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:

(a)
Insert sketch of graph

(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
${\begin{array}{rcl}\displaystyle {\frac {dy'}{d\theta }}&=&\displaystyle {{\frac {d}{d\theta }}{\bigg (}{\frac {2\cos(2\theta )\sin \theta +\sin(2\theta )\cos \theta }{2\cos(2\theta )\cos \theta -\sin(2\theta )\sin \theta }}{\bigg )}}\\&&\\&=&\displaystyle {\frac {(2\cos(2\theta )\cos \theta -\sin(2\theta )\sin \theta )(-4\sin(2\theta )\sin \theta +2\cos(2\theta )\cos(theta)+2\cos(2\theta )\cos \theta -\sin(2\theta )\sin \theta )}{(2\cos(2\theta )\cos \theta -\sin(2\theta )\sin \theta )^{2}}}\\&&\\&=&\displaystyle {\frac {2\cos ^{2}(2\theta )+2\sin ^{2}(2\theta )-3\sin \theta \cos(2\theta )+3\sin(2\theta )\cos \theta +\sin ^{2}\theta +\cos ^{2}\theta }{(2\cos(2\theta )\cos \theta -\sin(2\theta )\sin \theta )^{2}}}\\&&\\&=&\displaystyle {\frac {3-3\sin \theta \cos(2\theta )+3\sin(2\theta )\cos \theta }{(2\cos(2\theta )\cos \theta -\sin(2\theta )\sin \theta )^{2}}}\\\end{array}}$
since  $\sin ^{2}\theta +\cos ^{2}\theta =1$  and  $2\cos ^{2}(2\theta )+2\sin ^{2}(2\theta )=2.$

Step 2:
Now, using the resulting formula for   ${\frac {dy'}{d\theta }},$  we get
${\frac {d^{2}y}{dx^{2}}}={\frac {3-3\sin \theta \cos(2\theta )+3\sin(2\theta )\cos \theta }{(\cos(2\theta )-\sin \theta )^{3}}}.$

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)

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