Question

If $r={\left[2\phi +{\cos }^2\left(2\phi +\frac{\pi }{4}\right)\right]}^{\frac{1}{2}}$, then what is the value of the derivative of $\frac{\mathrm{d}r}{\mathrm{d}\phi }$ at $\phi =\frac{\pi }{4}$ ?

• A. $2{\left(\frac{1}{\pi +1}\right)}^{\frac{1}{2}}$
• B. $2{\left(\frac{2}{\pi +1}\right)}^2$
• C. ${\left(\frac{2}{\pi +1}\right)}^{\frac{1}{2}}$
• D. $2{\left(\frac{2}{\pi +1}\right)}^{\frac{1}{2}}$

Answer 1

The correct option is D

$2{\left(\frac{2}{\pi +1}\right)}^{\frac{1}{2}}$

Explanation for the correct option.

Step 1: Differentiate $\mathit{r}$ with respect to $\mathit{\phi }$.

Since $r={\left[2\phi +{\cos }^2\left(2\phi +\frac{\pi }{4}\right)\right]}^{\frac{1}{2}}$, then

$\begin{array}{rcl}\frac{dr}{d\phi }& =& \frac{1}{2}{\left[2\phi +{\cos }^2\left(2\phi +\frac{\mathrm{\pi }}{4}\right)\right]}^{-\frac{1}{2}}\left[2-2\sin \left(2\phi +\frac{\mathrm{\pi }}{4}\right)\cos \left(2\phi +\frac{\mathrm{\pi }}{4}\right)\right]\end{array}$

As per the trigonometric identity,

$\mathbf{2}\mathbf{sin}\left(x\right)\mathbf{cos}\left(x\right)\mathbf{=}\mathbf{sin}\left(2x\right)$

$\begin{array}{rcl}\therefore \frac{dr}{d\phi }& =& \left[2\phi +{\cos }^2\left(2\phi +\frac{\mathrm{\pi }}{4}\right)\right]{}^{-\frac{1}{2}}\left[1-\sin \left(4\phi +\frac{\pi }{2}\right)\right]\\ & =& {\left[2\phi +{\cos }^2\left(2\phi +\frac{\mathrm{\pi }}{4}\right)\right]}^{-\frac{1}{2}}\left[1-\cos \left(4\phi \right)\right]\end{array}$

Step 2: Evaluate $\frac{\mathrm{d}r}{\mathrm{d}\phi }$at $\phi =\frac{\pi }{4}$

Put $\phi =\frac{\pi }{4}$ in $\frac{\mathrm{d}r}{\mathrm{d}\phi }$we have:

$\begin{array}{rcl}\frac{dr}{d\phi }& =& {\left[\frac{2\mathrm{\pi }}{4}+{\cos }^2\left(\frac{2\mathrm{\pi }}{4}+\frac{\mathrm{\pi }}{4}\right)\right]}^{-\frac{1}{2}}\left[1-\cos \left(\frac{4\mathrm{\pi }}{4}\right)\right]\\ & =& {\left[\frac{\mathrm{\pi }}{2}+{\cos }^2\left(\frac{3\pi }{4}\right)\right]}^{-\frac{1}{2}}\left[1-\cos \left(\pi \right)\right]\\ & =& {\left[\frac{\mathrm{\pi }}{2}+{\left(\frac{1}{\sqrt{2}}\right)}^2\right]}^{-\frac{1}{2}}\left[1-\left(-1\right)\right]\\ & =& {\left[\frac{\mathrm{\pi }}{2}+\frac{1}{2}\right]}^{-\frac{1}{2}}2\\ & =& {\left(\frac{2}{\mathrm{\pi }+1}\right)}^{\frac{1}{2}}2\end{array}$

Hence, option (D) is correct.