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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

Please offer your assistance. :)

chndn
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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??? – Mr. Math Sep 08 '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 08 '13 at 10:53
  • I think that there is some other way to find it – chndn Sep 08 '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. – Mr. Math Sep 08 '13 at 10:56
  • can you tell me book name and page number – Mr. Math Sep 08 '13 at 10:56
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    The answer specified by the OP is definetely correct and can be derived using the hint I gave in the answers below. – Mufasa Sep 08 '13 at 11:20
  • @Jam From the graph point of view, $\frac{d^2y}{dx^2}$ is the second derivative of $y$ with respect to $x$. Scaling the graph by a factor of $a$ will definitely affect the second derivative. – Tunococ Sep 08 '13 at 11:57

2 Answers2

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

Mikasa
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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: $\large\frac{dy}{dx}=\frac{\frac{dy}{d\theta}}{\frac{dx}{d\theta}}$

Mufasa
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