What point should the man row to reach his destination in the least amount of time?
A man is floating in a rowboat 1 mile from the (straight) shoreline of a large lake. A town is located on the shoreline 1 mile from the point on the shoreline closest to the man. He intends to row in a straight line to some point P (on the shore) and then walk the remaining distance to the town. To what point should he row in order to reach his destination in the least time if:
a) he can walk 5 mi/h and row 3 mi/h
b) he can walk 5 mi/h and row 4 mi/h
- M3Lv 71 decade agoFavorite Answer
.. |\ angle ACP = z.
..A. x P ...B
...<---- l --->
let speed ratio land:water = k
dist. over water = w•sec z,
dist. over land = l - w•tan z
time taken ∞ k•sec z +(l - tan z)
dT/dz = k•secz•tan z - sec^2 z
= (ksin z - 1) / cos^2 z
for minima, sin z = 1/k
which yields x = w /√(k^2 - 1)
all we have to now do is to plug in
(a) x = 1/√((5/3)^2 - 1) = 0.75 mi <-------
(b) x = 1/√((5/4)^2 - 1) = 1 1/3 mi which means, row directly !
note that the optimal value of angle z is solely dependent on k !
- 1 decade ago
The picture to draw is a right triangle with a little line segment pointing above it. The length of the hypotenuse of the right triangle is the distance he will row, and the length of the line segment is the distance he will walk. Let's let x be the distance from the point on the shore closest to his boat and P. Then the distance he will walk is 1 - x. The Pythagorean theorem tells us that the distance he will row is sqrt (1 + x^2).
Divide by his speeds to get the time it takes to row or walk and add up to get the total time. For a), we have
f(x) = 1/3 * sqrt(1 + x^2) + 1/5*(1 - x)
To minimize, we take a derivative:
f'(x) = 1/6 * (1 + x^2)^(-1/2) * 2x - 1/5
Set equal to 0:
0 = x / (3 * sqrt(1+x^2)) - 1/5
x / (3 * sqrt(1+x^2)) = 1/5
5x = 3 * sqrt( 1 + x^2 )
25x^2 = 9 ( 1 + x^2)
25x^2 = 9 + 9x^2
16x^2 = 9
x^2 = 9/16
x = +/- 3/4
Since x is a distance, throw out the negative value. This should give the minimum (you can do the 2nd derivative test to check for sure).
b) If you do the same thing here, you get x = 4/3. But this would overshoot the town! So in this, case, it would be fastest to just row straight to it.
- gooberLv 71 decade ago
let r be the distance to row and p be the distance from the point on the shoreline closest to the man to P.
r = sqrt(p^2 + 1)
the time to travel is
t = r/3 + (1-p)/5
t = (p^2 +1)^(1/2) / 3 + (1-p) / 5
dt/dp = (1/3) (1/2)(p^2 +1)^(-1/2) 2 p - 1/5 = 0
0 = (1/3) (p^2+1)^(-1/2) p - 1/5
(1/3) (p^2+1)^(-1/2) p = 1/5
5p = 3sqrt(p^2+1)
25p^2 = 9(p^2 +1)
16p^2 = 9
p = 3/4
for the case where rowing = 4 mi/h
25p^2 = 16(p^2+1)
9 p^2 = 16
p = 4/3 which is a point beyond the town.
so it would be faster to row directly to the town.