Parametric & Polar Calculus practice questions — AP Calculus BC

$r^{\mathrm{\prime }}(t)=⟨2t,3t^2⟩$, so $r^{\mathrm{\prime }}(1)=⟨2,3⟩$.

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

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

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

$$L={\int }_0^{2}5dt=10.$$

$$A={\displaystyle \frac{1}{2}}{\int }_0^{2\pi }{2}^{2}d\theta ={\displaystyle \frac{1}{2}}(4)(2\pi )=4\pi \mathrm{.}$$

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

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

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

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

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

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