Grade 12 Calculus & Vectors (MCV4U)
MCV4U introduces two of the most powerful tools in mathematics: differential calculus and vector geometry. The calculus strand develops the derivative from limits through to optimization and curve sketching, providing the foundation for university-level Calculus I. The vectors strand develops geometric reasoning in two and three dimensions, with applications to physics and linear algebra. MHF4U must be taken concurrently or previously.
1. Rates of Change and the Derivative from First Principles
Average and Instantaneous Rate of Change
Rates of Change
The average rate of change of $f$ over $[a, b]$ is the slope of the secant line: $$\text{AROC} = \frac{f(b) - f(a)}{b - a}$$
The instantaneous rate of change at $x = a$ is the slope of the tangent line, defined as: $$f'(a) = \lim_{h \to 0} \frac{f(a + h) - f(a)}{h}$$
Worked Example 1.1 — Derivative from First Principles
Find $f'(x)$ for $f(x) = 3x^2 - 5$ using the limit definition.
Step 1 Write the difference quotient: $$\frac{f(x+h) - f(x)}{h} = \frac{[3(x+h)^2 - 5] - [3x^2 - 5]}{h}$$
Step 2 Expand $(x+h)^2 = x^2 + 2xh + h^2$: $$= \frac{3x^2 + 6xh + 3h^2 - 5 - 3x^2 + 5}{h} = \frac{6xh + 3h^2}{h} = 6x + 3h$$
Step 3 Take the limit as $h \to 0$: $$f'(x) = \lim_{h \to 0}(6x + 3h) = 6x$$
2. Differentiation Rules
Summary of Differentiation Rules
- Constant rule: $\dfrac{d}{dx}[c] = 0$
- Power rule: $\dfrac{d}{dx}[x^n] = nx^{n-1}$
- Constant multiple: $\dfrac{d}{dx}[cf(x)] = cf'(x)$
- Sum/difference: $\dfrac{d}{dx}[f \pm g] = f' \pm g'$
- Product rule: $\dfrac{d}{dx}[fg] = f'g + fg'$
- Quotient rule: $\dfrac{d}{dx}\!\left[\dfrac{f}{g}\right] = \dfrac{f'g - fg'}{g^2}$
- Chain rule: $\dfrac{d}{dx}[f(g(x))] = f'(g(x)) \cdot g'(x)$
- Trig derivatives: $\dfrac{d}{dx}[\sin x] = \cos x$; $\dfrac{d}{dx}[\cos x] = -\sin x$; $\dfrac{d}{dx}[\tan x] = \sec^2 x$
- Exponential: $\dfrac{d}{dx}[e^x] = e^x$; $\dfrac{d}{dx}[a^x] = a^x \ln a$
- Logarithm: $\dfrac{d}{dx}[\ln x] = \dfrac{1}{x}$
Worked Example 2.1 — Product and Chain Rules
Differentiate $f(x) = x^2 \sin(3x)$.
Use the product rule with $u = x^2$ and $v = \sin(3x)$:
$u' = 2x$ and $v' = \cos(3x) \cdot 3 = 3\cos(3x)$ (chain rule for $v$).
$$f'(x) = 2x\sin(3x) + x^2 \cdot 3\cos(3x) = 2x\sin(3x) + 3x^2\cos(3x)$$
Worked Example 2.2 — Quotient Rule
Differentiate $g(x) = \dfrac{e^x}{x^2 + 1}$.
$u = e^x$, $v = x^2 + 1$, so $u' = e^x$, $v' = 2x$: $$g'(x) = \frac{e^x(x^2+1) - e^x \cdot 2x}{(x^2+1)^2} = \frac{e^x(x^2 - 2x + 1)}{(x^2+1)^2} = \frac{e^x(x-1)^2}{(x^2+1)^2}$$
Worked Example 2.3 — Chain Rule
Differentiate $h(x) = \ln(x^3 + 5x)^4$.
