# 009C Sample Final 1, Problem 10

A curve is given 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:

(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
$y={\frac {-4}{3}}{\bigg (}x-{\frac {3{\sqrt {2}}}{2}}{\bigg )}+2{\sqrt {2}}.$ (b) $y={\frac {-4}{3}}{\bigg (}x-{\frac {3{\sqrt {2}}}{2}}{\bigg )}+2{\sqrt {2}}$ 