# 009C Sample Final 3, Problem 10

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A curve is described parametrically by

${\displaystyle x=t^{2}}$
${\displaystyle y=t^{3}-t}$

(a) Sketch the curve for  ${\displaystyle -2\leq t\leq 2.}$

(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  ${\displaystyle m={\frac {dy}{dx}}={\frac {\frac {dy}{dt}}{\frac {dx}{dt}}}.}$

Solution:

(a)

(b)

Step 1:
First, we need to find the slope of the tangent line.
Since   ${\displaystyle {\frac {dy}{dt}}=3t^{2}-1}$   and   ${\displaystyle {\frac {dx}{dt}}=2t,}$  we have

${\displaystyle {\frac {dy}{dx}}={\frac {\frac {dy}{dt}}{\frac {dx}{dt}}}={\frac {3t^{2}-1}{2t}}.}$

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