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 $\displaystyle \frac{d^2 y}{dx^2}$
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 $\displaystyle \frac{d^2 y}{dx^2}$
$\displaystyle \frac{dy}{dx} = \frac{\frac{dy}{d \theta}}{\frac{dx}{d \theta}}$ (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.
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!!!