UNIT 9: PARAMETRIC EQUATIONS, POLAR COORDINATES, AND VECTOR-VALUED FUNCTIONS (BC ONLY)

1. INTRODUCTION

In our previous explorations of calculus, we have primarily dealt with functions in the form $y = f(x)$, where a single output $y$ depends on a single input $x$. While this rectangular coordinate system is powerful, it is often insufficient for describing the complexities of the physical world. Consider the path of a fly buzzing around a room, the motion of a planet orbiting a star, or the trajectory of a projectile launched into the wind. These paths often loop back on themselves, cross their own tracks, or depend on time as an independent variable.

Unit 9 introduces three interconnected frameworks that extend the power of calculus to these scenarios: Parametric Equations, Vector-Valued Functions, and Polar Coordinates. Parametric equations allow us to define $x$ and $y$ separately in terms of a third variable, usually time $t$. Vector-valued functions provide a robust language for physics, describing position, velocity, and acceleration in multi-dimensional space. Finally, Polar coordinates offer a shift in perspective, defining points based on distance and angle rather than horizontal and vertical shifts—essential for analyzing circular or spiral behaviors.

By the end of this unit, you will be able to calculate the speed of a particle at any moment, determine the length of a winding path, and find the area enclosed by exotic shapes like cardioids and limaçons. This is the bridge between single-variable calculus and the multi-variable calculus used in advanced engineering, physics, and robotics.

2. ALL KEY CONCEPTS, TERMS, FOUNDATIONAL KNOWLEDGE, and PRINCIPLES

Key Terms and Definitions

1. Parameter: An independent variable (usually $t$ or $\theta$) that defines the coordinates of a point.
2. Parametric Equations: A pair of functions $x = f(t)$ and $y = g(t)$ that define the coordinates of a curve simultaneously.
3. Vector-Valued Function: A function of the form $\mathbf{r}(t) = \langle x(t), y(t) \rangle$, where the output is a vector representing position at time $t$.
4. Velocity Vector: The derivative of the position vector, $\mathbf{v}(t) = \mathbf{r}'(t) = \langle x'(t), y'(t) \rangle$.
5. Acceleration Vector: The derivative of the velocity vector, $\mathbf{a}(t) = \mathbf{v}'(t) = \mathbf{r}''(t) = \langle x''(t), y''(t) \rangle$.
6. Speed: The magnitude of the velocity vector, $|\mathbf{v}(t)| = \sqrt{(x'(t))^2 + (y'(t))^2}$.
7. Polar Coordinates: A coordinate system $(r, \theta)$ where $r$ is the directed distance from the origin (pole) and $\theta$ is the directed angle from the positive x-axis (polar axis).
8. Pole: The origin in the polar coordinate system.
9. Polar Axis: The ray starting at the pole and extending in the positive x-direction.
10. Arc Length: The total distance along a curve between two points.

Foundational Principles

• Elimination of Parameter: The process of converting parametric equations into a single rectangular equation $y = f(x)$ by solving for $t$.
• The Chain Rule for Parametrics: To find the slope of a parametric curve, we use: $$\frac{dy}{dx} = \frac{dy/dt}{dx/dt}$$
• Total Distance Traveled: For a particle in motion, the integral of the speed: $$S = \int_{a}^{b} \sqrt{(x'(t))^2 + (y'(t))^2} dt$$
• Polar-to-Rectangular Conversion: $$x = r \cos \theta, \quad y = r \sin \theta$$ $$r^2 = x^2 + y^2, \quad \tan \theta = \frac{y}{x}$$

3. IN-DEPTH EXPLANATION of EVERY CONCEPT and PRINCIPLE

3.1 Parametric Equations and Calculus

In parametric form, a curve in the plane is defined by: $$x = x(t)$$ $$y = y(t)$$ As $t$ increases, the point $(x(t), y(t))$ traces out a path. This allows us to represent curves that are not functions (like circles or figure-eights).

