Shortest distance along the shore and into the lake

[brotherbobby]

Homework Statement

A man is on the shore of a lake at point A and has to get to point B on the lake in the shortest possible time (see Fig below). The distance from point B to the shore is ##BC = d## and the distance ##AC=s##. The man can swim in water with a speed of ##v_1## and run along the shore with a speed of ##v_2##, greater than ##v_1##. Which way should he use - swim from point A straight to B or run a certain distance along the shore and then swim to point B?

Relevant Equations

(1) Pythagorean theorem for a right angled triangle with hypotenuse length ##c## and sides ##a,b## : ##c^2=a^2+b^2##
(2) For a function ##f(x)##, minimum value of ##f## can be found by solving ##\dfrac{df}{dx}=0##. At the minimum, the calculated value of ##x##, say some ##x_0##, should have ##\left.\dfrac{d^2f}{dx^2}\right|_{x=x_0}\!\!\!\!\!\!\!\!\!>0##

Attempt : I copy and paste the problem as it appeared in the text below.

To be sure, of course the man cannot swim directly to B for minimum time of travel. He should run a distance (##x##) to some point D along the shore and then swim to B. I draw those details in the diagram to the right.
The time of travel ##\small{t=\dfrac{x}{v_2}+\dfrac{\sqrt{d^2+(s-x)^2}}{v_1}}##.
For minimum time, ##\small{\dfrac{dt}{dx}=\dfrac{1}{v_2}-\dfrac{s-x}{v_1\sqrt{d^2+(s-x)^2}}=0}##, which, after some elementary algebra, comes out to be ##\boxed{\boldsymbol{x=s-\dfrac{v_1d}{\sqrt{v_2^2-v_1^2}}}}\quad{\color{green}{\Large\checkmark}}\qquad(1)##.

I checked for myself that for this value of ##x##, ##\dfrac{dt^2}{dx^2}>0##

I put the checkmark (##\checkmark##) because the answer checks out with that of the text. Of course the authors have done it without using calculus and that is how one should do it first. I will attempt it presently.

However, if I took values of the different speeds and distances and found out ##x## from equation ##(1)## above, I get a negative answer.

Let the distances and speeds be the following : ##s=2\,\text{km}, \, d=6\,\text{km},\, v_1=4\,\text{km/h},\, v_2=8\,\text{km/h}##.

Then from ##(1)##, ##\small{x=2-\dfrac{4\times 6}{\sqrt{8^2-4^2}}=2-2\sqrt{3}=-\text{ve}\;!}##

This would mean the man should run a distance to the left from A along the shore and then swim to B which cannot be least time. Running the same distance right would lead to a shorter time owing to less distance of swimming.

Request - Where am I going wrong with the numbers?

 

[haruspex]

The thing to check is whether your first equation represents reality in all cases.
For x<0 it doesn't: it gives a negative time for running on the shore.
Consequently, if there is no local minimum with x>0 you get a crazy answer.

 

[brotherbobby]

The thing to check is whether your first equation represents reality in all cases.
For x<0 it doesn't: it gives a negative time for running on the shore.
Consequently, if there is no local minimum with x>0 you get a crazy answer.

Yes, I realise it now. The answer ##\boxed{x=s-\dfrac{v_1d}{\sqrt{v_2^2-v_1^2}}}## makes sense for as long as ##x>0##. Put differently, ##x=s-\dfrac{v_1d}{\sqrt{v_2^2-v_1^2}}\Rightarrow \dfrac{v_1d}{\sqrt{v_2^2-v_1^2}}=s-x\ge 0##. If however those v's and d's combine to give an s less than x (but s>0), the man should swim directly to A without running along the shore.
As it happened, for the values I took, s turns out to be less than x.

 

[kuruman]

I think your problem was that you measured the "jump in" distance from the starting point. If ##x_{opt}## is the optimum distance from C, the time will be a minimum no matter how far from C the man starts running as long as the starting point is at ##s>x_{opt}##. So it makes more sense to optimize ##x## as measured from point C.

