Question

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

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

$r=\sqrt{2\varphi +{cos}^2(2\varphi +\frac{\pi }{4})}$

$\varphi =\frac{\pi }{4}$

$\Rightarrow r=\sqrt{2\times \frac{\pi }{4}+{cos}^2(2\times \frac{\pi }{4}+\frac{\pi }{4})}$

$=\sqrt{\frac{\pi }{2}+{cos}^2\left(\frac{3\pi }{4}\right)}$

$r\left(\frac{\pi }{4}\right)=\sqrt{\frac{\pi }{2}+\frac{1}{2}}=\frac{\sqrt{\pi +1}}{\sqrt{2}}$

$r^2=2\varphi +{cos}^2(2\varphi +\frac{\pi }{4})$

Differentiate w.r.t $\varphi$

$\Rightarrow 2r\frac{dr}{d\varphi }=2+2\cos (2\varphi +\frac{\pi }{4})(-\sin (2\varphi +\frac{\pi }{4}))(+2)$

$2r\left(\frac{\pi }{4}\right)\frac{dr}{d\varphi }=2+2\cos \frac{3\pi }{4}(-\sin \left(\frac{3\pi }{4}\right))(+2)$

$=2+2\times \frac{-1}{\sqrt{2}}\times \frac{-1}{\sqrt{2}}\times (+2)$

$2\times \left(\sqrt{\frac{\pi +1}{2}}\right)\times \frac{dr}{d\varphi }=4$

$\frac{dr}{d\varphi }=\frac{2\sqrt{2}}{\sqrt{\pi +1}}$