First and Second Derivatives

To find the slope of the tangent line to a parametric curve at a specific value of $t$: $$\frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}}, \text{ provided } \frac{dx}{dt} \neq 0$$

To find the concavity of the curve, we need the second derivative $\frac{d^2y}{dx^2}$. This requires care, as we are differentiating the first derivative with respect to $x$, not $t$: $$\frac{d^2y}{dx^2} = \frac{d}{dx} \left( \frac{dy}{dx} \right) = \frac{\frac{d}{dt} \left( \frac{dy}{dx} \right)}{\frac{dx}{dt}}$$ Crucial Note: A common mistake is forgetting to divide by $dx/dt$ a second time.

Arc Length of a Parametric Curve

If a curve is traced exactly once as $t$ goes from $a$ to $b$, the arc length $L$ is: $$L = \int_{a}^{b} \sqrt{\left(\frac{dx}{dt}\right)^2 + \left(\frac{dy}{dt}\right)^2} dt$$

3.2 Vector-Valued Functions (Motion in the Plane)

A vector-valued function $\mathbf{r}(t) = \langle x(t), y(t) \rangle$ (or $\mathbf{r}(t) = x(t)\mathbf{i} + y(t)\mathbf{j}$) describes the position of a particle.

• Velocity: $\mathbf{v}(t) = \mathbf{r}'(t) = \langle x'(t), y'(t) \rangle$. This vector is always tangent to the path.
• Acceleration: $\mathbf{a}(t) = \mathbf{v}'(t) = \langle x''(t), y''(t) \rangle$.
• Speed: A scalar quantity representing the rate of change of distance with respect to time: $$\text{Speed} = |\mathbf{v}(t)| = \sqrt{(x'(t))^2 + (y'(t))^2}$$
• Displacement: The change in position from $t=a$ to $t=b$: $$\text{Displacement} = \mathbf{r}(b) - \mathbf{r}(a) = \left\langle \int_a^b x'(t) dt, \int_a^b y'(t) dt \right\rangle$$
• Distance Traveled: The integral of the speed (equivalent to arc length): $$\text{Distance} = \int_{a}^{b} \sqrt{(x'(t))^2 + (y'(t))^2} dt$$

3.3 Polar Coordinates

In the polar system, we locate points using $(r, \theta)$. Note that $r$ can be negative. If $r < 0$, the point is $|r|$ units from the pole in the direction $\theta + \pi$.

Derivatives in Polar Form

Given $r = f(\theta)$, we use the conversion $x = r \cos \theta$ and $y = r \sin \theta$. To find the slope of the tangent line ($dy/dx$): $$\frac{dy}{dx} = \frac{\frac{dy}{d\theta}}{\frac{dx}{d\theta}}$$ Using the product rule on $x = f(\theta)\cos\theta$ and $y = f(\theta)\sin\theta$: $$\frac{dy}{dx} = \frac{f'(\theta)\sin\theta + f(\theta)\cos\theta}{f'(\theta)\cos\theta - f(\theta)\sin\theta}$$

Area in Polar Coordinates

The area of a sector in a circle is $\frac{1}{2}r^2\theta$. For a polar curve $r = f(\theta)$, the area $A$ swept out between $\theta = \alpha$ and $\theta = \beta$ is: $$A = \frac{1}{2} \int_{\alpha}^{\beta} [r(\theta)]^2 d\theta$$

Arc Length in Polar Coordinates

Derived from the parametric formula by treating $\theta$ as the parameter: $$L = \int_{\alpha}^{\beta} \sqrt{r^2 + \left(\frac{dr}{d\theta}\right)^2} d\theta$$

4. EXAMPLES

Example 1: Parametric Slope

Find the slope of the curve $x = t^2$, $y = t^3 - 3t$ at $t = 2$. Solution: $\frac{dx}{dt} = 2t$, $\frac{dy}{dt} = 3t^2 - 3$. At $t=2$: $\frac{dx}{dt} = 4$, $\frac{dy}{dt} = 3(4)-3 = 9$. $$\frac{dy}{dx} = \frac{9}{4}$$

