f'(x) = undefined
• Oblique POI: f'(x) ≠ 0