009C Sample Final 1, Problem 7

A curve is given in polar coordinates by

$r=1+\sin \theta$

a) Sketch the curve.

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

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

Foundations:

Solution:

(a)

Step 1:
Insert sketch of graph

(b)

Step 1:
First, recall we have
$y'={\frac {dy}{dx}}={\frac {{\frac {dr}{d\theta }}\sin \theta +r\cos \theta }{{\frac {dr}{d\theta }}\cos \theta -r\sin \theta }}$.
Since $r=1+\sin \theta$, ${\frac {dr}{d\theta }}=\cos \theta$.
Hence, $y'={\frac {\cos \theta \sin \theta +(1+\sin \theta )\cos \theta }{\cos ^{2}\theta -(1+\sin \theta )\sin \theta }}$

Step 2:
Thus, we have ${\begin{array}{rcl}\displaystyle {y'}&=&\displaystyle {\frac {2\cos \theta \sin \theta +\cos \theta }{\cos ^{2}\theta -\sin ^{2}\theta -\sin \theta }}\\&&\\&=&\displaystyle {\frac {\sin(2\theta )+\cos \theta }{\cos(2\theta )-\sin \theta }}\\\end{array}}$

(c)

Step 1:

Step 2:

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

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