009C Sample Final 3, Problem 10

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A curve is described parametrically by

$x=t^{2}$ $y=t^{3}-t$ (a) Sketch the curve for  $-2\leq t\leq 2.$ (b) Find the equation of the tangent line to the curve at the origin.

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   ${\frac {dy}{dt}}=3t^{2}-1$ and   ${\frac {dx}{dt}}=2t,$ we have

${\frac {dy}{dx}}={\frac {\frac {dy}{dt}}{\frac {dx}{dt}}}={\frac {3t^{2}-1}{2t}}.$ Step 2:
Now, the origin corresponds to  $x=0$ and  $y=0.$ This gives us two equations. When we solve for  $t,$ we get  $t=0.$ Plugging in  $t=0$ into
${\frac {dy}{dx}}={\frac {\frac {dy}{dt}}{\frac {dx}{dt}}}={\frac {3t^{2}-1}{2t}},$ we see that  ${\frac {dy}{dx}}$ is undefined at  $t=0.$ So, there is no tangent line at the origin.