Difficult multivariate vector valued calculus problem

[slowerlower]

The problem is: An object begins at rest at the origin. At time t=0, it begins to move with velocity v(t) = <1, cos(pi*t) - pi*sin(pi*t), sin(pi*t) + pi*t*cos(pi*t)>. At what time T, point, and acute angle does the object leave the sphere (x^2 + y^2 + z^2 = 50)?


So I'm thinking that the "velocity" indicates that the equation above is a derivative of the position function. That position function should be P(t) = <t, t*cos(pi*t), t*sin(pi*t)>. For what it's worth, the acceleration is v'(t) = <0, -pi(2sin(pi*t) + pi*t*cos(pi*t), pi(2cos(pi*t) - pi(2cos(pi*t) - pi*t*sin(pi*t)> (I think, it was a lot). I'm thinking the next step might be to plug in this Position function into the sphere somehow to solve for when it leaves the sphere, but I'm not sure. Any thoughts, even general ideas?

 

[HallsofIvy]

The sphere \(\displaystyle x^2+ y^2+ z^2= 50\) has radius \(\displaystyle \sqrt{50}= 5\sqrt{2}\). The object will "leave the sphere" when \(\displaystyle \sqrt{x^2+ y^2+ z^2}> 5\sqrt{2}\) or, equivalently, when \(\displaystyle x^2+ y^2+ z^2> 50\)
Replace x, y, and z with the given functions of t and solve for t.

 

[slowerlower]

So I solved for t=5, point (5,-5,0) which has a magnitude 5 root 2 and is on the sphere. Now I need the acute angle at the time it leaves the sphere, which should be based on the velocity function (tangent line) but I'm not sure how to solve from here. I'm thinking...

 

[HallsofIvy]

The "acute angle it makes" with what? If you mean with the sphere, you will need to find the tangent plane to the sphere at that point.