Differentiation of vector function

[boombaby]

Homework Statement

suppose f is a differentiable mapping of R¹ into R³ such that |f(t)|=1 for every t. Prove that $$f'(t)\cdot f(t)=0$$.
I guess it is more proper to write $$(\nabla f)(t) \cdot f(t)=0$$, where $$(\nabla f)(t)$$ is the gradient of f ant t.

Homework Equations

The Attempt at a Solution

it is then equivalent to prove $$\sum\;(Df_{i})(t)\cdot f_{i}(t)=0$$, but I've no idea of how to use the |f(t)|=1 to deduce the disired results, although the equation has an easy geometrically interpretation
Any hint may help, thanks a lot.

 

[Dick]

The f'(t) notation is fine. f(t) is a vector. You generally use the gradient notation for scalar functions, not vectors. And |f(t)|=1 is the same as saying f(t).f(t)=1, right?

 

[boombaby]

just differentiate both side and it is done...Thanks!
and thanks for the explanation about gradient...I misunderstood it...
Here is why I thought it improper..f'(t) is a linear transformation from R^1->R^3, whose basis are different things from the basis of R^3, so I thought it is weird to apply the dot product to f'(t) and f(t)...

 

[Dick]

For functions of a single variable it's simpler just to think of f'(t) as an element of R^3 rather than as a linear transformation. If f(t)=(x(t),y(t),z(t)), f'(t)=(x'(t),y'(t),z'(t)).

 

[boombaby]

it took me years to open and refresh this forums...my connecting to some websites sucks...
Thinking of it as a vector in R3 is okay and easy. okay, I found it more sensible in this way.
take 1 as a basis of R^1. So f'(t)(1) = $$\sum^{3}_{1} (Df_{i})(t)e_{i}$$, where e_i is the basis of R^3, and dot product makes sense

 

[Dick]

Sure, the linear transformation is f'(t)(h)=(x'(t),y'(t),z'(t))*h. That can be identified with a vector in R^3 in a pretty obvious way. I would say you are just throwing the 'h' away altogether.

 

[HallsofIvy]

Saying |f(t)|= 1 is the same as saying $f(t)\cdot f(t)= 1$. Use the product rule.

 

[boombaby]

it's been solved, Thanks anyway:)