Find the equations of the tangent and the normal at the point `' t '` on the curve `x=a\ s in^3t` , `y=b\ cos^3t` .

Answer

`y=0`

Explanation

Diff. given curve w.r.t. ' `t`


`\frac{d x}{d t}=3 a \sin ^{2} t \cos t` and `\frac{d y}{d t}=-3 b \cos ^{2} t \sin t`


So, `\frac{d y}{d x}=\frac{d y / d t}{d x / d t}`


`=\frac{-3 b \cos ^{2} t \sin t}{3 a^{2} \sin ^{2} t \cos t}`


`=-\frac{b \cos t}{a \sin t}`


When `t=\frac{\pi}{2}`, then


`(\frac{d y}{d x})_{t=\frac{\pi}{2}}=-\frac{b \cos \frac{\pi}{2}}{{asin} \frac{\pi}{2}}=0`


So, at `t=\frac{\pi}{2}` then `x=` a and `y=0`


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