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I was corrected that the following calculation is not correct, maybe you can help me finding the mistake:

The mathematical (well known) task was:

A person on a lake (distance to coast: d) wants to reach a point P on the coast (the coast has a straight line!). The boat can move with speed $v_1$. By feet the person travels with speed $v_2$. Where is point Q on the coastline, so that the time needed to get to P is minimized?

Regarding $v_1 < v_2$ (other case is easy):

Given the equation

$t(x) = \frac{\sqrt{d^2+x^2}}{v_1} + \frac{(y-x)}{v_2}$, where $y$ is the distance between P and the point, where the person would reach the coastline if moving orthogonal to the coastline.

I got $t'(x) = \frac{x}{v_1 \cdot \sqrt{^2+x^2}} - \frac{1}{v_2}$.

And for $t'(x_E)=0$ I got $x_E = \frac{(dv_1)^2)}{\sqrt{v_2^2 - v_1^2}}$.

Now I have just put this point $x_E$ into the second derivative:

$t''(x_E) = -\frac{d^2v_1}{(v_2^2-v_1^2)\cdot (d^2 + \frac{d^2v_1^2}{v_2^2-v_1^2})^{3/2}} + \frac{1}{v_1\cdot \sqrt{d^2 + \frac{d^2v_1^2}{v_2^2-v_1^2}}}$

And setting $t''(x_E) > 0$ I end with

$ 1 > \frac{v_1^2}{v_1^2+(v_2^2-v_1^2)} = \frac{v_1^2}{v_2^2}$.

So that should be it. But as I said above, I must have made a (huge) mistake? Maybe I am not allowed to "put in $x_E$ simply like that", and if so, I do not immediately see why.

Vazrael
  • 2,281

2 Answers2

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Yes, you can just plug $x_E$ into the second derivative, $t''(x_E).$ After that step, however, you have explained things backwards: you want to show that $t''(x_E) > 0,$ not "set" $t''(x_E) > 0.$ Whatever steps you took to get from $t''(x_E) > 0$ to $v_1^2/v_2^2 < 1,$ hopefully they are valid in the opposite direction too so that you can follow the steps from the given $v_1 < v_2$ all the way to $t''(x_E) > 0.$

Having done that, however, you still will have shown only that $x_E$ is a local minimum, not that it is the optimal answer for all possible values of $x.$ To show a global minimum I think it is enough to show that $t'(x)$ is defined everywhere and that it is negative whenever $x < x_E$ and positive whenever $x > x_E.$ Then the mean value theorem will show that $t(x) > t(x_E)$ whenever $x < x_E,$ and a second application of the theorem will show that $t(x) > t(x_E)$ whenever $x > x_E.$

David K
  • 98,388
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You introduced too many quantities. Denote by $p>0$ the distance from the nearest shore point $O$ to $P$, and put ${v_1\over v_2}=:\lambda<1$. Then we have to minimize the objective function $$T(x):={1\over v_1}\left(\sqrt{d^2+x^2}+\lambda(p-x)\right)\qquad(0\leq x\leq p)\ .$$ To this end we set up a candidate list consisting of $0$, $p$, and the zeros of $T'$ in $\>]0,p[\>$. There seems to be just one zero $x_*$ of $T'$ in the interval $\>]0,p[\>$. It follows that $$\min_{0\leq x\leq p}T(x)=\min\{T(0),T(p),T(x_*)\}\ .$$ There is no need to compute second derivatives for this problem.

Christian Blatter
  • 226,825