Example 2: Horizontal Tangents

Find the points $(x,y)$ where the curve $x = \cos t$, $y = \sin(2t)$ has horizontal tangents for $0 \leq t < 2\pi$. Solution: Horizontal tangents occur when $dy/dt = 0$ and $dx/dt \neq 0$. $\frac{dy}{dt} = 2\cos(2t) = 0 \implies 2t = \frac{\pi}{2}, \frac{3\pi}{2}, \frac{5\pi}{2}, \frac{7\pi}{2} \implies t = \frac{\pi}{4}, \frac{3\pi}{4}, \frac{5\pi}{4}, \frac{7\pi}{4}$. Check $dx/dt = -\sin t$. At these values, $\sin t \neq 0$. Points: $t=\pi/4 \implies (\frac{\sqrt{2}}{2}, 1)$ $t=3\pi/4 \implies (-\frac{\sqrt{2}}{2}, -1)$ $t=5\pi/4 \implies (-\frac{\sqrt{2}}{2}, 1)$ $t=7\pi/4 \implies (\frac{\sqrt{2}}{2}, -1)$

Example 3: Parametric Second Derivative

Find $\frac{d^2y}{dx^2}$ for $x = e^t$, $y = t e^{-t}$. Solution: $\frac{dx}{dt} = e^t$, $\frac{dy}{dt} = e^{-t} - te^{-t} = e^{-t}(1-t)$. $\frac{dy}{dx} = \frac{e^{-t}(1-t)}{e^t} = e^{-2t}(1-t)$. Now differentiate $y'$ with respect to $t$: $\frac{d}{dt}(e^{-2t}(1-t)) = -2e^{-2t}(1-t) + e^{-2t}(-1) = e^{-2t}(-2 + 2t - 1) = e^{-2t}(2t-3)$. Divide by $dx/dt$: $$\frac{d^2y}{dx^2} = \frac{e^{-2t}(2t-3)}{e^t} = e^{-3t}(2t-3)$$

Example 4: Parametric Arc Length

Find the length of the path $x = \sin(3t)$, $y = \cos(3t)$ from $t=0$ to $t=\pi$. Solution: $x' = 3\cos(3t)$, $y' = -3\sin(3t)$. $(x')^2 + (y')^2 = 9\cos^2(3t) + 9\sin^2(3t) = 9$. $$L = \int_0^{\pi} \sqrt{9} dt = \int_0^{\pi} 3 dt = 3\pi$$

Example 5: Vector Velocity and Acceleration

A particle moves with position $\mathbf{r}(t) = \langle t^3, t^2 \rangle$. Find the velocity and acceleration vectors at $t=1$. Solution: $\mathbf{v}(t) = \langle 3t^2, 2t \rangle \implies \mathbf{v}(1) = \langle 3, 2 \rangle$. $\mathbf{a}(t) = \langle 6t, 2 \rangle \implies \mathbf{a}(1) = \langle 6, 2 \rangle$.

Example 6: Speed Calculation

Find the speed of a particle at $t=2$ if $\mathbf{r}(t) = \langle \ln(t), t^2 \rangle$. Solution: $\mathbf{v}(t) = \langle 1/t, 2t \rangle$. At $t=2$, $\mathbf{v}(2) = \langle 1/2, 4 \rangle$. $$\text{Speed} = \sqrt{(1/2)^2 + 4^2} = \sqrt{0.25 + 16} = \sqrt{16.25} = \frac{\sqrt{65}}{2}$$

Example 7: Distance Traveled (Calculator Active)

A particle moves such that $x'(t) = \sqrt{1+t^2}$ and $y'(t) = \sin(t^2)$. Find the total distance traveled from $t=0$ to $t=2$. Solution: $$\text{Dist} = \int_0^2 \sqrt{(\sqrt{1+t^2})^2 + (\sin(t^2))^2} dt = \int_0^2 \sqrt{1+t^2 + \sin^2(t^2)} dt$$ Using numeric integration: $\approx 2.731$.

