Practice Problems – Parametric, Polar & Vector – AP Calculus BC

Below are 20 practice problems covering all topics from Module 05: Parametric, Polar & Vector Functions (BC). Topics include: parametric equations, derivatives of parametric curves, arc length, polar coordinates, polar derivatives and area, and vector-valued functions. Each problem includes a hidden solution — click the button to reveal the step-by-step answer.

Problem 01 Parametric

Eliminate the parameter: $x = t^2$, $y = t + 1$.

Solution:
$t = y - 1$, so $x = (y - 1)^2$. This is a parabola opening to the right.

Problem 02 Parametric Derivative

Find $\frac{dy}{dx}$ for $x = 2t$, $y = t^2$.

Solution:
$\frac{dx}{dt} = 2$, $\frac{dy}{dt} = 2t$, so $\frac{dy}{dx} = \frac{2t}{2} = t$.

Problem 03 Parametric Tangent

Find the tangent line to $x = \cos t$, $y = \sin t$ at $t = \pi/4$.

Solution:
Point: $(\sqrt{2}/2, \sqrt{2}/2)$
$\frac{dx}{dt} = -\sin t$, $\frac{dy}{dt} = \cos t$, slope $= \frac{\cos t}{-\sin t} = -\cot t = -1$ at $t=\pi/4$
Tangent: $y - \sqrt{2}/2 = -1(x - \sqrt{2}/2)$ → $y = -x + \sqrt{2}$

Problem 04 Arc Length

Find the arc length of $x = 3t$, $y = 4t$ from $t=0$ to $t=2$.

Solution:
$\frac{dx}{dt}=3$, $\frac{dy}{dt}=4$, $L = \int_0^2 \sqrt{3^2+4^2} dt = \int_0^2 5 dt = 10$

Problem 05 Polar Conversion

Convert $(r, \theta) = (6, \pi/3)$ to rectangular coordinates.

Solution:
$x = 6\cos(\pi/3) = 6(1/2)=3$, $y = 6\sin(\pi/3) = 6(\sqrt{3}/2)=3\sqrt{3}$

Problem 06 Polar Conversion

Convert $(-2, 2)$ to polar coordinates.

Solution:
$r = \sqrt{4+4} = 2\sqrt{2}$, $\tan\theta = -1$, $\theta = 3\pi/4$ (Quadrant II), so $(2\sqrt{2}, 3\pi/4)$

Problem 07 Polar Equation

Convert $r = 2\cos\theta$ to rectangular form.

Solution:
Multiply by $r$: $r^2 = 2r\cos\theta$ → $x^2 + y^2 = 2x$ → $(x-1)^2 + y^2 = 1$ (circle centered at (1,0))

Problem 08 Polar Derivative

Find $\frac{dy}{dx}$ for $r = 2$ at $\theta = \pi/2$.

Solution:
$r=2$ constant, $dr/d\theta=0$
$\frac{dy}{dx} = \frac{0\cdot\sin\theta + 2\cos\theta}{0\cdot\cos\theta - 2\sin\theta} = \frac{2\cos\theta}{-2\sin\theta} = -\cot\theta$
At $\theta=\pi/2$, $\cot(\pi/2)=0$, so slope $= 0$ (horizontal tangent)

Problem 09 Polar Area

Find the area enclosed by $r = 3$.

Solution:
$A = \frac{1}{2} \int_0^{2\pi} 9 d\theta = \frac{1}{2} \cdot 9 \cdot 2\pi = 9\pi$

Problem 10 Polar Area

Find the area enclosed by $r = 2(1 + \cos\theta)$.

Solution:
$A = \frac{1}{2} \int_0^{2\pi} 4(1+\cos\theta)^2 d\theta = 2 \int_0^{2\pi} (1 + 2\cos\theta + \cos^2\theta) d\theta = 2 \int_0^{2\pi} (\frac{3}{2} + 2\cos\theta + \frac{1}{2}\cos2\theta) d\theta = 2 \cdot 3\pi = 6\pi$

Problem 11 Vector-Valued

For $\mathbf{r}(t) = \langle t^2, t^3 \rangle$, find $\mathbf{v}(t)$ and $\mathbf{a}(t)$.

