# 009C Sample Final 3, Problem 7

A curve is given in polar coordinates by

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

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

(b) Sketch the curve.

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

Foundations:
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   ${\displaystyle y'}$   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:
First, we need to convert this Cartesian point into polar.
We have
${\displaystyle {\begin{array}{rcl}\displaystyle {r}&=&\displaystyle {\sqrt {x^{2}+y^{2}}}\\&&\\&=&\displaystyle {\sqrt {{\frac {2}{4}}+{\frac {2}{4}}}}\\&&\\&=&\displaystyle {\sqrt {1}}\\&&\\&=&\displaystyle {1.}\end{array}}}$
Also, we have
${\displaystyle {\begin{array}{rcl}\displaystyle {\tan \theta }&=&\displaystyle {\frac {y}{x}}\\&&\\&=&\displaystyle {1.}\end{array}}}$
So,  ${\displaystyle \theta ={\frac {\pi }{4}}.}$
Now, this point in polar is  ${\displaystyle {\bigg (}1,{\frac {\pi }{4}}{\bigg )}.}$
Step 2:
Now, we plug in  ${\displaystyle \theta ={\frac {\pi }{4}}}$  into our polar equation.
We get
${\displaystyle {\begin{array}{rcl}\displaystyle {r}&=&\displaystyle {1+\cos ^{2}{\bigg (}{\frac {2\pi }{4}}{\bigg )}}\\&&\\&=&\displaystyle {1+(0)^{2}}\\&&\\&=&\displaystyle {1.}\end{array}}}$
So, the point  ${\displaystyle (x,y)}$  belongs to the curve.
(b)

(c)

Step 1:
Since  ${\displaystyle r=1+\cos ^{2}(2\theta ),}$

${\displaystyle {\frac {dr}{d\theta }}=-4\cos(2\theta )\sin(2\theta ).}$

Since

${\displaystyle y'={\frac {dy}{dx}}={\frac {{\frac {dr}{d\theta }}\sin \theta +r\cos \theta }{{\frac {dr}{d\theta }}\cos \theta -r\sin \theta }},}$

we have

${\displaystyle {\begin{array}{rcl}\displaystyle {\frac {dy}{dx}}&=&\displaystyle {{\frac {-4\cos(2\theta )\sin(2\theta )\sin \theta +(1+\cos ^{2}(2\theta ))\cos \theta }{-4\cos(2\theta )\sin(2\theta )\cos \theta -(1+\cos ^{2}(2\theta ))\sin \theta }}.}\\\end{array}}}$

Step 2:
Now, recall from part (a) that the given point in polar coordinates is  ${\displaystyle {\bigg (}1,{\frac {\pi }{4}}{\bigg )}.}$
Therefore, the slope of the tangent line at this point is
${\displaystyle {\begin{array}{rcl}\displaystyle {m}&=&\displaystyle {\frac {-4\cos({\frac {\pi }{2}})\sin({\frac {\pi }{2}})\sin({\frac {\pi }{4}})+(1+\cos ^{2}({\frac {\pi }{2}}))\cos({\frac {\pi }{4}})}{-4\cos({\frac {\pi }{2}})\sin({\frac {\pi }{2}})\cos({\frac {\pi }{4}})-(1+\cos ^{2}({\frac {\pi }{2}}))\sin({\frac {\pi }{4}})}}\\&&\\&=&\displaystyle {\frac {0+(1)({\frac {\sqrt {2}}{2}})}{0-(1)({\frac {\sqrt {2}}{2}})}}\\&&\\&=&\displaystyle {-1.}\end{array}}}$
Therefore, the equation of the tangent line at the point  ${\displaystyle (x,y)}$  is
${\displaystyle y=-1{\bigg (}x-{\frac {\sqrt {2}}{2}}{\bigg )}+{\frac {\sqrt {2}}{2}}.}$

(c)     ${\displaystyle y=-1{\bigg (}x-{\frac {\sqrt {2}}{2}}{\bigg )}+{\frac {\sqrt {2}}{2}}}$