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# Calculus and Vectors MCV4U

This course builds on students' previous experience with functions and their developing understanding of rates of change. Students will solve problems involving geometric and algebraic representations of vectors and representations of lines and planes in three dimensional space; broaden their understanding of rates of change to include the derivatives of polynomial, sinusoidal, exponential, rational, and radical functions; and apply these concepts and skills to the modelling of real-world relationships. Students will also refine their use of the mathematical processes necessary for success in senior mathematics. This course is intended for students who choose to pursue careers in fields such as science, engineering, economics, and some areas of business, including those students who will be required to take a university-level calculus, linear algebra, or physics course. Co-rerequisite: Grade 12 Advanced Functions MHF4U

# Vectors

## A sailboat undergoes a force vector of 50N [N 40˚ W] and a second force vector of 30N [E 10˚ N]. # DOT PRODUCT

## Calculate the dot product of the following given the magnitudes and angle.

$\vec{a}•\vec{b} = |\vec{a}| |\vec{b}|\cosθ$

# Cross Product

## Given the vectors below. ## The area of a triangle is half the area of a parallelogram. Complete the following using the points below, assuming they all lie on the same plane. A(1, -2, -8/3), B(4, 0, -4), C(2, 2, -2)

# Vector Equations of Planes

## Given the following vector equations of planes:

\begin{align} \vec{r_1} & = (1, 2, 3) + s\langle -3, -2, -2\rangle + t\langle 0, 0, -2\rangle \\ \\ \vec{r_2} & = (-5, -2, -5) + s\langle -3, -2, -2\rangle + t\langle -6, -4, -8\rangle \end{align}

# Limits

## For the limit:

$\lim\limits_{ x\rightarrow 0 }{ \dfrac { \sqrt { x+1 } -1 }{ x } }$

## Determine the values for the cubic function below (where a, b, c ∈ R and a, b, c ≠ 0).

$f(x) = a{ x }^{ 3 }+b{ x }^{ 2 }+(2a + b)x + c$ Given the following information: \begin{align} f(0) = \ 2, \\ \\ \\ \lim\limits_{ x\rightarrow 1 }{ f(x) } = \ 6, \\ \\ \\ \lim\limits_{ x\rightarrow -4 }{ f(x) } = \ 3 \end{align}

## The amount of bacteria, b, in hundreds, on a piece of moldy cheese, in time, t hours, can be represented by the following piecewise function:

\begin{align} & b(t)=1 + t^2 , \quad if \quad 0 \le t \le 5 \\ \\ & b(t)=3 + \frac{ 3 }{ 2 }t^2, \quad if \quad 5 < t \le 10 \\ \\ \end{align}

## Using the following statements for the questions below.

1. $\lim\limits_{ x\rightarrow a }{ k } = k$, for any constant k
2. $\lim\limits_{ x\rightarrow a }{ \left[ f(x)\pm g(x) \right] } = \lim\limits_{ x\rightarrow a }{ f(x) } \pm \lim\limits_{ x\rightarrow a }{ g(x) }$
3. $\lim\limits_{ x\rightarrow a }{ \left[ c·f(x) \right] = } c·a\left[ \lim\limits_{ x\rightarrow a }{ f(x) } \right]$ for any constant c

# Rate of Change

## Linear density is a measurement of the change in mass of a material, over a length.

$Linear\ Density = \dfrac{\Delta\ mass}{\Delta\ length}$ d(x) represents the mass (in kg) of a water pipe along its length (in m), over 0 < x < 50 m, $d(x) = \sqrt{0.25x + 0.125}$

## Given the position function, where s(t) is position in meters, at time 't' in seconds.

$s(t) = \dfrac{1}{4}t^2 - 5t + 7$

# Modeling Derivatives (Motion, Optimization)

## An object is moving in a straight line. Its velocity, v is shown on the graph below as it varies with time t. ## The 1-dimensional motion of an object is tracked on the three graphs: position s(t), velocity v(t), and acceleration a(t). The motion is mapped over the phases A - F, and the points I - VII. [The positive direction is defined as 'to the right', and the position is comprised of piecewise, quadratic motions]. ## Displacement (d) is the change in position (∆s).

\begin{align} \vec{d} & = \Delta \vec{s} \\ \\ & = \vec{s_2} - \vec{s_1} \end{align}

