How to find the second derivative for a parametric function?

Question

A curve is defined parametrically by $x(t)$ and $y(t)$ with $\frac {dx}{dt} = 3\sin(t)$ and $\frac {dy}{dt} = 6\cos(t)$. Find $\frac {d^2y}{dx^2}$ at $\frac {\pi}{6}$

The answer choices to this were $-8$, $8$, $-\frac {1}{12}$, $\frac {3}{16}$, or $-\frac {16}{3}$. The first time I've evaluated this, I got $-8$, but that was apparently incorrect. I'm relatively new to these kinds of problems, and I'm not even exactly sure if I did it correctly. Can someone evaluate this and help to explain it?

Answer 1

Try to see how you can write $\frac{d^2y}{dx^2}$ in terms of the parameter $t$.

$$\frac{d^2y}{dx^2} = \frac{d}{dx} \left( \frac{dy}{dx} \right)$$

Now, try to introduce $dt$ by using the chain rule:

$$= \frac{dt}{dx} \frac{d}{dt} \left( \frac{dy}{dx} \right) $$

Now,

$$\frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}} = \frac{6\cos t}{3\sin t} =2\cot t $$

$$\therefore \frac{d^2y}{dx^2} = \frac{\frac{d(2\cot t)}{dt}}{\frac{dx}{dt}} = \frac{-2\csc^2 t}{3\sin t}$$

$$at \ \frac{\pi}{6} : =\frac{-2\times 4 \times 2}{3} = \color{green}{-\frac{16}{3}}$$

Answer 2

HINT So $$x'(t)^2 + \frac{y'(t)}{2} = 9.$$ Can you solve for $y'(x)$ and compute $y''(x)$?

Answer 3

HINT:

$$\frac{dy}{dx}=\frac{\frac{dy}{dt}}{\frac{dx}{dt}}=\frac{6\cos (t)}{3\sin (t)}=2\cot (t)$$ $$\frac{d^2y}{dx^2}=\frac{d}{dx}\left(\frac{dy}{dx}\right)=\frac{dt}{dx}\frac{d}{dt}\left(\frac{dy}{dx}\right)$$