009C Sample Final 1, Problem 10

A curve is given in polar parametrically by

$x(t)=3\sin t$

$y(t)=4\cos t$

$0\leq t\leq 2\pi$

(a) Sketch the curve.

(b) Compute the equation of the tangent line at   $t_{0}={\frac {\pi }{4}}$.

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

Solution:

(a)
Insert sketch of curve

(b)

Step 1:
First, we need to find the slope of the tangent line.
Since   ${\frac {dy}{dt}}=-4\sin t$   and   ${\frac {dx}{dt}}=3\cos t,$  we have
${\frac {dy}{dx}}={\frac {\frac {dy}{dt}}{\frac {dx}{dt}}}=-{\frac {4\sin t}{3\cos t}}.$
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 above.
(b)     $y=-{\frac {4}{3}}{\bigg (}x-{\frac {3{\sqrt {2}}}{2}}{\bigg )}+2{\sqrt {2}}$

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