009C Sample Final 3, Problem 7 Detailed Solution

A curve is given in polar coordinates by

$r=1+\cos ^{2}(2\theta ).$

(a) Show that the point with Cartesian coordinates  $(x,y)={\bigg (}{\frac {\sqrt {2}}{2}},{\frac {\sqrt {2}}{2}}{\bigg )}$  belongs to the curve.

(b) Sketch the curve.

(c) In Cartesian coordinates, find the equation of the tangent line at  ${\bigg (}{\frac {\sqrt {2}}{2}},{\frac {\sqrt {2}}{2}}{\bigg )}.$

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Background Information:
1. What two pieces of information do you need to write the equation of a line?
You need the slope of the line and a point on the line.
2. 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)

Step 1:
First, we need to convert this Cartesian point into polar.
We have
Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle {\begin{array}{rcl}\displaystyle {r}&=&\displaystyle {\sqrt {x^{2}+y^{2}}}\\&&\\&=&\displaystyle {\sqrt {{\frac {2}{4}}+{\frac {2}{4}}}}\\&&\\&=&\displaystyle {\sqrt {1}}\\&&\\&=&\displaystyle {1.}\end{array}}}
Also, we have
${\begin{array}{rcl}\displaystyle {\tan \theta }&=&\displaystyle {\frac {y}{x}}\\&&\\&=&\displaystyle {1.}\end{array}}$
So,  $\theta ={\frac {\pi }{4}}.$
Now, this point in polar is  ${\bigg (}1,{\frac {\pi }{4}}{\bigg )}.$

Step 2:
Now, we plug in  Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle \theta ={\frac {\pi }{4}}}   into our polar equation.
We get
${\begin{array}{rcl}\displaystyle {r}&=&\displaystyle {1+\cos ^{2}{\bigg (}{\frac {2\pi }{4}}{\bigg )}}\\&&\\&=&\displaystyle {1+(0)^{2}}\\&&\\&=&\displaystyle {1.}\end{array}}$
So, the point  $(x,y)$  belongs to the curve.

(b)
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(c)

Step 1:
Since  Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle r=1+\cos ^{2}(2\theta ),}
${\frac {dr}{d\theta }}=-4\cos(2\theta )\sin(2\theta ).$
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
${\begin{array}{rcl}\displaystyle {\frac {dy}{dx}}&=&\displaystyle {{\frac {-4\cos(2\theta )\sin(2\theta )\sin \theta +(1+\cos ^{2}(2\theta ))\cos \theta }{-4\cos(2\theta )\sin(2\theta )\cos \theta -(1+\cos ^{2}(2\theta ))\sin \theta }}.}\\\end{array}}$

Step 2:
Now, recall from part (a) that the given point in polar coordinates is  ${\bigg (}1,{\frac {\pi }{4}}{\bigg )}.$
Therefore, the slope of the tangent line at this point is
Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle {\begin{array}{rcl}\displaystyle {m}&=&\displaystyle {\frac {-4\cos({\frac {\pi }{2}})\sin({\frac {\pi }{2}})\sin({\frac {\pi }{4}})+(1+\cos ^{2}({\frac {\pi }{2}}))\cos({\frac {\pi }{4}})}{-4\cos({\frac {\pi }{2}})\sin({\frac {\pi }{2}})\cos({\frac {\pi }{4}})-(1+\cos ^{2}({\frac {\pi }{2}}))\sin({\frac {\pi }{4}})}}\\&&\\&=&\displaystyle {\frac {0+(1)({\frac {\sqrt {2}}{2}})}{0-(1)({\frac {\sqrt {2}}{2}})}}\\&&\\&=&\displaystyle {-1.}\end{array}}}
Therefore, the equation of the tangent line at the point  $(x,y)$  is
Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle y=-1{\bigg (}x-{\frac {\sqrt {2}}{2}}{\bigg )}+{\frac {\sqrt {2}}{2}}.}

Final Answer:
(a)     See above.
(b)     See above.
(c)     $y=-1{\bigg (}x-{\frac {\sqrt {2}}{2}}{\bigg )}+{\frac {\sqrt {2}}{2}}$

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