Example 8: Finding Position from Velocity

If $\mathbf{v}(t) = \langle 2t, 3 \rangle$ and $\mathbf{r}(0) = \langle 1, 5 \rangle$, find $\mathbf{r}(2)$. Solution: $x(2) = x(0) + \int_0^2 2t dt = 1 + [t^2]_0^2 = 1+4 = 5$. $y(2) = y(0) + \int_0^2 3 dt = 5 + [3t]_0^2 = 5+6 = 11$. $\mathbf{r}(2) = \langle 5, 11 \rangle$.

Example 9: Rectangular to Polar

Convert the point $(x,y) = (-1, 1)$ to polar coordinates with $r > 0$ and $0 \leq \theta < 2\pi$. Solution: $r^2 = (-1)^2 + (1)^2 = 2 \implies r = \sqrt{2}$. $\tan \theta = 1/(-1) = -1$. Since the point is in Quadrant II, $\theta = 3\pi/4$. Polar: $(\sqrt{2}, 3\pi/4)$.

Example 10: Polar to Rectangular Curve

Convert $r = 4\sin\theta$ to rectangular form. Solution: Multiply both sides by $r$: $r^2 = 4r\sin\theta$. Substitute $r^2 = x^2 + y^2$ and $y = r\sin\theta$: $x^2 + y^2 = 4y \implies x^2 + (y-2)^2 = 4$. (A circle centered at $(0,2)$ with radius 2).

Example 11: Slope of a Polar Curve

Find the slope of $r = 2$ at $\theta = \pi/4$. Solution: $r = 2$ is a circle. The slope should be perpendicular to the radial line. $x = 2\cos\theta \implies dx/d\theta = -2\sin\theta$. $y = 2\sin\theta \implies dy/d\theta = 2\cos\theta$. $\frac{dy}{dx} = \frac{2\cos(\pi/4)}{-2\sin(\pi/4)} = \frac{\sqrt{2}}{-\sqrt{2}} = -1$.

Example 12: Polar Area (Full Circle)

Find the area of $r = 3$. Solution: $$A = \frac{1}{2} \int_0^{2\pi} 3^2 d\theta = \frac{1}{2} [9\theta]_0^{2\pi} = \frac{18\pi}{2} = 9\pi$$ (Matches $\pi r^2$).

Example 13: Area of One Petal of a Rose Curve

Find the area of one petal of $r = \cos(2\theta)$. Solution: A petal of a 4-petal rose occurs when $r$ goes from positive to zero to positive. $\cos(2\theta) = 0 \implies 2\theta = -\pi/2, \pi/2 \implies \theta = -\pi/4, \pi/4$. $$A = \frac{1}{2} \int_{-\pi/4}^{\pi/4} \cos^2(2\theta) d\theta = \frac{1}{2} \int_{-\pi/4}^{\pi/4} \frac{1 + \cos(4\theta)}{2} d\theta$$ $$A = \frac{1}{4} \left[ \theta + \frac{\sin(4\theta)}{4} \right]_{-\pi/4}^{\pi/4} = \frac{1}{4} [(\pi/4 + 0) - (-\pi/4 + 0)] = \frac{\pi}{8}$$

Example 14: Area of a Cardioid

Find the area of $r = 1 + \sin\theta$. Solution: The cardioid is traced for $0 \leq \theta \leq 2\pi$. $$A = \frac{1}{2} \int_0^{2\pi} (1 + \sin\theta)^2 d\theta = \frac{1}{2} \int_0^{2\pi} (1 + 2\sin\theta + \sin^2\theta) d\theta$$ $$A = \frac{1}{2} \int_0^{2\pi} \left(1 + 2\sin\theta + \frac{1-\cos(2\theta)}{2}\right) d\theta = \frac{1}{2} [ \frac{3}{2}\theta - 2\cos\theta - \frac{\sin(2\theta)}{4} ]_0^{2\pi}$$ $$A = \frac{1}{2} [ (3\pi - 2 - 0) - (0 - 2 - 0) ] = \frac{3\pi}{2}$$

