- #1

brotherbobby

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- 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.

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?

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