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Q.
The $2^{nd}$ derivative of $a \,sin^3t$ with respect to $a \,cos^3t$ at $t=\frac{\pi}{4}$ is
Continuity and Differentiability
Solution:
Let $y= asin^3t$ and $x = a \,cos^3\,t$
On differentiating $w.r.t. t$, we get
$\frac{dy}{dt}=3a\,sin^{2}\,t\,cos\,t$ and
$\frac{dx}{dt}=3a\,cos^{2}\,t\left(-sin\,t\right)$
$\therefore \frac{dy}{dx}=\frac{\left(\frac{dy}{dt}\right)}{\left(\frac{dx}{dt}\right)}$
$=\frac{3a\,sin^{2}\,t\,cos\,t}{3a\,cos^{2}\,t\left(-sin\,t\right)}$
$=-tan\,t$
Again differentiating $w.r.t. x$, we get
$\frac{d^{2}\,y}{dx^{2}}=-sec^{2}\,t \frac{dt}{dx}$
$=-\frac{-sec^{2}\,t}{3a\,cos^{2}\,t\left(-sin\,t\right)}$
$=\frac{1}{3a}\left(\frac{sec^{4}\,t}{sin\,t}\right)$
$\therefore \left(\frac{d^{2}y}{dx^{2}}\right)_{t=\frac{\pi}{4}}=\frac{1}{3a}\cdot\frac{4}{\frac{1}{\sqrt{2}}}$
$=\frac{4\sqrt{2}}{3a}$