009C Sample Final 2, Problem 9 Detailed Solution

A curve is given in polar coordinates by

$r=\sin(2\theta ).$

(a) Sketch the curve.

(b) Compute  $y'={\frac {dy}{dx}}.$

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

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Background Information:
How do you calculate  $y'$  for a polar curve  $r=f(\theta )?$
Since  $x=r\cos(\theta )$  and  $y=r\sin(\theta ),$  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:

(a)
400px

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

Final Answer:
(a)    See above
(b)    $y'={\frac {2\cos ^{2}\theta \sin \theta -\sin ^{3}\theta }{\cos ^{3}\theta -2\sin ^{2}\theta \cos \theta }}$
(c)
${\frac {d^{2}y}{dx^{2}}}={\frac {(\cos ^{3}\theta -2\sin ^{2}\theta \cos \theta )(-4\cos \theta \sin ^{2}\theta +2\cos ^{3}\theta -3\sin ^{2}\theta \cos \theta )-(2\cos ^{2}\theta \sin \theta -\sin ^{3}\theta )(-3\cos ^{2}\theta \sin \theta -4\sin \theta \cos ^{2}\theta +2\sin ^{3}\theta )}{(\cos ^{3}\theta -2\sin ^{2}\theta \cos \theta )^{2}(2\cos(2\theta )\cos \theta -\sin(2\theta )\sin \theta )}}$

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