prove that $f(t)$ is orthogonal to $f'(t)$ for som all $t \in I$

Question

Suppose that $I$ is nonempty open interval and that $f: I \to \Bbb R^m$ is differentiable on $I$

If $f(I) \subseteq \partial B_r(0) $ for some fixed $r>0$, prove that $f(t)$ is orthogonal to $f'(t)$ for som all $t \in I$

I guess I need to use the chain rule. Sorry, I cannot say any idea about the question.

But what is orthogonal?

How can I link between orhogonal and the chain rule of partial derivatives ?

I am glad to show me the proof step by step. Thanks for helping:)

Answer 1

Edit I am just adding some details to the approach suggested by @David Mitra in the above comments.

Hint: by definition,

$$f(t)=(x(t),y(t))$$

with $x^2(t)+y^2(t)=r^2$ for all $t\in I$. The trick, as pointed out by @David Mitra is to note that

$$\|f(t)\|^2=x^2(t)+y^2(t)=r^2,$$

i.e. $\|f(t)\|$ is constant for all $t\in I$. Try now to compute the derivative

$$\frac{d}{dt}\|f(t)\|^2 $$

using the definition of $\|f(t)\|^2$. You should arrive at

$$\frac{d}{dt}\|f(t)\|^2=2\langle f(t),f'(t)\rangle,$$

where $f'(t)=(x'(t),y'(t))$. This is sufficient to finish the problem (why?).