Question:

If $\displaystyle x = cos \theta + \theta sin \theta$ , $\displaystyle y = sin \theta - \theta cos \theta $ , then show that

$\displaystyle \frac{dy}{dx} = tan \theta$

Also find $\displaystyle \frac{d^2 y}{dx^2}$

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- Dec 5th 2009, 12:45 AMzorroDifferentiate and find the square of dy/dx
Question:

If $\displaystyle x = cos \theta + \theta sin \theta$ , $\displaystyle y = sin \theta - \theta cos \theta $ , then show that

$\displaystyle \frac{dy}{dx} = tan \theta$

Also find $\displaystyle \frac{d^2 y}{dx^2}$ - Dec 5th 2009, 01:51 AMmr fantastic
$\displaystyle \frac{dy}{dx} = \frac{\frac{dy}{d \theta}}{\frac{dx}{d \theta}}$ (which I'm sure will be somewhere in yur class notes or textbook).

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

If you need more help, please show all your working and state specifically where you are stuck. - Dec 5th 2009, 04:14 AMHallsofIvy
"yur"? (Worried)

- Dec 6th 2009, 07:50 PMzorroIs this correct?

I am getting the $\displaystyle \frac{d^2 y }{dx^2} = sec^2 \theta .\frac{1}{ \theta cos \theta}$

$\displaystyle

\frac{d}{d \theta} \left( \frac{dy}{dx} \right) = \frac{d}{d \theta} (tan \theta) = sec^2 \theta

$

$\displaystyle

\frac{d \theta}{dx} = \frac{1}{\frac{dx}{d \theta}} = \frac{1}{\theta cos \theta}

$

therefore

$\displaystyle

\frac{d^2 y}{dx^2} = sec^2 \theta . \frac{1}{ \theta cos \theta}

$

Is this Right!!! - Dec 7th 2009, 12:31 AMmr fantastic
- Dec 7th 2009, 05:46 PMzorroSo after simplifying
- Dec 8th 2009, 12:51 AMmr fantastic
- Dec 8th 2009, 02:19 AMHallsofIvy
By the way, Zorro, $\displaystyle \frac{d^2y}{dx^2}$ is NOT "the square of dy/dx"!

- Dec 8th 2009, 02:31 AMzorroThanks u every one for helping me
Thank u Mr fantastic for ur help (Bow)