# Parametric Equations: Find $\frac{\mathrm d^2y}{\mathrm dx^2}$.

Find $\dfrac{\mathrm d^2y}{\mathrm dx^2}$, as a function of $t$, for the given the parametric equations: \begin{align}x&=3-3\cos(t)\\y&=3+\cos^4(t)\end{align} $\displaystyle\dfrac{\mathrm d^2y}{\mathrm dx^2}=\ldots$

I don't really understand this section that I am learning at all, is there any useful website I can look over to help me understand this concept better? Thanks!

• From what I skimmed, this gives a pretty simple breakdown of what the question requires. mathcentre.ac.uk/resources/uploaded/mc-ty-parametric-2009-1.pdf
– Zhoe
Commented Nov 21, 2013 at 17:20
• @Zhoe This is perfect! Thanks! Commented Nov 21, 2013 at 17:37
– Zhoe
Commented Nov 21, 2013 at 17:43

It is known that $$\dfrac {\mathrm{d}y}{\mathrm{d}x} = \dfrac {\frac{\mathrm{d}y}{\mathrm{d}t}}{\frac{\mathrm{d}x}{\mathrm{d}t}}.$$Do you know how to find $y'(t)$ and $x'(t)$? Do this to get $\frac{\mathrm{d}y}{\mathrm{d}x}$ and then differentiate again. I am not finishing the whole thing for you since you haven't shown your work, but I think this hint should help you enough to finish.

• Thanks! Just looking for a brief idea! Commented Nov 21, 2013 at 17:37
• You're welcome! Accept, please? :P Commented Nov 21, 2013 at 17:41

You're given $x(t), y(t)$. So find $\dfrac{dx}{dt},\;\dfrac{dy}{dt}$.

$$\text{Then note that}\;\frac{dy}{dx} = \dfrac{dy/dt}{dx/dt}$$

Now you need to find $\dfrac{d^2y}{dx^2}$.

• Beat you! $\mathrm{}\\$ +1, nevertheless. Commented Nov 21, 2013 at 17:19
• @amWhy: Needs another TU +1 Commented Nov 22, 2013 at 1:03
• Hi, dear friend, @B.S.! 8) Commented Dec 5, 2013 at 15:34
• @amWhy: Hiiiii. Hope you are doing well. :-) But, I have not been so good today. Many of questions are in especial fields. :( Commented Dec 5, 2013 at 15:54