# Linear map derivative proof

Let $f:\mathbb{R} \rightarrow \mathbb{R^2}$ be a differentiable function such that $|f(t)|$ = 1 for all $t\in \mathbb{R}$. Show that $$f'(t) * f(t) = 0$$for all $t\in \mathbb{R}.$

It's my understanding that $|f(t)|$ is of the form $\sqrt{(f_1(t)^2) + (f_2(t)^2)}$ (not sure if this is relevant, or needed). I also believe $f'(t)$ is of the form $(f_1'(t), f_2'(t))$. So, since $|f(t)| = 1$ for all $t$, surely $f_1'(t) = f_2'(t) = 0$. Is it this simple, or have I made some mistake?

• Could you be a bit more precise on the "surely"? If you want a hint to solve the problem, then: $1=|f(t)|^2=f(t)\cdot f(t)$. What do you get when you differentiate? – mickep May 7 '17 at 19:29
• $f(t)$ could be $(\sin(t), \cos(t))$ – WW1 May 7 '17 at 19:30
• @mickep I suppose the "surely" means my proof is a little too hand-wavey to be technical - if $|f(t)|$ is constant, then $f'(t)$ must be zero. Well, differentiating that gives $2 * f(t) * f'(t)$. Since the derivative of $1$ is $0$, does that complete the proof? I only ask this because linear maps confuse me a bit - sometimes I'm not sure how to treat derivatives when something like $f(t)$ has multiple components. – mizichael May 7 '17 at 19:32
• As @WW1 showed, $f'$ does not have to be zero. You say that linear maps confuse you. What is the linear map here? – mickep May 7 '17 at 19:34
• @mickep I guess I misspoke - the entirety of Chapter 9 in baby Rudin confuses me a bit. Any function or linear map with the range in multiple dimensions. – mizichael May 7 '17 at 19:35

Note that as you have written, the $|f(t)|=1\implies |f(t)|^2=1$ condition gives us $$f_1(t)^2+f_2(t)^2=1\implies2(f_1(t)f_1'(t)+f_2(t)f_2'(t))=0\implies f(t)\cdot f'(t)=0$$ as required.