Rewrite using log power law: $h(x) = 4\ln(x^3 + 5x)$.
$$h'(x) = 4 \cdot \frac{1}{x^3 + 5x} \cdot (3x^2 + 5) = \frac{4(3x^2 + 5)}{x^3 + 5x}$$
3. Derivatives and Applications: Curve Sketching and Optimization
First and Second Derivative Tests
Increasing/Decreasing and Concavity
- $f'(x) > 0$ on an interval: $f$ is increasing there
- $f'(x) < 0$ on an interval: $f$ is decreasing there
- A local maximum occurs at $x = c$ if $f'(c) = 0$ and $f'$ changes from positive to negative
- A local minimum occurs at $x = c$ if $f'(c) = 0$ and $f'$ changes from negative to positive
- $f''(x) > 0$: $f$ is concave up; $f''(x) < 0$: concave down
- An inflection point is where $f''$ changes sign
Worked Example 3.1 — Full Curve Sketch
Analyze and sketch $f(x) = x^3 - 3x^2 - 9x + 5$.
Critical points $f'(x) = 3x^2 - 6x - 9 = 3(x^2 - 2x - 3) = 3(x-3)(x+1) = 0$. Critical points: $x = 3$ and $x = -1$.
Sign of $f'$ On $(-\infty, -1)$: $f' > 0$ (increasing). On $(-1, 3)$: $f' < 0$ (decreasing). On $(3, \infty)$: $f' > 0$ (increasing).
Local extrema Local max at $x = -1$: $f(-1) = -1 - 3 + 9 + 5 = 10$. Local min at $x = 3$: $f(3) = 27 - 27 - 27 + 5 = -22$.
Inflection $f''(x) = 6x - 6 = 0 \Rightarrow x = 1$. $f(1) = 1 - 3 - 9 + 5 = -6$. Inflection point at $(1, -6)$.
Optimization
Worked Example 3.2 — Optimization Problem
A farmer has 200 m of fencing to enclose a rectangular field. One side is along a river and needs no fence. Find the dimensions that maximize the area.
Setup Let $x$ = length parallel to river (no fence), $y$ = width. Constraint: $x + 2y = 200 \Rightarrow x = 200 - 2y$. Objective: maximize $A = xy = (200 - 2y)y = 200y - 2y^2$.
Optimize $\dfrac{dA}{dy} = 200 - 4y = 0 \Rightarrow y = 50$ m.
Answer $x = 200 - 2(50) = 100$ m. Maximum area $= 100 \times 50 = 5000$ m$^2$. Confirmed by $\dfrac{d^2A}{dy^2} = -4 < 0$ (maximum).
4. Introduction to Integration
The Antiderivative
$F(x)$ is an antiderivative of $f(x)$ if $F'(x) = f(x)$. The general antiderivative (indefinite integral) is: $$\int f(x)\,dx = F(x) + C$$ where $C$ is the constant of integration.
Basic antiderivative rules:
- $\displaystyle\int x^n\,dx = \frac{x^{n+1}}{n+1} + C$, for $n \neq -1$
- $\displaystyle\int e^x\,dx = e^x + C$
- $\displaystyle\int \sin x\,dx = -\cos x + C$
- $\displaystyle\int \cos x\,dx = \sin x + C$
- $\displaystyle\int \frac{1}{x}\,dx = \ln|x| + C$
Worked Example 4.1 — Antiderivatives
Find: (a) $\displaystyle\int (4x^3 - 6x + 2)\,dx$ and (b) $\displaystyle\int \frac{3}{\sqrt{x}}\,dx$.
(a) $\dfrac{4x^4}{4} - \dfrac{6x^2}{2} + 2x + C = x^4 - 3x^2 + 2x + C$.
(b) Rewrite: $3x^{-1/2}$. Then $\displaystyle\int 3x^{-1/2}\,dx = \dfrac{3x^{1/2}}{1/2} + C = 6\sqrt{x} + C$.
Worked Example 4.2 — Initial Value Problem
Given that $f'(x) = 6x^2 - 4x$ and $f(1) = 3$, find $f(x)$.
Step 1 Integrate: $f(x) = 2x^3 - 2x^2 + C$.
Step 2 Use the initial condition: $f(1) = 2 - 2 + C = C = 3$, so $C = 3$.
$f(x) = 2x^3 - 2x^2 + 3$.
