# 009C Sample Final 2, Problem 9

A curve is given in polar coordinates by

${\displaystyle r=\sin(2\theta ).}$

(a) Sketch the curve.

(b) Compute  ${\displaystyle y'={\frac {dy}{dx}}.}$

(c) Compute  ${\displaystyle y''={\frac {d^{2}y}{dx^{2}}}.}$

Foundations:
How do you calculate   ${\displaystyle y'}$   for a polar curve  ${\displaystyle r=f(\theta )?}$

Since   ${\displaystyle x=r\cos(\theta ),~y=r\sin(\theta ),}$  we have

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

Solution:

(a)

(b)

Step 1:
Since  ${\displaystyle r=\sin(2\theta ),}$

${\displaystyle {\frac {dr}{d\theta }}=2\cos(2\theta ).}$

Step 2:
Since

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

we have

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

since
${\displaystyle \sin(2\theta )=2\sin \theta \cos \theta ,~\cos(2\theta )=\cos ^{2}\theta -\sin ^{2}\theta .}$

(c)

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

${\displaystyle {\begin{array}{rcl}\displaystyle {\frac {dy'}{d\theta }}&=&\displaystyle {{\frac {d}{d\theta }}{\bigg (}{\frac {2\cos ^{2}\theta \sin \theta -\sin ^{3}\theta }{\cos ^{3}\theta -2\sin ^{2}\theta \cos \theta }}{\bigg )}}\\&&\\&=&\displaystyle {{\frac {(\cos ^{3}\theta -2\sin ^{2}\theta \cos \theta )(-4\cos \theta \sin ^{2}\theta +2\cos ^{3}\theta -3\sin ^{2}\theta \cos \theta )-(2\cos ^{2}\theta \sin \theta -\sin ^{3}\theta )(-3\cos ^{2}\theta \sin \theta -4\sin \theta \cos ^{2}\theta +2\sin ^{3}\theta )}{(\cos ^{3}\theta -2\sin ^{2}\theta \cos \theta )^{2}}}.}\end{array}}}$

Step 2:
Now, using the resulting formula for   ${\displaystyle {\frac {dy'}{d\theta }},}$  we get

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

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