Question

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

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

Answer 1

Answer: (D)

Given $r={\left[2\varphi +{\cos }^2(2\varphi +\pi /4)\right]}^{1/2}$
$\mathrm{dr}/d\varphi =(1/2){\left[2\varphi +{\cos }^2(2\varphi +\pi /4)\right]}^{-1/2}[2-2\sin (4\varphi +\pi /2)]$
$={\left[2\varphi +{\cos }^2(2\varphi +\pi /4)\right]}^{-1/2}[1-\cos (4\varphi )]$
dr/d\phiat $\varphi =\pi /4$
$={\left[2\pi /4+{\cos }^2(2\pi /4+\pi /4)\right]}^{-1/2}[1-\cos (4\pi /4)]$
$={\left[\pi /2+{\cos }^2(\pi /2+\pi /4)\right]}^{-1/2}[1-\cos (\pi )]$
$={\left[\pi /2+{\cos }^2(3\pi /4)\right]}^{-1/2}[1+1]$
$={\left[\pi /2+{\cos }^2(3\pi /4)\right]}^{-1/2}\times 2$
$=[\pi /2+1/2{]}^{-1/2}\times 2$
$=2[(\pi +1)/2{]}^{-1/2}$
$=2\sqrt{2}/\sqrt{(}\pi +1)$
$=2(2/\pi +1{)}^{1/2}$