Difference between revisions of "009C Sample Final 3, Problem 7"

Revision as of 18:51, 2 December 2017

A curve is given in polar coordinates by

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)}

(a) Show that the point with Cartesian coordinates 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)=\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 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(\frac{\sqrt{2}}{2},\frac{\sqrt{2}}{2}\bigg).}

Solution

Detailed Solution

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@@ Line 9: / Line 9: @@
 <span class="exam">(c) In Cartesian coordinates, find the equation of the tangent line at &nbsp;<math style="vertical-align: -15px">\bigg(\frac{\sqrt{2}}{2},\frac{\sqrt{2}}{2}\bigg).</math>
-{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
+<hr>
-!Foundations: &nbsp;
+[[009C Sample Final 3, Problem 7 Solution|'''<u>Solution</u>''']]
-|-
-|'''1.''' What two pieces of information do you need to write the equation of a line?
-|-
-|
-&nbsp; &nbsp; &nbsp; &nbsp;You need the slope of the line and a point on the line.
-|-
-|'''2.''' How do you calculate &nbsp; <math style="vertical-align: -5px">y'</math> &nbsp; for a polar curve &nbsp;<math style="vertical-align: -5px">r=f(\theta)?</math>
-|-
-|
-&nbsp; &nbsp; &nbsp; &nbsp;Since &nbsp; <math style="vertical-align: -5px">x=r\cos(\theta),~y=r\sin(\theta),</math>&nbsp; we have
-|-
-|
-&nbsp; &nbsp; &nbsp; &nbsp;<math>y'=\frac{dy}{dx}=\frac{\frac{dr}{d\theta}\sin\theta+r\cos\theta}{\frac{dr}{d\theta}\cos\theta-r\sin\theta}.</math>
-|}
-'''Solution:'''
+[[009C Sample Final 3, Problem 7 Detailed Solution|'''<u>Detailed Solution</u>''']]
-'''(a)'''
-{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
-!Step 1: &nbsp;
-|-
-|First, we need to convert this Cartesian point into polar.
-|-
-|We have
-|-
-|&nbsp; &nbsp; &nbsp; &nbsp; <math>\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}</math>
-|-
-|Also, we have
-|-
-|&nbsp; &nbsp; &nbsp; &nbsp; <math>\begin{array}{rcl}
-\displaystyle{\tan \theta } & = & \displaystyle{\frac{y}{x}}\\
-&&\\
-& = & \displaystyle{1.}
-\end{array}</math>
-|-
-|So, &nbsp;<math style="vertical-align: -15px">\theta=\frac{\pi}{4}.</math>
-|-
-|Now, this point in polar is &nbsp;<math style="vertical-align: -15px">\bigg(1,\frac{\pi}{4}\bigg).</math>
-|}
-{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
-!Step 2: &nbsp;
-|-
-|Now, we plug in &nbsp;<math style="vertical-align: -15px">\theta=\frac{\pi}{4}</math>&nbsp; into our polar equation.
-|-
-|We get
-|-
-|&nbsp; &nbsp; &nbsp; &nbsp; <math>\begin{array}{rcl}
-\displaystyle{r} & = & \displaystyle{1+\cos^2\bigg(\frac{2\pi}{4}\bigg)}\\
-&&\\
-& = & \displaystyle{1+(0)^2}\\
-&&\\
-& = & \displaystyle{1.}
-\end{array}</math>
-|-
-|So, the point &nbsp;<math style="vertical-align: -5px">(x,y)</math>&nbsp; belongs to the curve.
-|}
-{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
-!(b) &nbsp;
-|-
-|Insert graph
-|-
-|
-|-
-|
-|}
-'''(c)'''
-{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
-!Step 1: &nbsp;
-|-
-|Since &nbsp;<math style="vertical-align: -5px">r=1+\cos^2(2\theta),</math>
-|-
-|
-&nbsp; &nbsp; &nbsp; &nbsp;<math>\frac{dr}{d\theta}=-4\cos(2\theta)\sin(2\theta).</math>
-|-
-|Since
-|-
-|
-&nbsp; &nbsp; &nbsp; &nbsp;<math>y'=\frac{dy}{dx}=\frac{\frac{dr}{d\theta}\sin\theta+r\cos\theta}{\frac{dr}{d\theta}\cos\theta-r\sin\theta},</math>
-|-
-|we have
-|-
-|
-&nbsp; &nbsp; &nbsp; &nbsp; <math>\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}</math>
-|-
-|
-|}
-{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
-!Step 2: &nbsp;
-|-
-|Now, recall from part (a) that the given point in polar coordinates is &nbsp;<math style="vertical-align: -15px">\bigg(1,\frac{\pi}{4}\bigg).</math>
-|-
-|Therefore, the slope of the tangent line at this point is
-|-
-|&nbsp; &nbsp; &nbsp; &nbsp; <math>\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}</math>
-|-
-|Therefore, the equation of the tangent line at the point &nbsp;<math style="vertical-align: -5px">(x,y)</math>&nbsp; is
-|-
-|&nbsp; &nbsp; &nbsp; &nbsp;<math>y=-1\bigg(x-\frac{\sqrt{2}}{2}\bigg)+\frac{\sqrt{2}}{2}.</math>
-|}
-{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
-!Final Answer: &nbsp;
-|-
-|&nbsp; &nbsp; '''(a)''' &nbsp; &nbsp; See above.
-|-
-|&nbsp; &nbsp; '''(b)''' &nbsp; &nbsp; See above.
-|-
-|&nbsp; &nbsp; '''(c)''' &nbsp; &nbsp; <math>y=-1\bigg(x-\frac{\sqrt{2}}{2}\bigg)+\frac{\sqrt{2}}{2}</math>
-|}
 [[009C_Sample_Final_3|'''<u>Return to Sample Exam</u>''']]

Difference between revisions of "009C Sample Final 3, Problem 7"

Revision as of 18:51, 2 December 2017

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