Question

If $x=\sin t\cos 2t$ and $y=\cos t\sin 2t$, then at $t=\frac{\mathrm{\pi }}{4}$, the value of $\frac{dy}{dx}$ is equal to

• A. $-\frac{1}{2}$
• B. $\frac{1}{2}$
• C. $-2$
• D. $2$

Answer 1

The correct option is B

$\frac{1}{2}$

Explanation for the correct option.

Step 1. Find the value of $\frac{dx}{dt}$ at $t=\frac{\mathrm{\pi }}{4}$.

Differentiate $x=\sin t\cos 2t$ with respect to $t$ using product rule of differentiation.

$\begin{array}{rcl}\frac{dx}{dt}& =& \frac{d}{dt}\left(\sin t\cos 2t\right)\\ & =& \cos t\cos 2t+\sin t(-2\sin 2t)\\ & =& \cos t\cos 2t-2\sin t\sin 2t\end{array}$

Now substitute $\frac{\mathrm{\pi }}{4}$ for $t$ to find the value of $\frac{dx}{dt}$ at $t=\frac{\mathrm{\pi }}{4}$.

$\begin{array}{rcl}{\overline{)\frac{dx}{dt}}}_{t=\frac{\mathrm{\pi }}{4}}& =& \cos \frac{\mathrm{\pi }}{4}\cos \left(2\times \frac{\mathrm{\pi }}{4}\right)-2\sin \frac{\mathrm{\pi }}{4}\sin \left(2\times \frac{\mathrm{\pi }}{4}\right)\\ & =& \frac{1}{\sqrt{2}}\times 0-2\times \frac{1}{\sqrt{2}}\times 1\mathbf{[}\mathbf{\because }\mathbf{}\mathbf{c}\mathbf{o}\mathbf{s}\left(\frac{\pi }{4}\right)\mathbf{=}\mathbf{s}\mathbf{i}\mathbf{n}\left(\frac{\pi }{4}\right)\mathbf{=}\frac{\mathbf{1}}{\sqrt{\mathbf{2}}}\mathbf{,}\mathbf{}\mathbf{c}\mathbf{o}\mathbf{s}\frac{\mathbf{\pi }}{\mathbf{2}}\mathbf{=}\mathbf{0}\mathbf{,}\mathbf{}\mathbf{s}\mathbf{i}\mathbf{n}\frac{\mathbf{\pi }}{\mathbf{2}}\mathbf{=}\mathbf{1}\mathbf{]}\\ & =& 0-\frac{2}{\sqrt{2}}\\ & =& -\sqrt{2}\end{array}$

Step 2. Find the value of $\frac{dy}{dt}$ at $t=\frac{\mathrm{\pi }}{4}$.

Differentiate $y=\cos t\sin 2t$ with respect to $t$ using product rule of differentiation.

$\begin{array}{rcl}\frac{dy}{dt}& =& \frac{d}{dt}\left(\cos t\sin 2t\right)\\ & =& -\sin t\sin 2t+\cos t\left(2\cos 2t\right)\\ & =& -\sin t\sin 2t+2\cos t\cos 2t\end{array}$

Now substitute $\frac{\mathrm{\pi }}{4}$ for $t$ to find the value of $\frac{dy}{dt}$ at $t=\frac{\mathrm{\pi }}{4}$.

$\begin{array}{rcl}{\overline{)\frac{dy}{dt}}}_{t=\frac{\mathrm{\pi }}{4}}& =& -\sin \frac{\mathrm{\pi }}{4}\sin \left(2\times \frac{\mathrm{\pi }}{4}\right)+2\cos \frac{\mathrm{\pi }}{4}\cos \left(2\times \frac{\mathrm{\pi }}{4}\right)\\ & =& -\frac{1}{\sqrt{2}}\times 1+2\times \frac{1}{\sqrt{2}}\times 0\left[\because \cos \left(\frac{\pi }{4}\right)=\sin \left(\frac{\pi }{4}\right)=\frac{1}{\sqrt{2}},\cos \frac{\pi }{2}=0,\sin \frac{\pi }{2}=1\right]\\ & =& -\frac{1}{\sqrt{2}}+0\\ & =& -\frac{1}{\sqrt{2}}\end{array}$

Step 3. Find the value of $\frac{dy}{dx}$ at $t=\frac{\mathrm{\pi }}{4}$.

The value of $\frac{dy}{dx}$ at $t=\frac{\mathrm{\pi }}{4}$ is given as: ${\overline{)\frac{dy}{dx}}}_{t=\frac{\mathrm{\pi }}{4}}={\overline{)\frac{dy}{dt}}}_{t=\frac{\mathrm{\pi }}{4}}\times {\overline{)\frac{dt}{dx}}}_{t=\frac{\mathrm{\pi }}{4}}$.

Substitute the found values:

$\begin{array}{rcl}{\overline{)\frac{dy}{dx}}}_{t=\frac{\mathrm{\pi }}{4}}& =& -\frac{1}{\sqrt{2}}\times \frac{1}{-\sqrt{2}}\\ & =& \frac{1}{2}\end{array}$

Hence, the correct option is B.