2

I am self-studying multivariable calculus using MIT's publicly available materials, and I have been stumped by this exercise from Chapter 14.4 of the first volume of Apostol's calculus text:

A vector-valued function $F$, which is never zero and has a continuous derivative $F'(t)$ for all $t$, is always parallel to its derivative. Prove that there is a constant vector $A$ and a positive real-valued function $u$ such that $F(t) = u(t)A$ for all $t$.

This is what I have so far: By hypothesis, we have \begin{align} F(t) = s(t)F'(t) \end{align} for all $t$, where $s(t)$ is a real-valued function. Since $F(t) \neq 0$, we know that $s(t) \neq 0$, $F'(t) \neq 0$. Moreover, we know that since $F$ is differentiable, $s(t)$ and $F'(t)$ are both differentiable, and that therefore \begin{align} F'(t) & = s(t)F''(t) + s'(t)F'(t) \end{align}

This, unfortunately, is where I run out of steam. I would very much appreciate a gentle hint to get me going -- not, if it can be avoided, a complete solution. I suspect that I'm missing something obvious...

Thanks in advance.

Muphrid
  • 19,902
solitaireartist
  • 980
  • 2
    An easy way is to solve for the individual components of $F(t)$. Find $x(t)$ s.t. $dx/dt=v(t)x$ - what do you get? – lemon Oct 07 '14 at 02:13
  • I'm sure I'm being dense, here, but I'm not sure how this helps. Let $x(t)$ be an arbitrary component of $F(t)$, as you suggest. Then we have $x(t) = s(t)x\prime(t)$, don't we? That is, I don't see how you arrived at $x\prime(t) = x\prime(t)x$. Now, it's true that we can make assumptions about $s(t)$ and come up with some $x(t)$ that work -- for example, if $s(t) = c \in \mathbb{R}$, $c \neq 0$, we have $x(t) = e^{\frac{t}{c}}$. But what if $s(t) = x^2 + 1$? Or $s(t) = \sin t + 2$? Or, really, any real function guaranteed to be non-zero, which seems to me to be all we know about $s(t)$? – solitaireartist Oct 07 '14 at 12:30
  • 1
    $F$ and $F'$ are parallel so you can write $F(t)=s(t)F'(t)$ or, equally, you can write $F'(t)=v(t)F(t)$ where $v(t)=1/s(t)$. Either way, it's the same outcome. When you solve $x'=vx$ you get $x(t)=A_x\exp\int v(t) dt$ for a constant $A_x$. Does that help? – lemon Oct 07 '14 at 13:01
  • Indeed it does. Thank you. – solitaireartist Oct 07 '14 at 13:03

1 Answers1

0

With great thanks to lemon above, here's the solution:

Since $F$ and $F\prime$ are parallel and $F$ is always non-zero, we can write $F\prime(t) = s(t)F(t)$ where $s(t)$ is a continuous (and therefore integrable) real-valued function such that $s(t) \neq 0$ for all $t$. Then if $f_i(t)$ is an arbitrary component of $F(t)$, we have \begin{align} \frac{df_i}{dt} & = sf_i\Rightarrow\\ \frac{df_i}{f_i} & = s(t)dt\Rightarrow\\ \ln(f_i) & = \int s(t)dt + C\Rightarrow\\ f_i & = e^Ce^{\int s(t)dt} \end{align} where $C \in \mathbb{R}$. Let $A_i = e^C$, $u(t) = e^{\int s(t)dt}$; then $f_i(t) = A_iu(t)$. So we have \begin{align} F(t) & = (A_1u(t), \ldots, A_nu(t))\\ & = u(t)(A_1, \ldots, A_n) \end{align} Observing that $u(t)$ is indeed positive for all $t$, and letting $A = (A_1, \ldots, A_n)$, we have the result.

solitaireartist
  • 980