Question

find d2y/dx2 of x=acos^3t y=asin^3t

Answer 1

Hope this will help you.

Answer 2

Answer:

$\frac{d^{2} y}{d x^{2}}=\frac{1}{3 a \sin t \cos ^{4} t}$

Step-by-step explanation:

Let us take,

$\begin{array}{l}{x=a \cos ^{3} t} \\ {y=a \sin ^{3} t}\end{array}$

Now, on differentiation of both sides with respect to t, we get

$\frac{d x}{d t}=a \times 3 \times \cos ^{2} t(-\sin t)$

By simplifying the above step, we get,  

$\begin{array}{l}{\frac{d x}{d t}=-3 a \sin t \cos ^{2} t} \\ {x=a \sin ^{3} t}\end{array}$

Now, on differentiation of both sides with respect to t, we get,  

$\frac{d y}{d t}=a \times 3 \times \sin ^{2} t(\cos t)$

Now, rearranging the above step in general format, we get

$\frac{d y}{d t}=3 a \sin ^{2} t \cos t$

Now we get,

$\frac{d y}{d x} \ \text{is obtained on dividing} \ \frac{d y}{d t} \text { by } \frac{d t}{d x^{\prime}}$

Thus we get,

$\therefore \frac{d y}{d x}=\frac{\frac{d y}{d t}}{\frac{d t}{d x}}=\frac{3 a \sin ^{2} t \cos t}{-3 a \sin t \cos ^{2} t}=-\tan t$

$\therefore \frac{d^{2} y}{d x^{2}}=\frac{d}{d x}\left(\frac{d y}{d x}\right)$

$=\frac{d}{d x}(-\tan t)$

$=\frac{d}{d t}(-\tan t) \times \frac{d t}{d x}$

$=-\sec ^{2} t\left(-\frac{1}{3 a \sin t \cos ^{2} t}\right)$

$=-\frac{1}{\cos ^{2} t}\left(-\frac{1}{3 a \sin t \cos ^{2} t}\right)$

$\Rightarrow \frac{d^{2} y}{d x^{2}}=\frac{1}{3 a \sin t \cos ^{4} t}$