5. Vectors in 2D and 3D
Vector Operations
A vector $\vec{v} = (v_1, v_2, v_3)$ in 3D has:
- Magnitude: $|\vec{v}| = \sqrt{v_1^2 + v_2^2 + v_3^2}$
- Addition: $\vec{u} + \vec{v} = (u_1 + v_1, u_2 + v_2, u_3 + v_3)$
- Scalar multiplication: $k\vec{v} = (kv_1, kv_2, kv_3)$
- Unit vector: $\hat{v} = \dfrac{\vec{v}}{|\vec{v}|}$
Dot Product
Dot Product
$\vec{u} \cdot \vec{v} = u_1v_1 + u_2v_2 + u_3v_3 = |\vec{u}||\vec{v}|\cos\theta$
where $\theta$ is the angle between the vectors. Two vectors are perpendicular (orthogonal) if and only if $\vec{u} \cdot \vec{v} = 0$.
Worked Example 5.1 — Angle Between Vectors
Find the angle between $\vec{a} = (2, -1, 3)$ and $\vec{b} = (1, 4, -2)$.
$\vec{a} \cdot \vec{b} = (2)(1) + (-1)(4) + (3)(-2) = 2 - 4 - 6 = -8$.
$|\vec{a}| = \sqrt{4 + 1 + 9} = \sqrt{14}$, $|\vec{b}| = \sqrt{1 + 16 + 4} = \sqrt{21}$.
$\cos\theta = \dfrac{-8}{\sqrt{14}\sqrt{21}} = \dfrac{-8}{\sqrt{294}}$, so $\theta = \cos^{-1}\!\left(\dfrac{-8}{\sqrt{294}}\right) \approx 117.8°$.
Cross Product
Cross Product
For $\vec{u} = (u_1, u_2, u_3)$ and $\vec{v} = (v_1, v_2, v_3)$: $$\vec{u} \times \vec{v} = (u_2v_3 - u_3v_2,\; u_3v_1 - u_1v_3,\; u_1v_2 - u_2v_1)$$
The cross product is perpendicular to both $\vec{u}$ and $\vec{v}$, and $|\vec{u} \times \vec{v}| = |\vec{u}||\vec{v}|\sin\theta$ gives the area of the parallelogram spanned by $\vec{u}$ and $\vec{v}$.
Worked Example 5.2 — Cross Product
Find $\vec{a} \times \vec{b}$ where $\vec{a} = (3, -1, 2)$ and $\vec{b} = (1, 2, -4)$.
$$\vec{a} \times \vec{b} = \begin{vmatrix} \vec{i} & \vec{j} & \vec{k} \\ 3 & -1 & 2 \\ 1 & 2 & -4 \end{vmatrix}$$
$= \vec{i}[(-1)(-4) - (2)(2)] - \vec{j}[(3)(-4) - (2)(1)] + \vec{k}[(3)(2) - (-1)(1)]$
$= \vec{i}[4 - 4] - \vec{j}[-12 - 2] + \vec{k}[6 + 1]$
$= (0, 14, 7)$
6. Lines and Planes in Space
Equations of Lines in 3D
A line through point $P_0(x_0, y_0, z_0)$ with direction vector $\vec{d} = (a, b, c)$:
Vector equation: $(x, y, z) = (x_0, y_0, z_0) + t(a, b, c)$
Parametric equations: $x = x_0 + ta$, $y = y_0 + tb$, $z = z_0 + tc$
Symmetric equations: $\dfrac{x - x_0}{a} = \dfrac{y - y_0}{b} = \dfrac{z - z_0}{c}$ (when $a, b, c \neq 0$)
Equations of Planes
A plane with normal vector $\vec{n} = (A, B, C)$ through $P_0(x_0, y_0, z_0)$: $$A(x - x_0) + B(y - y_0) + C(z - z_0) = 0$$ or equivalently $Ax + By + Cz + D = 0$ where $D = -Ax_0 - By_0 - Cz_0$.
Worked Example 6.1 — Equation of a Plane
Find the equation of the plane containing points $A(1, 0, 2)$, $B(3, -1, 1)$, and $C(0, 2, 4)$.
Step 1 Find two vectors in the plane: $\vec{AB} = (2, -1, -1)$ and $\vec{AC} = (-1, 2, 2)$.
Step 2 The normal vector is $\vec{n} = \vec{AB} \times \vec{AC}$: $$\vec{n} = ((-1)(2) - (-1)(2),\; (-1)(-1) - (2)(2),\; (2)(2) - (-1)(-1)) = (0, -3, 3)$$
Simplify by dividing by $-3$: use $\vec{n} = (0, 1, -1)$.