Also note that if I define ##~~\sin\theta\equiv \dfrac{x}{\sqrt{d^2+x^2}},~~## where ##\theta## is the angle opposite to right side CD in triangle BCD (see diagram on the right), your equation $$\frac{dt}{dx}=\frac{1}{v_2}-\frac{s-x}{v_1\sqrt{d^2+(s-x)^2}}=0$$becomes $$\frac{1}{v_2}-\dfrac{\sin\theta}{v_1}=0\implies \sin\theta=\frac{v_1}{v_2}.$$ Thus, from the right triangle $$\frac{\sin\theta}{\sqrt{1-\sin^2\theta}}=\tan\theta =\frac{x_{opt}}{d}\implies x_{opt}=\frac{v_1}{\sqrt{v_2^2-v_1^2}}d.$$You might also recognize the path followed by the man as the path of light incident at grazing incidence on a dielectric interface from an optically rarer to an optically denser medium. According to Fermat's principle, when light travels from point A to point B it does so in the least possible time. Angle ##\theta## in this case is the critical angle for total internal reflection.

On edit
I corrected the defining equation of ##\sin\theta## to make it consistent with the text, where it is stated that ##\theta## is the angle opposite to right side CD in triangle BCD and that ##x## is measured from point C. I also added a diagram to make the situation clearer. My thanks to @haruspex for pointing out the inconsistency in post #5. The corrected version does not lead to a solution to the left of A. With the definition of ##x## as the distance from C point A becomes irrelevant. I apologize for the confusion.

 

[haruspex]

If ##x_{opt}## is the optimum distance from C, the time will be a minimum no matter how far from C the man starts running as long as the starting point is at ##s>x_{opt}##. So it makes more sense to optimize ##x## as measured from point C.

Also note that if I define ##~~\sin\theta\equiv \dfrac{s-x}{\sqrt{d^2+(s-x)^2}},~~## where ##\theta## is the angle opposite to right side CD in triangle BCD, your equation $$\frac{dt}{dx}=\frac{1}{v_2}-\frac{s-x}{v_1\sqrt{d^2+(s-x)^2}}=0$$becomes $$\frac{1}{v_2}-\dfrac{\sin\theta}{v_1}=0\implies \sin\theta=\frac{v_1}{v_2}.$$ Thus, from the right triangle $$\frac{\sin\theta}{\sqrt{1-\sin^2\theta}}=\tan\theta =\frac{x_{opt}}{d}\implies x_{opt}=\frac{v_1}{\sqrt{v_2^2-v_1^2}}d.$$

Neat, but it would still have led to an answer that the place to jump in is to the left of A. I would guess @brotherbobby would have been just as puzzled.
Either way, the resolution of the puzzlement is that the time of travel is ##\small{t=\dfrac{|x|}{v_2}+\dfrac{\sqrt{d^2+(s-x)^2}}{v_1}}##.

 

[Hill]

Either way, the resolution of the puzzlement is that the time of travel is ##\small{t=\dfrac{|x|}{v_2}+\dfrac{\sqrt{d^2+(s-x)^2}}{v_1}}##.

Also, ##x## should be redefined: instead of

a distance (x)

(which is positive by definition) it is a coordinate of ##D## with the origin in ##A##.

 

[kuruman]

Neat, but it would still have led to an answer that the place to jump in is to the left of A. I would guess @brotherbobby would have been just as puzzled.
Either way, the resolution of the puzzlement is that the time of travel is ##\small{t=\dfrac{|x|}{v_2}+\dfrac{\sqrt{d^2+(s-x)^2}}{v_1}}##.

I messed up the definition of ##\sin\theta## because I copied and pasted from @brotherbobby's expression without changing ##s-x## to ##x## in the numerator. The original equation is inconsistent with the text. See edit note and added figure in post #4. Thanks for the catch.