009C Sample Final 1, Problem 7 Detailed Solution

A curve is given in polar coordinates by

$r=1+\sin \theta$

(a) Sketch the curve.

(b) Compute  $y'={\frac {dy}{dx}}$.

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

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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
Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle y'={\frac {dy}{dx}}={\frac {{\big (}{\frac {dr}{d\theta }}{\big )}\sin \theta +r\cos \theta }{{\big (}{\frac {dr}{d\theta }}{\big )}\cos \theta -r\sin \theta }}.}

Solution:

(a)
400px

(b)

Step 1:
First, recall we have
Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle y'={\frac {dy}{dx}}={\frac {{\big (}{\frac {dr}{d\theta }}{\big )}\sin \theta +r\cos \theta }{{\big (}{\frac {dr}{d\theta }}{\big )}\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:
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
${\begin{array}{rcl}\displaystyle {\frac {dy'}{d\theta }}&=&\displaystyle {{\frac {d}{d\theta }}{\bigg (}{\frac {\sin(2\theta )+\cos \theta }{\cos(2\theta )-\sin \theta }}{\bigg )}}\\&&\\&=&\displaystyle {\frac {(\cos(2\theta )-\sin \theta )(2\cos(2\theta )-\sin \theta )-(\sin(2\theta )+\cos \theta )(-2\sin(2\theta )-\cos \theta )}{(\cos(2\theta )-\sin \theta )^{2}}}\\&&\\&=&\displaystyle {\frac {2\cos ^{2}(2\theta )+2\sin ^{2}(2\theta )-3\sin \theta \cos(2\theta )+3\sin(2\theta )\cos \theta +\sin ^{2}\theta +\cos ^{2}\theta }{(\cos(2\theta )-\sin \theta )^{2}}}\\&&\\&=&\displaystyle {\frac {3-3\sin \theta \cos(2\theta )+3\sin(2\theta )\cos \theta }{(\cos(2\theta )-\sin \theta )^{2}}}\\\end{array}}$
since  $\sin ^{2}\theta +\cos ^{2}\theta =1$  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)     ${\frac {\sin(2\theta )+\cos \theta }{\cos(2\theta )-\sin \theta }}$
(c)     ${\frac {3-3\sin \theta \cos(2\theta )+3\sin(2\theta )\cos \theta }{(\cos(2\theta )-\sin \theta )^{3}}}$

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