# 009C Sample Final 2, Problem 9

A curve is given in polar coordinates by

$r=\sin(2\theta ).$ (a) Sketch the curve.

(b) Compute  $y'={\frac {dy}{dx}}.$ (c) Compute  $y''={\frac {d^{2}y}{dx^{2}}}.$ Foundations:
How do you calculate   $y'$ for a polar curve  $r=f(\theta )?$ Since   $x=r\cos(\theta ),~y=r\sin(\theta ),$ we have

$y'={\frac {dy}{dx}}={\frac {{\frac {dr}{d\theta }}\sin \theta +r\cos \theta }{{\frac {dr}{d\theta }}\cos \theta -r\sin \theta }}.$ Solution:

(b)

Step 1:
Since  $r=\sin(2\theta ),$ ${\frac {dr}{d\theta }}=2\cos(2\theta ).$ Step 2:
Since

$y'={\frac {dy}{dx}}={\frac {{\frac {dr}{d\theta }}\sin \theta +r\cos \theta }{{\frac {dr}{d\theta }}\cos \theta -r\sin \theta }},$ we have

${\begin{array}{rcl}\displaystyle {y'}&=&\displaystyle {\frac {2\cos(2\theta )\sin \theta +\sin(2\theta )\cos \theta }{2\cos(2\theta )\cos \theta -\sin(2\theta )\sin \theta }}\\&&\\&=&\displaystyle {\frac {2\cos ^{2}\theta \sin \theta -\sin ^{3}\theta }{\cos ^{3}\theta -2\sin ^{2}\theta \cos \theta }}\end{array}}$ since
$\sin(2\theta )=2\sin \theta \cos \theta ,~\cos(2\theta )=\cos ^{2}\theta -\sin ^{2}\theta .$ (c)

Step 1:
We have   ${\frac {d^{2}y}{dx^{2}}}={\frac {\frac {dy'}{d\theta }}{{\frac {dr}{d\theta }}\cos \theta -r\sin \theta }}.$ So, first we need to find   ${\frac {dy'}{d\theta }}.$ We have

${\begin{array}{rcl}\displaystyle {\frac {dy'}{d\theta }}&=&\displaystyle {{\frac {d}{d\theta }}{\bigg (}{\frac {2\cos ^{2}\theta \sin \theta -\sin ^{3}\theta }{\cos ^{3}\theta -2\sin ^{2}\theta \cos \theta }}{\bigg )}}\\&&\\&=&\displaystyle {{\frac {(\cos ^{3}\theta -2\sin ^{2}\theta \cos \theta )(-4\cos \theta \sin ^{2}\theta +2\cos ^{3}\theta -3\sin ^{2}\theta \cos \theta )-(2\cos ^{2}\theta \sin \theta -\sin ^{3}\theta )(-3\cos ^{2}\theta \sin \theta -4\sin \theta \cos ^{2}\theta +2\sin ^{3}\theta )}{(\cos ^{3}\theta -2\sin ^{2}\theta \cos \theta )^{2}}}.}\end{array}}$ Step 2:
Now, using the resulting formula for   ${\frac {dy'}{d\theta }},$ we get

${\frac {d^{2}y}{dx^{2}}}={\frac {(\cos ^{3}\theta -2\sin ^{2}\theta \cos \theta )(-4\cos \theta \sin ^{2}\theta +2\cos ^{3}\theta -3\sin ^{2}\theta \cos \theta )-(2\cos ^{2}\theta \sin \theta -\sin ^{3}\theta )(-3\cos ^{2}\theta \sin \theta -4\sin \theta \cos ^{2}\theta +2\sin ^{3}\theta )}{(\cos ^{3}\theta -2\sin ^{2}\theta \cos \theta )^{2}(2\cos(2\theta )\cos \theta -\sin(2\theta )\sin \theta )}}.$ (b)    $y'={\frac {2\cos ^{2}\theta \sin \theta -\sin ^{3}\theta }{\cos ^{3}\theta -2\sin ^{2}\theta \cos \theta }}$ ${\frac {d^{2}y}{dx^{2}}}={\frac {(\cos ^{3}\theta -2\sin ^{2}\theta \cos \theta )(-4\cos \theta \sin ^{2}\theta +2\cos ^{3}\theta -3\sin ^{2}\theta \cos \theta )-(2\cos ^{2}\theta \sin \theta -\sin ^{3}\theta )(-3\cos ^{2}\theta \sin \theta -4\sin \theta \cos ^{2}\theta +2\sin ^{3}\theta )}{(\cos ^{3}\theta -2\sin ^{2}\theta \cos \theta )^{2}(2\cos(2\theta )\cos \theta -\sin(2\theta )\sin \theta )}}$ 