Question

The $2^{nd}$ derivative of $asi{n}^{3}t$ with respect to $aco{s}^{3}t$ at $t=\frac{\pi }{4}$ is

• A. $\frac{4\sqrt{2}}{3a}$
• B. $2$
• C. $\frac{1}{12a}$
• D. None of these

Answer 1

Answer: (A)

Let $y=asi{n}^{3}t$ and $x=aco{s}^{3}t$
On differentiating $w.r.t.t$, we get
$\frac{dy}{dt}=3asi{n}^{2}tcost$ and
$\frac{dx}{dt}=3aco{s}^{2}t\left(-sint\right)$
$\therefore \frac{dy}{dx}=\frac{\left(\frac{dy}{dt}\right)}{\left(\frac{dx}{dt}\right)}$
$=\frac{3asi{n}^{2}tcost}{3aco{s}^{2}t\left(-sint\right)}$
$=-tant$
Again differentiating $w.r.t.x$, we get
$\frac{d^{2}y}{d{x}^2}=-se{c}^{2}t\frac{dt}{dx}$
$=-\frac{-se{c}^{2}t}{3aco{s}^{2}t\left(-sint\right)}$
$=\frac{1}{3a}\left(\frac{se{c}^{4}t}{sint}\right)$
$\therefore {\left(\frac{d^{2}y}{d{x}^2}\right)}_{t=\frac{\pi }{4}}=\frac{1}{3a}\cdot \frac{4}{\frac{1}{\sqrt{2}}}$
$=\frac{4\sqrt{2}}{3a}$