Step 3 Plane equation through $A(1, 0, 2)$: $0(x-1) + 1(y - 0) + (-1)(z - 2) = 0 \Rightarrow y - z + 2 = 0$.
7. Practice Problems
Problem 1 — First Principles
Use the limit definition to find $f'(x)$ for $f(x) = \dfrac{1}{x}$.
Show Solution
$\dfrac{f(x+h)-f(x)}{h} = \dfrac{\frac{1}{x+h} - \frac{1}{x}}{h} = \dfrac{\frac{x - (x+h)}{x(x+h)}}{h} = \dfrac{-h}{hx(x+h)} = \dfrac{-1}{x(x+h)}$.
As $h \to 0$: $f'(x) = \dfrac{-1}{x^2}$.
Problem 2 — Differentiation Rules
Differentiate $y = \dfrac{\sin x}{e^x + 1}$.
Show Solution
Using the quotient rule with $u = \sin x$, $v = e^x + 1$, $u' = \cos x$, $v' = e^x$:
$$y' = \frac{\cos x(e^x + 1) - \sin x \cdot e^x}{(e^x + 1)^2} = \frac{e^x(\cos x - \sin x) + \cos x}{(e^x + 1)^2}$$
Problem 3 — Chain Rule
Differentiate $f(x) = \sqrt{4x^2 - 3x + 1}$.
Show Solution
$f(x) = (4x^2 - 3x + 1)^{1/2}$. By the chain rule:
$f'(x) = \dfrac{1}{2}(4x^2 - 3x + 1)^{-1/2} \cdot (8x - 3) = \dfrac{8x - 3}{2\sqrt{4x^2 - 3x + 1}}$
Problem 4 — Optimization
Find the dimensions of a closed cylindrical can with volume 1000 cm$^3$ that minimizes the total surface area.
Show Solution
Let radius $= r$, height $= h$. Constraint: $\pi r^2 h = 1000 \Rightarrow h = \dfrac{1000}{\pi r^2}$.
Surface area: $SA = 2\pi r^2 + 2\pi r h = 2\pi r^2 + \dfrac{2000}{r}$.
$\dfrac{dSA}{dr} = 4\pi r - \dfrac{2000}{r^2} = 0 \Rightarrow 4\pi r^3 = 2000 \Rightarrow r^3 = \dfrac{500}{\pi} \Rightarrow r = \sqrt[3]{\dfrac{500}{\pi}} \approx 5.42$ cm.
$h = \dfrac{1000}{\pi(5.42)^2} \approx 10.84$ cm (note $h = 2r$, confirming the optimal can is as tall as its diameter).
Problem 5 — Antiderivative
Evaluate $\displaystyle\int \left(3\cos x - \frac{2}{x} + 5e^x\right)dx$.
Show Solution
$3\sin x - 2\ln|x| + 5e^x + C$.
Problem 6 — Dot Product
Find the value of $k$ such that $\vec{a} = (3, k, -2)$ and $\vec{b} = (k, 4, 6)$ are perpendicular.
Show Solution
$\vec{a} \cdot \vec{b} = 3k + 4k - 12 = 7k - 12 = 0 \Rightarrow k = \dfrac{12}{7}$.
Problem 7 — Cross Product Application
Find the area of the triangle with vertices $A(1, 0, 0)$, $B(0, 2, 0)$, and $C(0, 0, 3)$.
Show Solution
$\vec{AB} = (-1, 2, 0)$, $\vec{AC} = (-1, 0, 3)$.
$\vec{AB} \times \vec{AC} = (2 \cdot 3 - 0 \cdot 0,\; 0 \cdot (-1) - (-1) \cdot 3,\; (-1)(0) - (2)(-1)) = (6, 3, 2)$.
$|\vec{AB} \times \vec{AC}| = \sqrt{36 + 9 + 4} = \sqrt{49} = 7$.
Area of triangle $= \dfrac{1}{2} \times 7 = \dfrac{7}{2}$ sq units.
Problem 8 — Lines and Planes
Find the parametric equations of the line through $P(2, -1, 3)$ and $Q(5, 1, -2)$.
Show Solution
Direction vector: $\vec{PQ} = (3, 2, -5)$. Using point $P$:
$x = 2 + 3t$, $y = -1 + 2t$, $z = 3 - 5t$, $t \in \mathbb{R}$.