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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!

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  • $\begingroup$ 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 $\endgroup$
    – Zhoe
    Commented Nov 21, 2013 at 17:20
  • $\begingroup$ @Zhoe This is perfect! Thanks! $\endgroup$ Commented Nov 21, 2013 at 17:37
  • $\begingroup$ Your Welcome @Christopher23 $\endgroup$
    – Zhoe
    Commented Nov 21, 2013 at 17:43

2 Answers 2

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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.

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  • $\begingroup$ Thanks! Just looking for a brief idea! $\endgroup$ Commented Nov 21, 2013 at 17:37
  • $\begingroup$ You're welcome! Accept, please? :P $\endgroup$ Commented Nov 21, 2013 at 17:41
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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 $\dfrac{d^2y}{dx^2}$.

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

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