Question

Find a vector function that represents the curve of intersection of the paraboloid z=3x2+2y2 and the cylinder y=6x2. Use the variable t for the parameter.

Answer 1

Answer:

$\vec{r(t)}=t\hat{i}+6t^2\hat{j}+(3t^2+72t^4)\hat{k} \text{ For }t \in (-\infty, \infty)$

Step-by-step explanation:

First plug the equation of y into the equation of z, so that we get their intersection:

We plug  

$y=6x^2$

Into: $z=3x^2+2y^2$

So, we get:

$z=3x^2+2(6x^2)^2\\z=3x^2+2(36x^4)\\z=3x^2+72x^4$

Then we set  

$x=t$  

And plug that in the equation $y=6x^2$ and the one we just got for z.

So that we get:

$y=6t^2, z=3t^2+72t^4$

Therefore, the vector function that represents the curve of intersection is:

$\vec{r(t)}=t\hat{i}+6t^2\hat{j}+(3t^2+72t^4)\hat{k} \text{ For }t \in (-\infty, \infty)$