# Vector-valued function tangent

1. Sep 13, 2010

### Jonnyb42

1. The problem statement, all variables and given/known data

If a curve has the property that the position vector r(t) is always perpendicular to the tangent vector r'(t), show that the curve lies on the sphere with center at the origin.

2. Relevant equations

I know dot product might help:

r(t) . r'(t) = 0

and the equation of a sphere in 3-space:

r2 = x2 + y2 + z2

3. The attempt at a solution

if I write out the components of the dot product...

r(t) . r'(t) = fx(t)*fx'(t) + fy(t)*fy'(t) + fz(t)*fz'(t) = 0

From there, I am not sure what to do, if that even is the right way to start.

2. Sep 13, 2010

### Staff: Mentor

What if you integrate both sides of your last equation?

3. Sep 13, 2010

### Jonnyb42

Wow, how did you think of that?

It seems to work. The one thing I need help with is integrating the right side of 0, I think it's my lack of calculus knowledge. Does it become a constant?

4. Sep 13, 2010

### Staff: Mentor

I don't know - it just occurred to me because of those terms fx fx'.
Yes.

5. Sep 13, 2010

### Jonnyb42

Thank you very much, just for completion's sake, I'll show the rest of the work:

It is easier to write functions with different letters, so from before, fx(t) will now be f(t), fy(t) will now be g(t), and fz(t) is now h(t).

$$\int f(t)df(t)$$ + $$\int g(t)dg(t)$$ + $$\int h(t)dh(t)$$ = $$\int 0dt$$

$$\stackrel{1}{2}$$ f2(t) + $$\stackrel{1}{2}$$ g2(t) + $$\stackrel{1}{2}$$ h2(t) = C

f2(t) + g2(t) + h2(t) = r2 <-- form of a sphere.

Last edited: Sep 13, 2010