Question

If x= asec^3 theta and y=a tan^3theta find d^2y/dx^2 at theta =π/3

Answer 1

Answer: The answer is $\dfrac{1}{48\sqrt 3~a}.$

Step-by-step explanation:  We are given that

$x=a\sec^3 \theta,~~~y=a\tan^3 \theta.\\\\\\\left(\dfrac{d^2y}{dx^2}\right)_{\theta=\frac{\pi}{3}}=?$

Now, we have after differentiating 'y' with respect to 'x',

$\dfrac{dy}{dx}\\\\\\=\dfrac{\frac{dy}{d\theta}}{\frac{dx}{d\theta}}\\\\\\=\dfrac{\frac{d}{d\theta}a\tan^3\theta}{\frac{d}{d\theta}a\sec^3\theta}\\\\\\=\dfrac{3a\tan^2\theta\sec^2\theta}{3a\sec^2\theta\sec \theta\tan \theta}\\\\\\=\dfrac{\tan \theta}{\sec \theta}=\sin \theta.$

Therefore,

again differentiating with respect to 'x', we have

$\dfrac{d^2y}{dx^2}\\\\\\=\dfrac{d}{dx}\sin \theta\\\\\\=\dfrac{d}{d\theta}\sin \theta\dfrac{\d\theta}{dx}\\\\\\=\dfrac{\cos\theta}{3a\sec^2\theta\sec\theta\tan\theta}\\\\\\=\dfrac{1}{3a}\dfrac{\cos^5\theta}{\sin \theta}.$

Hence,

$\left(\dfrac{d^2y}{dx^2}\right)_{\theta=\frac{\pi}{3}}\\\\\\=\dfrac{1}{3a}\dfrac{cos^5\dfrac{\pi}{3}}{\sin\frac{\pi}{3}}\\\\\\=\dfrac{1}{3a}\dfrac{(\frac{1}{2})^5}{\frac{\sqrt 3}{2}}\\\\\\=\dfrac{1}{48\sqrt 3~a}.$

Hence solved.

Answer 2

Answer:

Required answer is 1/48√3.

Step-by-step explanation:

Given:

$x=a\,sec^3 \,\theta\:\:\:and\:\:\:y=a\,tan^3 \,\theta$

To find: $(\frac{\mathrm{d^2}y}{\mathrm{d}x^2})_{\theta=\frac{\pi}{3}}$

first to get $\frac{\mathrm{d}y}{\mathrm{d}x}$ by derivating y and x by $\theta$

We get,

$\frac{\mathrm{d}y}{\mathrm{d}\theta}=\frac{\mathrm{d}(a\,tan^3\,\theta)}{\mathrm{d}\theta}=a\times3\,tan^2\theta\times\frac{\mathrm{d}(tan\,\theta)}{\mathrm{d}\theta}=3a\,tan^2\,\theta\:sec^2\theta$

$\frac{\mathrm{d}x}{\mathrm{d}\theta}=\frac{\mathrm{d}(a\,sec^3\,\theta)}{\mathrm{d}\theta}=a\times3\,sex^2\theta\times\frac{\mathrm{d}(sec\,\theta)}{\mathrm{d}\theta}=3a\,sec^2\,\theta\:sec\theta\,tan\,\theta$

Now,

$\frac{\mathrm{d}y}{\mathrm{d}x}=\frac{\frac{\mathrm{d}y}{\mathrm{d}\theta}}{\frac{\mathrm{d}x}{\mathrm{d}\theta}}=\frac{3a\,tan^2\,\theta\:sec^2\theta}{3a\,sec^2\,\theta\:sec\theta\,tan\,\theta}=\frac{tan\,\theta}{sec\theta}=\frac{\frac{sin\,\theta}{cos\,\theta}}{\frac{1}{cos\,\theta}}=sin\,\theta$

Finally,

$\frac{\mathrm{d^2}y}{\mathrm{d}x^2}=\frac{\mathrm{d}}{\mathrm{d}x}(\frac{\mathrm{d}y}{\mathrm{d}x})=\frac{\mathrm{d}sin\,\theta}{\mathrm{d}x}=cos\,\theta\times\frac{\mathrm{d}\,\theta}{\mathrm{d}x}=cos\,\theta\times\frac{1}{3a\,sec^2\,\theta\:sec\theta\,tan\,\theta}=\frac{cos\,\theta}{3a\,sec^3\,\theta\,tan\,\theta}=\frac{cos^5\,\theta}{3a\:sin\,\theta}$

Value of $(\frac{\mathrm{d^2}y}{\mathrm{d}x^2})_{\theta=\frac{\pi}{3}}=\frac{cos^5\,(\frac{\pi}{3})}{3a\:sin\,(\frac{\pi}{3})}=\frac{(\frac{1}{2})^5}{3a(\frac{\sqrt{3}}{2})}=\frac{1}{48\sqrt{3}}$

Therefore, Required answer is 1/48√3.