009C Sample Final 1, Problem 10 Detailed Solution

A curve is given in polar parametrically 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 x(t)=3\sin t}

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(t)=4\cos t}

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 0\leq t \leq 2\pi}

(a) Sketch the curve.

(b) Compute 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 t_0=\frac{\pi}{4}} .

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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. What is the slope of the tangent line of a parametric curve?
The slope 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 m=\frac{dy}{dx}=\frac{\big(\frac{dy}{dt}\big)}{\big(\frac{dx}{dt}\big)}.}

Solution:

(a)
500px

 (b)

Step 1:
First, we need to find the slope of the tangent line.
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 \frac{dy}{dt}=-4\sin t}   and  ${\frac {dx}{dt}}=3\cos t,$  we have
Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle {\frac {dy}{dx}}={\frac {\frac {dy}{dt}}{\frac {dx}{dt}}}={\frac {-4\sin t}{3\cos t}}.}
So, at  $t_{0}={\frac {\pi }{4}},$  the slope of the tangent line is
$m={\frac {-4\sin {\bigg (}{\frac {\pi }{4}}{\bigg )}}{3\cos {\bigg (}{\frac {\pi }{4}}{\bigg )}}}={\frac {-4}{3}}.$

Step 2:
Since we have the slope of the tangent line, we just need a find a point on the line in order to write the equation.
If we plug in  $t_{0}={\frac {\pi }{4}}$  into the equations for  $x(t)$  and  $y(t),$  we get
$x{\bigg (}{\frac {\pi }{4}}{\bigg )}=3\sin {\bigg (}{\frac {\pi }{4}}{\bigg )}={\frac {3{\sqrt {2}}}{2}}$
and
$y{\bigg (}{\frac {\pi }{4}}{\bigg )}=4\cos {\bigg (}{\frac {\pi }{4}}{\bigg )}=2{\sqrt {2}}.$
Thus, the point  ${\bigg (}{\frac {3{\sqrt {2}}}{2}},2{\sqrt {2}}{\bigg )}$  is on the tangent line.

Step 3:
Using the point found in Step 2, the equation of the tangent line at  $t_{0}={\frac {\pi }{4}}$  is
$y={\frac {-4}{3}}{\bigg (}x-{\frac {3{\sqrt {2}}}{2}}{\bigg )}+2{\sqrt {2}}.$

Final Answer:
(a)     See Step 1 above for the graph.
(b)     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{-4}{3}\bigg(x-\frac{3\sqrt{2}}{2}\bigg)+2\sqrt{2}}

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