## A bird flies in a straight line with position s in meters after t seconds represented by the equation:

$s(t)=\frac{1}{3} t^2 - 2t + 5$

## For a moving object where s is position, v is velocity, and k and h are constants such that k, h ∈ℝ. Given the equations for displacement and velocity:

\begin{align} & s(t) =k{ t }^{ 2 }+(2h+1)t+5 \\ \\ & v(t)=7t+4 \\ \\ \end{align}

## A ball is thrown vertically from the ground of an unknown planet. Its position, s in meters above the ground at t seconds is modeled by the following function:

$s(t)=-3.5{ t }^{ 2 }+16t+2.1$

## The height 's(t)' of a projectile over time 't' is modeled by the function below.

$s(t) = -2t^2 + 15t + 1$

# Curve Sketching (Differentiation)

## Given the graph of the function ## Complete the questions for the sketch of some, fictitious function below.

 f'(x) = 0 f'(x) = D.N.E. f'(x) = Undefined notes: Critical Point ✓ ✓ the most general term (also covers vertical POI) Stationary Point ✓ same or opposite slope on either side (more of a general term) Turning Point ✓ opposite slope on either side Horizontal Point of Inflection (POI) ✓ opposite concavity on either side, where f''(x) = 0 Vertical Point of Inflection (POI) ✓ opposite concavity on either side, where f''(x) = undefined Cusp or Corner ✓ different derivative (or limit) on either side: f'(x-) ≠ f'(x+) ## For the function...

$f(x)={ x }^{ 5 }-3{ x }^{ 4 }+5{ x }^{ 3 }$

## For the function.

$f(x)=\dfrac { { x }^{ 2 }-9 }{ x+1 }$

## For the function:

$f(x)=\dfrac { 3{ x }^{ 2 }+8 }{ { x }^{ 2 }-9 }$

## Determine each of the properties for the function:

$f(x) = \dfrac{x + 4}{(x-2)(x+2)}$

# Derivatives (Trig, Implicit, Log, Exp) & Related Rates

## Given:

\begin{align} f(x)=\sec{ x } \\ \\ g(x)=\csc{ x } \end{align}

## Given the following information.

\begin{align} & \lim _{ \theta \rightarrow 0 }{ \frac { \sin { \theta } }{ \theta } =1 } \\ \\ & \lim _{ \theta \rightarrow 0 }{ \frac { \cos { \theta } -1 }{ \theta } =0 } \\ \\ & \sin { (A\pm B)=\sin { A\cos { B\pm \sin { B\cos { A } } } } } \\ \\ & \cos { (A\pm B)=\cos { A\cos { B\mp \sin { A\sin { B } } } } } \\ \\ \end{align}

## Use implicit differentiation.

$3x^2 + y^2 = 25$

## Remember the log laws to use when differentiating the questions below.

\begin{align} & \ln(ab) = \ln(a) + \ln(b) \\ \\ & \ln\left(\frac{a}{b}\right) = \ln(a) - \ln(b) \\ \\ & \ln(a^n) = n · \ln(a) \\ \\ & \ln(e^a) = a \\ \\ & e^{\ln(a)} = a \\ \\ \end{align}

## For the function...

\begin{align} f(x) & = \big[ \ln(3x) \big]^3 \end{align}

## For the function:

$f(x)=\dfrac { { x }^{ 3 }+\sqrt { x } -3 }{ { e }^{ 3{ x }^{ 2 }+1 } }$

## Given:

$f(x)={ e }^{ 2{ x }^{ 2 }+5x+3 }$

## A particle moves in a straight line such that its displacement, “s,” in a time “t” can be modeled by the following function:

$s(t)=3\sin { t } +4\cos { 2t } -{ e }^{ 0.3t }$

# Related Rates

## Solve the related rate problems. For some of the problems, remember that for uniform motion:

\begin{align} distance & = (speed)(time) \\ \\ displacement & = (velocity)(time) \end{align}

# (AP) Integration (Anti-Differentiation)

## Evaluate the following indefinite integrals by inspection.

$\displaystyle\int{\left( kx^n \right)}\, dx = \dfrac{kx^{n+1}}{n + 1} + c$

## Evaluate the following indefinite integrals by inspection.

$\displaystyle\int{\left( ax + b \right)^n}\, dx = \dfrac{(ax + b)^{n+1}}{a(n+1)} + c$

