# Calculus III Parametric Equations

1. Jul 8, 2008

### ae4jm

1. The problem statement, all variables and given/known data
I've uploaded a scan of the questions. Questions 4, 5, and 6 are given in the 3 files uploaded. They all come from the given information from the first scan of the problem.

2. Relevant equations

3. The attempt at a solution
I've worked everything I could on paper; what am I missing. How should I go about finishing the problem and am I correct so far?

Any good links for parametric equations also?

Thanks for the help!

File size:
71.9 KB
Views:
57
File size:
34.9 KB
Views:
58
File size:
32.6 KB
Views:
55
2. Jul 8, 2008

### HallsofIvy

Staff Emeritus
For part 3, you have correctly shown that the curve passes through (3, 0) for $t=\sqrt{3}$ and for $t= -\sqrt{3}$. I confess that I don't understand the rather complicated way you do that! It is sufficient to show that $y= t^3- 3t= 0$ has roots 0, $\sqrt{3}$ and $-\sqrt{3}$ and that only the last two also satisfy $x= t^2= 3$.

You have shown that the tangent line at $t= \sqrt{3}$ is $y= \sqrt{3}x- 3\sqrt{3}$ but you don't seem to have completed the problem for $t= -\sqrt{3}$.

For (4), asking where the tangent lines are horizontal or vertical, again, you have again written a complicated calculation starting with writing $x^2+ y^2$ in terms of t. I can see no reason for that. Horizontal tangents occur where $y'= 3t2-1= 0$ and vertical tangents occur where $x'= 2t= 0$.

3. Jul 8, 2008

### ae4jm

Thanks HallsofIvy, once again, for the great help. These forums are great to get the wheels turning!