Proof of r'(t) x r''(t) = ||r'(t)||^2 · (T(t) x T'(t))

[issisoccer10]

Homework Statement

We know that T(t) = r'(t)/||r'(t)||, or equivalently, r'(t) = ||r'(t)||·T(t). Differentiate this equation to find r''(t), then show that

r'(t) x r''(t) = ||r'(t)||^2 · (T(t) x T'(t)).

Homework Equations

r''(t) = ||r'(t)||'·T(t) + T'(t)·||r'(t)||

The Attempt at a Solution

I was able to get as far the differentiation above but from there I got stuck. I attempted to cross both sides, which I'm pretty sure needs to be done, but I'm confused how to get the right side of the equation to simplify to the desired form. Any help would be greatly appreciated...Thanks a lot

 

[Defennder]

Actually I don't see any puzzle here. Are you sure this is what the question is asking for? All you have to do is to differentiate r'(t)=|r'(t)|T(t) to get r''(t)=|r'(t)|T'(t). Then simply do the cross product for r'(t) and r''(t) from the above 2 expressions. You should get the answer as shown.

 

[ozymandias]

Use the fact that the cross product of a vector with itself is zero.

Assaf
"www.physicallyincorrect.com"[/URL]

 

[HallsofIvy]

Actually I don't see any puzzle here. Are you sure this is what the question is asking for? All you have to do is to differentiate r'(t)=|r'(t)|T(t) to get r''(t)=|r'(t)|T'(t). Then simply do the cross product for r'(t) and r''(t) from the above 2 expressions. You should get the answer as shown.

No. r''(t) is not |r'(t)|T'(t). Since |r'(t)| is a function of t itself you have to use the product rule.

 

[Defennder]

Oh, yeah. I kept thinking |r'(t)| could be treated as a constant.