009C Sample Final 3, Problem 10 Detailed Solution

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.

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Background Information:
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)
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(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
${\frac {dy}{dx}}={\frac {3t^{2}-1}{2t}},$
we see that  ${\frac {dy}{dx}}$  is undefined at  $t=0.$
So, there is no tangent line at the origin.

Final Answer:
(a)    See above
(b)    There is no tangent line at the origin.

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