# 009C Sample Final 1, Problem 7

A curve is given in polar coordinates by

${\displaystyle r=1+\sin \theta }$
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)

Step 1:

(b)

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

(c)

Step 1:
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
${\displaystyle {\begin{array}{rcl}\displaystyle {\frac {dy'}{d\theta }}&=&\displaystyle {{\frac {d}{d\theta }}{\bigg (}{\frac {\sin(2\theta )+\cos \theta }{\cos(2\theta )-\sin \theta }}{\bigg )}}\\&&\\&=&\displaystyle {\frac {(\cos(2\theta )-\sin \theta )(2\cos(2\theta )-\sin \theta )-(\sin(2\theta )+\cos \theta )(-2\sin(2\theta )-\cos \theta )}{(\cos(2\theta )-\sin \theta )^{2}}}\\&&\\&=&\displaystyle {\frac {2\cos ^{2}(2\theta )+2\sin ^{2}(2\theta )-3\sin \theta \cos(2\theta )+3\sin(2\theta )\cos \theta +\sin ^{2}\theta +\cos ^{2}\theta }{(\cos(2\theta )-\sin \theta )^{2}}}\\&&\\&=&\displaystyle {\frac {3-3\sin \theta \cos(2\theta )+3\sin(2\theta )\cos \theta }{(\cos(2\theta )-\sin \theta )^{2}}}\\\end{array}}}$
since ${\displaystyle \sin ^{2}\theta +\cos ^{2}\theta =1}$ and ${\displaystyle 2\cos ^{2}(2\theta )+2\sin ^{2}(2\theta )=2.}$
Step 2:
Now, using the resulting formula for ${\displaystyle {\frac {dy'}{d\theta }},}$ we get
${\displaystyle {\frac {d^{2}y}{dx^{2}}}={\frac {3-3\sin \theta \cos(2\theta )+3\sin(2\theta )\cos \theta }{(\cos(2\theta )-\sin \theta )^{3}}}.}$
(b) ${\displaystyle {\frac {\sin(2\theta )+\cos \theta }{\cos(2\theta )-\sin \theta }}}$
(c) ${\displaystyle {\frac {3-3\sin \theta \cos(2\theta )+3\sin(2\theta )\cos \theta }{(\cos(2\theta )-\sin \theta )^{3}}}}$