- #1

- 3,682

- 1,556

How do we prove by Calculus of Variations that the minimum time for boat crossing a river (perpendicular to the current for starters) with current ##v_r##, and boat velocity in still water ##v_b## that the path will be a straight line?

I tried, but I got lost.

Let ##x## be the distance from launch point A to point B (directly opposite A) on the other bank.

## \rightarrow^{x^+} \downarrow^{y^+}##

I get the following equations:

$$ \frac{dx}{dt} = v_b \cos \theta \tag{1}$$

$$ \frac{dy}{dt} = v_b \sin \theta - v_r \tag{2}$$

I believe we are supposed to minimize the following:

$$ T = \frac{1}{v_b} \int \frac{dx}{ \cos \theta} $$

I notice that using the Chain Rule on ##(2)##:

$$ \frac{dy}{dx} = \tan \theta - \frac{v_r}{v_b} \sec \theta $$

Thats implies that:

$$ 0 = \int dy = \int \left( \tan \theta - \frac{v_r}{v_b} \sec \theta \right) dx $$

Is it then not only the case that the integral on the RHS is zero if:

$$ \tan \theta - \frac{v_r}{v_b} \sec \theta = 0 \implies \sin \theta = \frac{v_r}{v_b} $$

Which means that ## \theta ## is a constant... i.e. the path is a straight line. I realize this doesn't show the time is minimized though.

What is the proper way to go about proving ##(1)## is minimized using the Calculus of Variations?