009C Sample Final 1, Problem 10

A curve is given parametrically by

${\displaystyle x(t)=3\sin t}$
${\displaystyle y(t)=4\cos t}$
${\displaystyle 0\leq t\leq 2\pi }$
a) Sketch the curve.
b) Compute the equation of the tangent line at ${\displaystyle t_{0}={\frac {\pi }{4}}}$.
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 ${\displaystyle m={\frac {dy}{dx}}={\frac {\frac {dy}{dt}}{\frac {dx}{dt}}}.}$

Solution:

(a)

(b)

Step 1:
First, we need to find the slope of the tangent line.
Since ${\displaystyle {\frac {dy}{dt}}=-4\sin t}$ and ${\displaystyle {\frac {dx}{dt}}=3\cos t,}$ we have
${\displaystyle {\frac {dy}{dx}}={\frac {\frac {dy}{dt}}{\frac {dx}{dt}}}={\frac {-4\sin t}{3\cos t}}.}$
So, at ${\displaystyle t_{0}={\frac {\pi }{4}},}$ the slope of the tangent line is
${\displaystyle m={\frac {-4\sin {\bigg (}{\frac {\pi }{4}}{\bigg )}}{3\cos {\bigg (}{\frac {\pi }{4}}{\bigg )}}}={\frac {-4}{3}}.}$
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 ${\displaystyle t_{0}={\frac {\pi }{4}}}$ into the equations for ${\displaystyle x(t)}$ and ${\displaystyle y(t),}$ we get
${\displaystyle x{\bigg (}{\frac {\pi }{4}}{\bigg )}=3\sin {\bigg (}{\frac {\pi }{4}}{\bigg )}={\frac {3{\sqrt {2}}}{2}}}$ and
${\displaystyle y{\bigg (}{\frac {\pi }{4}}{\bigg )}=4\cos {\bigg (}{\frac {\pi }{4}}{\bigg )}=2{\sqrt {2}}.}$
Thus, the point ${\displaystyle {\bigg (}{\frac {3{\sqrt {2}}}{2}},2{\sqrt {2}}{\bigg )}}$ is on the tangent line.
Step 3:
Using the point found in Step 2, the equation of the tangent line at ${\displaystyle t_{0}={\frac {\pi }{4}}}$ is
${\displaystyle y={\frac {-4}{3}}{\bigg (}x-{\frac {3{\sqrt {2}}}{2}}{\bigg )}+2{\sqrt {2}}.}$
(b) ${\displaystyle y={\frac {-4}{3}}{\bigg (}x-{\frac {3{\sqrt {2}}}{2}}{\bigg )}+2{\sqrt {2}}}$