Example 15: Area Between Two Polar Curves

Find the area inside $r = 3\sin\theta$ and outside $r = 1 + \sin\theta$. Solution: Intersection: $3\sin\theta = 1 + \sin\theta \implies 2\sin\theta = 1 \implies \sin\theta = 1/2$. $\theta = \pi/6, 5\pi/6$. $$A = \frac{1}{2} \int_{\pi/6}^{5\pi/6} [(3\sin\theta)^2 - (1+\sin\theta)^2] d\theta$$ $$A = \frac{1}{2} \int_{\pi/6}^{5\pi/6} (9\sin^2\theta - 1 - 2\sin\theta - \sin^2\theta) d\theta = \frac{1}{2} \int_{\pi/6}^{5\pi/6} (8\sin^2\theta - 2\sin\theta - 1) d\theta$$ Calculating this integral yields $\pi$.

Example 16: Arc Length of a Polar Curve

Find the length of $r = e^\theta$ from $\theta = 0$ to $\theta = 1$. Solution: $r' = e^\theta$. $$L = \int_0^1 \sqrt{(e^\theta)^2 + (e^\theta)^2} d\theta = \int_0^1 \sqrt{2e^{2\theta}} d\theta = \int_0^1 \sqrt{2} e^\theta d\theta$$ $$L = \sqrt{2}[e^\theta]_0^1 = \sqrt{2}(e - 1)$$

Example 17: Velocity Vector Integration

If $\mathbf{a}(t) = \langle \cos t, \sin t \rangle$, $\mathbf{v}(0) = \langle 0, 1 \rangle$, find $\mathbf{v}(t)$. Solution: $\mathbf{v}(t) = \langle \int \cos t dt, \int \sin t dt \rangle = \langle \sin t + C_1, -\cos t + C_2 \rangle$. At $t=0$: $0 + C_1 = 0 \implies C_1=0$; $-1 + C_2 = 1 \implies C_2=2$. $\mathbf{v}(t) = \langle \sin t, 2 - \cos t \rangle$.

Example 18: Vector Particle Crossing x-axis

A particle moves along $\mathbf{r}(t) = \langle t-2, t^2-4 \rangle$. When does it cross the x-axis? Solution: Crosses x-axis when $y(t) = 0 \implies t^2 - 4 = 0 \implies t = 2$ (assuming $t \geq 0$). Position at $t=2$: $(0, 0)$.

Example 19: Speed in Polar

Given a particle moving along $r = 3\theta$ where $\theta = t^2$, find the speed at $t=1$. Solution: $x = r\cos\theta = 3\theta\cos\theta$, $y = r\sin\theta = 3\theta\sin\theta$. At $t=1, \theta=1$. $d\theta/dt = 2t = 2$. $dx/dt = \frac{dx}{d\theta}\frac{d\theta}{dt} = (3\cos\theta - 3\theta\sin\theta)(2)$. $dy/dt = \frac{dy}{d\theta}\frac{d\theta}{dt} = (3\sin\theta + 3\theta\cos\theta)(2)$. At $\theta=1$: $dx/dt = 6\cos(1) - 6\sin(1)$, $dy/dt = 6\sin(1) + 6\cos(1)$. $\text{Speed} = \sqrt{(6\cos1 - 6\sin1)^2 + (6\sin1 + 6\cos1)^2} = \sqrt{36(\cos^2 1 - 2\sin1\cos1 + \sin^2 1 + \sin^2 1 + 2\sin1\cos1 + \cos^2 1)}$ $\text{Speed} = \sqrt{36(1+1)} = \sqrt{72} = 6\sqrt{2}$.

