# Find the second derivative of the given function

If $$x=a(\cos \theta + \theta \sin \theta)$$$$y=a(\sin \theta- \theta \cos \theta)$$ prove that $$\frac{d^2y}{dx^2}= \frac{\sec^3 \theta}{a \theta}$$

Can you solve this for me?

I tried finding $\frac{dy}{dx}$ by dividing $\frac{dy}{dt}$ by $\frac{dx}{dt}$ but failed to get the required answer

$$\frac{\frac{d}{d \theta} \frac{ \cos \theta + \theta \sin \theta}{\theta \cos \theta - \sin \theta}}{\frac{dx}{d \theta}}=\frac{1+\theta^2}{a(\theta \cos \theta- \sin \theta)}$$ I am stuck here

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Question is clearly wrong where did you get this from. You will not have "a" in the final answer that leads me to believe question is wrong where did you get question what's the source??? – MRK Sep 8 '13 at 10:45
It is from a reputed book. I have many similar questions with answers containing a which will cancel out in the method we followed – chndn Sep 8 '13 at 10:53
I think that there is some other way to find it – chndn Sep 8 '13 at 10:54
i am 100% certain there is no a in the answer think of it from a graph point of view the "a" will have no influence on second derivative let alone the first derivative. – MRK Sep 8 '13 at 10:56
The answer specified by the OP is definetely correct and can be derived using the hint I gave in the answers below. – Mufasa Sep 8 '13 at 11:20

Hint: $$y_t=at\sin t,~~x_t=at\cos t,~~ y'=\frac{y_t}{x_t}=\tan t,t\neq (2k+1)\pi/2,~~y''=\frac{\sec^2 t}{x_t}$$

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"Spot on" Babak! +1 – amWhy Sep 9 '13 at 0:10

HINT: $\large\frac{d^2y}{dx^2}=\frac{d}{dx}(\frac{dy}{dx})=\frac{d}{d\theta}(\frac{dy}{dx})\times\frac{d\theta}{dx}=\frac{\frac{d}{d\theta}(\frac{dy}{dx})}{\frac{dx}{d\theta}}$

And of course: $\large\frac{dy}{dx}=\frac{\frac{dy}{d\theta}}{\frac{dx}{d\theta}}$

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Very nice hints. +1 – DonAntonio Sep 8 '13 at 11:39
can someone post the answer. I still couldnt reach the answer – chndn Sep 8 '13 at 14:58
@chndn -- It would be better if you posted the work you have done so far in trying to get to the answer so that we can help spot where you may have made a mistake. – Mufasa Sep 8 '13 at 15:25
okay i am gonna edit it – chndn Sep 8 '13 at 15:26
@chndn -- can you please show how you got your values for $dy/d\theta$ and $dx/d\theta$? – Mufasa Sep 8 '13 at 15:38