Browse Questions

# If $x=a \bigg( \cos\: \theta +\log \tan\large \frac{\theta}{2} \bigg)$ and $y = a\sin\: \theta,$ find the value of $\large\frac{d^2y}{dx^2}$$\: at \: \theta = \large\frac{\pi}{4}. Can you answer this question? ## 1 Answer 0 votes Toolbox: • y=f(x) • \large\frac{dy}{dx}$$=f'(x)$
• $\large\frac{d^2y}{dx^2}=\frac{d}{dx}\big(\frac{dy}{dx}\big)$
• $\large\frac{dy}{dx}=\frac{dy}{d\theta}\times \frac{d\theta}{dx}$
Step 1:
$x=a(\cos \theta+\log\tan\large\frac{\theta}{2})$ and $y=a\sin\theta$
Let us differentiate w.r.t $\theta$ we get,
$\large\frac{dx}{d\theta}$$=a[-\sin \theta+\large\frac{1}{\tan\large\frac{\theta}{2}}$$.\sec^2\large\frac{\theta}{2}\times \frac{1}{2}]$
$\;\;\;\quad=a[-\sin \theta+\large\frac{\cos\theta/2}{\sin\theta/2}\times \frac{1}{\cos^2\theta/2}\times \frac{1}{2}]$
$\;\;\;\quad=a[-\sin \theta+\large\frac{1}{2\sin\theta/2\cos\theta/2}]$
But $2\sin\theta/2\cos\theta/2=\sin\theta$
$\;\;\;\quad=a[-\sin \theta+\large\frac{1}{\sin\theta}]$
$\;\;\;\quad=a[\large\frac{1-\sin^2\theta}{\sin\theta}]$
$1-\sin^2\theta=\cos^2\theta$
$\;\;\;\quad=a[\large\frac{\cos^2\theta}{\sin\theta}]$
Step 2:
$\large\frac{dy}{d\theta}$$=a\cos\theta \large\frac{dy}{dx}=\frac{dy}{d\theta}\times \frac{d\theta}{dx} \qquad=\large\frac{a\cos\theta\times \sin\theta}{a\cos^2\theta} \large\frac{dy}{dx}=\frac{\sin\theta}{\cos\theta} \;\;\quad=$$\tan\theta$
Step 3:
Differentiating with respect to $\theta$
$\large\frac{d^2y}{dx^2}=\frac{d}{d\theta}$$(\tan\theta)\large\frac{d\theta}{dx} \qquad=\sec^2\theta\times \large\frac{\sin \theta}{a\cos^2\theta} \qquad=\large\frac{\sec^4\theta.\sin\theta}{a} Step 4: \large\frac{d^2y}{dx^2} at \theta=\large\frac{\pi}{4} (\large\frac{d^2y}{dx^2})_{\theta=\pi/4}=\large\frac{\sec^4\pi/4.\sin\pi/4}{a} \qquad\qquad\;\;=\large\frac{\sqrt 2^4\large\frac{1}{\sqrt 2}}{a} \qquad\qquad\;\;=\large\frac{1}{a}$$2\sqrt 2$