Example 20: Vertical Tangent in Parametric

Find $t$ such that $x = t^3 - 3t, y = t^2$ has a vertical tangent. Solution: $dx/dt = 3t^2 - 3 = 0 \implies t = \pm 1$. Check $dy/dt = 2t$. At $t=1, dy/dt = 2 \neq 0$. At $t=-1, dy/dt = -2 \neq 0$. Vertical tangents at $t=1, t=-1$.

Example 21: Area inside Inner Loop of a Limaçon

Find the area inside the inner loop of $r = 1 - 2\sin\theta$. Solution: Inner loop exists when $r < 0$. $1 - 2\sin\theta = 0 \implies \sin\theta = 1/2 \implies \theta = \pi/6, 5\pi/6$. Between $\pi/6$ and $5\pi/6$, $r$ is negative. $$A = \frac{1}{2} \int_{\pi/6}^{5\pi/6} (1 - 2\sin\theta)^2 d\theta$$ $$A = \frac{1}{2} \int_{\pi/6}^{5\pi/6} (1 - 4\sin\theta + 4\sin^2\theta) d\theta = \frac{1}{2} \int_{\pi/6}^{5\pi/6} (1 - 4\sin\theta + 2(1-\cos 2\theta)) d\theta$$ $$A = \frac{1}{2} [ 3\theta + 4\cos\theta - \sin(2\theta) ]_{\pi/6}^{5\pi/6} = \pi - \frac{3\sqrt{3}}{2}$$

Example 22: Derivative of Arc Length

If $s(t)$ is the distance traveled along $x=f(t), y=g(t)$, find $s'(t)$. Solution: $s(t) = \int_0^t \sqrt{f'(u)^2 + g'(u)^2} du$. By Fundamental Theorem of Calculus: $s'(t) = \sqrt{(f'(t))^2 + (g'(t))^2}$, which is speed.

Example 23: Polar Equation of a Line

Find the polar equation of the vertical line $x = 3$. Solution: $r\cos\theta = 3 \implies r = 3\sec\theta$.

Example 24: Vector Acceleration Direction

If $\mathbf{r}(t) = \langle \cos(t^2), \sin(t^2) \rangle$, show acceleration is not always toward the origin. Solution: $\mathbf{v}(t) = \langle -2t\sin(t^2), 2t\cos(t^2) \rangle$. $\mathbf{a}(t) = \langle -2\sin(t^2) - 4t^2\cos(t^2), 2\cos(t^2) - 4t^2\sin(t^2) \rangle$. Since $\mathbf{a}(t)$ is not a simple scalar multiple of $\mathbf{r}(t)$, it doesn't point directly toward the origin (centripetal) unless the $2\sin$ and $2\cos$ terms are negligible.

Example 25: Point of Intersection

Where do $r = 2\sin\theta$ and $r = 2\cos\theta$ intersect (other than the pole)? Solution: $2\sin\theta = 2\cos\theta \implies \tan\theta = 1 \implies \theta = \pi/4$. $r = 2\sin(\pi/4) = \sqrt{2}$. Point: $(\sqrt{2}, \pi/4)$.

Example 26: Concavity of Polar Curve

Determine if $r = 1/\theta$ is concave up or down at $\theta = \pi$. Solution: This requires $\frac{d^2y}{dx^2}$. $x = \frac{\cos\theta}{\theta}, y = \frac{\sin\theta}{\theta}$. $x' = \frac{-\theta\sin\theta - \cos\theta}{\theta^2}, y' = \frac{\theta\cos\theta - \sin\theta}{\theta^2}$. At $\theta = \pi$: $x' = 1/\pi^2, y' = -1/\pi$. $\frac{dy}{dx} = \frac{-1/\pi}{1/\pi^2} = -\pi$. Differentiate $y'$ with respect to $\theta$ and divide by $x'$. (Advanced calculation shows concavity is positive/up).