## Evaluate the following indefinite integrals by inspection.

$\displaystyle\int{\left( \dfrac{n}{ax + b} \right)}\, dx = \dfrac{n}{a} \ln|ax+b| + c$

## Evaluate the following indefinite integrals by inspection.

$\displaystyle\int{\left( e^{\ ax + b} \right)}\, dx = \dfrac{1}{a} e^{ax + b} + c$

## Evaluate the following indefinite integrals by inspection.

\begin{align} \displaystyle\int{\Big( \sin(ax + b) \Big)}\, dx = \dfrac{-\cos(ax + b)}{a} + c \\ \\ \displaystyle\int{\Big( \cos(ax + b) \Big)}\, dx = \dfrac{\sin(ax + b)}{a} + c \end{align}

## For the sake of comparison, and for your edification, using the following integral...

$\displaystyle\int{x^7 \sqrt{5x^4 + 2\pi} \,dx}$

## Solve the following integrals by using substitution. [This formula is a general guide, but not a precise procedure. Your steps will differ based on the question.]

\begin{align} & \displaystyle\int{f(u) \, u’ }\,dx \\ \\ = & \displaystyle\int{f(u) \dfrac{du}{\cancel{dx}} }\,\cancel{dx} \\ \\ = & \displaystyle\int{f(u)}·\,du \\ \\ \end{align}

## Solve the following integrals by parts. The derivation of 'integration by parts' is based on product rule, shown below. Generally, you want to use 'integration by parts' after 'integration by substitution' doesn't work. Try to understand all the different types.

\begin{align} \dfrac{d}{d\,x} (u \cdot v) = & u’\,v + u\,v’ \\ \\ \displaystyle\int{\dfrac{d}{d\,x} (u \cdot v)} = & \displaystyle\int{\left(u’\,v + u\,v’\right)} \\ \\ u \cdot v = & \displaystyle\int{u’\,v} + \displaystyle\int{u\,v’} \\ \\ \displaystyle\int{u\,v’} = & u \cdot v - \displaystyle\int{u’\,v} \end{align}

# Motion/Kinematics

## Given the formula map below....

$\begin{array}{lcl} & & \xleftarrow{\color{DeepPink}{\displaystyle\int{\bigg( v(t) \bigg)} \, dt}} & & \xleftarrow{\color{DeepPink}{\displaystyle\int{\bigg( a(t) \bigg)} \, dt}} & & \\ \color{RoyalBlue}{s(t)} & & & \quad \color{RoyalBlue}{v(t)} \quad & & & \color{RoyalBlue}{a(t)} \\ & & \xrightarrow[\color{Lime}{\displaystyle{s\,’(t)}}]{} & & \xrightarrow[\color{Lime}{\displaystyle{v\,’(t)}}]{} & & \\ \end{array}$

## Given the velocity function below, in m/s...

$v(t) = 3t - 2$

## An object travels with the acceleration below, in m/s.

\begin{align} a(t) = \sin(2t) \\ \\ \end{align}

### Calculate the displacement from 4s to 7s if the initial position (position zero) is 2m when time equals $\pi$ seconds. Solution Position is the integral of velocity... \begin{align} s(t) & = \displaystyle\int{\bigg( v(t) \bigg)} \, dt \\ \\ & = \displaystyle\int{\bigg( \dfrac{-\cos(2t)}{2} \bigg)} \, dt \\ \\ & = \dfrac{-\sin(2t)}{4} + c \\ \\ \end{align} Plug in the givens to get the equation, \begin{align} s(t) & = \dfrac{-\sin(2t)}{4} + c \\ \\ 2 & = \dfrac{-\sin(2\pi)}{4} + c \\ \\ 2 & = 0 + c \\ \\ c & = 2 \\ \\ s(t) & = \dfrac{-\sin(2t)}{4} + 2 \\ \\ \end{align} Displacement is, \begin{align} \text{displacement} & = s(t_2) - s(t_1) \\ \\ & = \dfrac{-\sin(2(7))}{4} + 2 - \left[ \dfrac{-\sin(2(4))}{4} + 2 \right] \\ \\ & = \dfrac{-\sin(14)}{4} + \dfrac{\sin(8)}{4} \\ \\ & = -0.0003\ m \\ \\ \end{align} (With calculator in radians) Percent complete:
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