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
$y'={\frac {2\cos(2\theta )\sin \theta +\sin(2\theta )\cos \theta }{2\cos(2\theta )\cos \theta -\sin(2\theta )\sin \theta }}.$

(c)

Step 1:

Step 2:

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

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