Solution:
$\mathbf{v}(t) = \langle 2t, 3t^2 \rangle$, $\mathbf{a}(t) = \langle 2, 6t \rangle$

Problem 12 Speed

Find the speed of $\mathbf{r}(t) = \langle \cos t, \sin t \rangle$ at $t = \pi/2$.

Solution:
$\mathbf{v}(t) = \langle -\sin t, \cos t \rangle$, speed $= \sqrt{(-\sin t)^2 + (\cos t)^2} = 1$ (constant)

Problem 13 Vector Tangent

Find the tangent line to $\mathbf{r}(t) = \langle t, t^2 \rangle$ at $t=2$.

Solution:
$\mathbf{r}(2) = (2,4)$, $\mathbf{r}'(t) = \langle 1, 2t \rangle$, $\mathbf{r}'(2) = \langle 1, 4 \rangle$
Tangent line: $(x, y) = (2, 4) + s\langle 1, 4 \rangle$ → parametric: $x = 2 + s$, $y = 4 + 4s$
Cartesian: $y = 4x - 4$

Problem 14 Projectile Motion

A projectile is launched with $v_0 = 30$ m/s at $\theta = 60^\circ$. Find the position vector.

Solution:
$v_{0x} = 30\cos60^\circ = 15$, $v_{0y} = 30\sin60^\circ = 15\sqrt{3}$
$\mathbf{r}(t) = \langle 15t, 15\sqrt{3}t - 4.9t^2 \rangle$

Problem 15 Polar Area

Find the area inside $r = 2$ and outside $r = 1$.

Solution:
$A = \frac{1}{2} \int_0^{2\pi} (2^2 - 1^2) d\theta = \frac{1}{2} \int_0^{2\pi} 3 d\theta = \frac{1}{2} \cdot 3 \cdot 2\pi = 3\pi$

Problem 16 Parametric Tangent

For $x = t^2$, $y = t^3 - 3t$, find where the tangent is horizontal.

Solution:
$\frac{dy}{dx} = \frac{3t^2 - 3}{2t} = 0$ → $3t^2 - 3 = 0$ → $t = \pm 1$
Points: $t=1$ → $(1, -2)$, $t=-1$ → $(1, 2)$

Problem 17 Parametric Tangent

For $x = t^2$, $y = t^3 - 3t$, find where the tangent is vertical.

Solution:
Vertical tangent when $dx/dt = 0$ and $dy/dt \neq 0$
$dx/dt = 2t = 0$ → $t=0$, $dy/dt = -3 \neq 0$
Point: $(0,0)$

Problem 18 Polar Derivative

For $r = 1 + \cos\theta$, find $\frac{dy}{dx}$ at $\theta = \pi/3$.

Solution:
$dr/d\theta = -\sin\theta$
At $\theta=\pi/3$: $r=1.5$, $dr/d\theta=-\sqrt{3}/2$
Numerator = $(-\sqrt{3}/2)(\sqrt{3}/2) + 1.5(0.5) = -3/4 + 0.75 = 0$
Denominator = $(-\sqrt{3}/2)(0.5) - 1.5(\sqrt{3}/2) = -\sqrt{3}/4 - 3\sqrt{3}/4 = -\sqrt{3}$
$\frac{dy}{dx} = 0$ (horizontal tangent)

Problem 19 Vector-Valued

For $\mathbf{r}(t) = \langle t^2, \ln t \rangle$, find $\mathbf{a}(t)$.

Solution:
$\mathbf{v}(t) = \langle 2t, 1/t \rangle$, $\mathbf{a}(t) = \langle 2, -1/t^2 \rangle$

Problem 20 Polar Area

Find the area of one petal of $r = \cos(2\theta)$.

Solution:
One petal occurs from $-\pi/4$ to $\pi/4$ (or $0$ to $\pi/4$ by symmetry).
$A = \frac{1}{2} \int_{-\pi/4}^{\pi/4} \cos^2(2\theta) d\theta = \int_{0}^{\pi/4} \cos^2(2\theta) d\theta = \int_{0}^{\pi/4} \frac{1+\cos4\theta}{2} d\theta = \frac{1}{2} \left[\theta + \frac{\sin4\theta}{4}\right]_0^{\pi/4} = \frac{1}{2} \cdot \frac{\pi}{4} = \frac{\pi}{8}$

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