Parametric & Polar Calculus — AP Calculus BC practice questions

Question 1

For $r(t) = \langle t^{2},\, t^{3}\rangle$, the speed at $t = 1$ is:

Accepted Answer

$r'(t) = \langle 2t,\, 3t^{2}\rangle$, so $r'(1) = \langle 2,\, 3\rangle$.

Speed $= \sqrt{2^{2} + 3^{2}} = \sqrt{4 + 9} = \sqrt{13}$.

Answer

$\sqrt{5}$

Answer

$\sqrt{13}$

Answer

$5$

Answer

$13$

Question 2

The arc length of $x = 3t$, $y = 4t$ for $0 \le t \le 2$ is:

Accepted Answer

$\dfrac{dx}{dt} = 3$, $\dfrac{dy}{dt} = 4$.

Integrand: $\sqrt{3^{2} + 4^{2}} = \sqrt{25} = 5$ (constant).

$$L = \int_{0}^{2} 5\,dt = 10.$$

Answer

$5$

Answer

$7$

Answer

$10$

Answer

$14$

Question 3

The area enclosed by $r = 2$ for $	heta \in [0, 2\pi]$ is:

Accepted Answer

$$A = \dfrac{1}{2}\int_{0}^{2\pi} 2^{2}\,d	heta = \dfrac{1}{2}(4)(2\pi) = 4\pi.$$

Consistent with $\pi r^{2} = \pi (2)^{2} = 4\pi$ — the curve $r = 2$ is a circle of radius $2$.

Answer

$2\pi$

Answer

$\pi$

Answer

$4\pi$

Answer

$8\pi$

Question 4

The polar point $(r,	heta) = \left(2,\, 	frac{\pi}{3}ight)$ has Cartesian coordinates:

Accepted Answer

$x = 2\cos\dfrac{\pi}{3} = 2 \cdot \dfrac{1}{2} = 1$.

$y = 2\sin\dfrac{\pi}{3} = 2 \cdot \dfrac{\sqrt{3}}{2} = \sqrt{3}$.

So $(x, y) = (1, \sqrt{3})$.

Answer

$(1,\, \sqrt{3})$

Answer

$(\sqrt{3},\, 1)$

Answer

$(2,\, \sqrt{3})$

Answer

$(\sqrt{3},\, 2)$

Question 5

A curve is given by $x = t^{2}$, $y = t^{3}$. The slope $\dfrac{dy}{dx}$ at $t = 2$ equals:

Accepted Answer

$\dfrac{dx}{dt} = 2t$ and $\dfrac{dy}{dt} = 3t^{2}$, so $\dfrac{dy}{dx} = \dfrac{3t^{2}}{2t} = \dfrac{3t}{2}$.

At $t = 2$: $\dfrac{dy}{dx} = \dfrac{3(2)}{2} = 3$.

Answer

$\dfrac{3}{2}$

Answer

$2$

Answer

$3$

Answer

$6$

Parametric & Polar Calculus practice questions — AP Calculus BC