So, the problem is this:
Find all values of t such that r'(t) is parallel to the xy-plane.
And my equation is:
r(t)=(Squareroot(t+1) , cos(t), t4-8t2)
Well, I will definately have to know how to take the dirivative of the given vector valued function, which I know how to do. I think An important thing to know would be: the cross-product of the two vector-valued functions. But I can't really think of any pertinant formulae off hand that would help with this problem .... (maybe that's why I'm confused).
The Attempt at a Solution
Okay, so as I said I need to take the derivative and then I thought take the cross-product of the derivative and the original function.
So I did that and got: r'(t)= (1/2(t+1)-1/2, -sin(t), 4t3- 16t)
Wasn't much of a problem there, but now I'm kinda clueless as to how I would find ALL values of t for r'(t) parallel to the xy-plane. So I was thinking about the cross-product because that will tell me if the vector valued function and its derivative are parallel (if I get 0 as my answer when I do the cross-product) then maybe I can figure out a parametric equation for a plane that contains the vector function (the first one not the derivative), and all values in that plane would be parallel to the derivative because the plane is parallel to the derivative. >.< That sounds kind of complicated and somewhat off base to me. I'm speculating here because I really am not sure what to do. Although, that really wouldn't help me much considering I have to deal with the xy-plane. Hm,
So does anyone think they could help steer me in the right direction conceptually? I think I'm stuck conceptually on how I would go about finding these values of t ... I mean I know what they mean by a Vector Valued Function I understand what a function is and what a vector is I know about derivatives and understand all of that and I know when two vectors are parallel when given the corrodinates, but I just don't get this problem. Could someone help me at least start thinking in the right direction.