Question

MVC:
Given the following equations for the motion of a particle:
x=t-1
y=4e^(-t)
z=2-root(t)
where t>0
Find the acute angle between the velocity vector and the normal line to the surface (x^2/4)+y^2+z^2=1 at the points where the particle collides with the surface.

Thank you so much in advance!

Answer 1

Alright, we're given (x,y,z) as a function of t. Or, better, we can write a position vector r as the vector sum of xi,yj, and zk. Do you follow?

Answer 2

r = (t-1)i + (4e^(-t))j + (2-root(t))k, right?

Answer 3

Yup. The velocity vector is the differential of this. I think your difficult is in finding the angle?

Answer 4

so v = i + (-4e^(-t))j + (-t^(-1/2))k
if I derived correctly

Answer 5

The derivative of 2-t^(1/2) is -(1/2)t^(-1/2).

Answer 6

Yep, of course. That was silly. So then do I plug the parametric equations into the surface function to get a solution for where the particles hit the surface?

Answer 7

Yep.

Answer 8

Remember the normal line is orthogonal to the surface.

Answer 9

Position into surface function to find times. Times into velocity for magnitude. Find orthogonal vector to surface.

Answer 10

I'm headed out. Good luck!

Answer 11

You're awesome. Thank you so much!