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   Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle x=r\cos(\theta),~y=r\sin(\theta),}   we have
Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\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
Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\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
Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \begin{array}{rcl} \displaystyle{\tan \theta } & = & \displaystyle{\frac{y}{x}}\\ &&\\ & = & \displaystyle{1.} \end{array}}
So,  Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \theta=\frac{\pi}{4}.}
Now, this point in polar is  Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \bigg(1,\frac{\pi}{4}\bigg).}

Step 2:
Now, we plug in  Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \theta=\frac{\pi}{4}}   into our polar equation.
We get
Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\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  Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle (x,y)}   belongs to the curve.

(b)
441px

(c)

Step 1:
Since  Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle r=1+\cos^2(2\theta),}
Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \frac{dr}{d\theta}=-4\cos(2\theta)\sin(2\theta).}
Since
Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\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
Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\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  Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \bigg(1,\frac{\pi}{4}\bigg).}
Therefore, the slope of the tangent line at this point is
Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\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  Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle (x,y)}   is
Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle y=-1\bigg(x-\frac{\sqrt{2}}{2}\bigg)+\frac{\sqrt{2}}{2}.}

Final Answer:
(a)     See above.
(b)     See above.
(c)     Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle y=-1\bigg(x-\frac{\sqrt{2}}{2}\bigg)+\frac{\sqrt{2}}{2}}

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