Example 27: Maximum x-value

Find the maximum $x$-coordinate of $x = -t^2 + 4t, y = t^3$. Solution: $dx/dt = -2t + 4 = 0 \implies t = 2$. $x(2) = -4 + 8 = 4$. Since $x(t)$ is a downward parabola, this is the global maximum.

Example 28: Speed of a Circular Orbit

A particle moves $\mathbf{r}(t) = \langle R\cos(\omega t), R\sin(\omega t) \rangle$. Find its speed. Solution: $\mathbf{v}(t) = \langle -R\omega\sin(\omega t), R\omega\cos(\omega t) \rangle$. $\text{Speed} = \sqrt{R^2\omega^2\sin^2(\omega t) + R^2\omega^2\cos^2(\omega t)} = \sqrt{R^2\omega^2} = |R\omega|$.

Example 29: Integral of Vector Function

Evaluate $\int_0^1 \langle t^2, e^t \rangle dt$. Solution: $\langle [\frac{1}{3}t^3]_0^1, [e^t]_0^1 \rangle = \langle \frac{1}{3}, e - 1 \rangle$.

Example 30: Symmetry in Polar

Show $r = 2\cos(3\theta)$ is symmetric about the x-axis. Solution: Replace $\theta$ with $-\theta$: $r = 2\cos(3(-\theta)) = 2\cos(-3\theta) = 2\cos(3\theta)$. Since the equation is unchanged, it is symmetric about the polar axis ($x$-axis).

Example 31: Area of a Circle via Polar Integral

Find the area of $r = a$. Solution: $$A = \frac{1}{2} \int_0^{2\pi} a^2 d\theta = \frac{a^2}{2} [2\pi - 0] = \pi a^2$$.

Example 32: Slope of Cardioid at the Pole

Find the slope of $r = 1 + \cos\theta$ as it approaches the pole. Solution: $r = 0 \implies 1 + \cos\theta = 0 \implies \theta = \pi$. The slope of the tangent at the pole is simply $\tan(\text{the value of } \theta \text{ where } r=0)$. Slope = $\tan(\pi) = 0$.

Example 33: Parametric Displacement

If $\mathbf{v}(t) = \langle \sin t, \cos t \rangle$, find the displacement from $t=0$ to $t=\pi$. Solution: $\Delta x = \int_0^\pi \sin t dt = [-\cos t]_0^\pi = -(-1) - (-1) = 2$. $\Delta y = \int_0^\pi \cos t dt = [\sin t]_0^\pi = 0 - 0 = 0$. Displacement = $\langle 2, 0 \rangle$.

Example 34: Tangent Line Equation

Find the equation of the tangent line to $x = t^2 + 1, y = 2t$ at $t = 1$. Solution: At $t=1$: $x = 2, y = 2$. $dx/dt = 2t = 2$, $dy/dt = 2$. Slope $m = 2/2 = 1$. Line: $y - 2 = 1(x - 2) \implies y = x$.

Example 35: Polar Arc Length of a Spiral

Find the length of $r = \theta^2$ for $0 \leq \theta \leq \sqrt{5}$. Solution: $r' = 2\theta$. $$L = \int_0^{\sqrt{5}} \sqrt{(\theta^2)^2 + (2\theta)^2} d\theta = \int_0^{\sqrt{5}} \sqrt{\theta^4 + 4\theta^2} d\theta = \int_0^{\sqrt{5}} \theta \sqrt{\theta^2 + 4} d\theta$$ Let $u = \theta^2 + 4, du = 2\theta d\theta$: $$L = \frac{1}{2} \int_4^9 \sqrt{u} du = \frac{1}{2} [\frac{2}{3}u^{3/2}]_4^9 = \frac{1}{3} [27 - 8] = \frac{19}{3}$$

End of Unit 9 content. This chapter covers the foundational and advanced calculus concepts for Parametric, Vector, and Polar functions as required for the AP Calculus BC curriculum.