A swimmer is @ a point 500m from the closest point on a straight shorline. She needs to reach a cottage?

Question

located 1800 m down shore from the closest point. If she swims @ 4m/s and she walks 6 m/s, how far fomr the cottage should she come ashore so as to arrive at the cottage in the shortest time?

Answer 1

Let

w = distance from cottage to come ashore and walk
s = distance swam
t = time

Given

swim rate = 4
walk rate = 6

w = 1800 - x
s = √(x² + 500²)

t = s/4 + w/6 = [√(x² + 500²)]/4 + (1800 - x)/6

dt/dx = x/[4√(x² + 500²)] - 1/6 = 0
x/[4√(x² + 500²)] = 1/6
6x = 4√(x² + 500²)
3x = 2√(x² + 500²)
9x² = 4(x² + 500²) = 4x² + 1000²
5x² = 1000²
x² = 1000²/5
x = 1000/√5 = 1000(√5)/5 = 200√5

w = 1800 - x = 1800 - 200√5 = 1352.7864 m

Answer 2

Steve was correct, so hats off to him. However, he gave no method.

She should swim to 447.2136...m down shore, or 1352.7864...m from the cottage. (447.2136 = 1000/sqrt(5).) The distance she swims is 670.82 metres (see below), which takes her I67.70... secs, and she walks for 225.46... secs, for a total of 393.17... secs or 6 mins 33.2 secs. I worked the problem this way, after correcting an algebraic error in my original attempt:

It's easiest to use the "downstream distance" of the arrival point as a variable. That makes the various expressions simplest to handle. Let the distance she swims be "D" metres; then, by Pythagoras,

D^2 = (500)^2 + z^2.

The remaining distance to walk is (1800 - z). She swims at 4m/s, and walks at 6m/s. Therefore, the total time she takes is:

t = {[(500^2) + z^2]^1/2} / 4 + (1800 - z) / 6.

To find the stationary value of t (presumably the minimum value) as z is varied, differentiate t wrt z:

dt/dz = z / {4 [(500^2) + z^2]^1/2} - 1/6.

This is zero when z / {4 [(500^2) + z^2]^1/2} = 1/6, i.e. (simplifying slightly, squaring and re-ordering) when:

9 z^2 = 4 [(500^2) + z^2] = (1000)^2 + 4 z^2, or:

5 z^2 = 1000^2, or sqrt(5) z = 1000.

So z = 447.2135...m (!)

z is the downstream distance at the end of her swim; at that point she is 1800 - z metres or 1352.7864... metres from the cottage. With this z, D works out as 670.82 metres

You could differentiate again to show that this extremum is a MINIMUM, but it's obvious when you think about it; if she swam several kilometres away from the cottage's direction, or several kilometres further downstream, and then walked to the cottage, she'd take a hell of a time!

It may also be interesting to note that the mathematical problem here is the same as trying to find the critical angle for a light ray to actually emerge from a medium (water again?) into a vacuum --- or air, to all intents and purposes --- where the speed of light in the medium is only 4/6 c or 2/3 c. (Refractive index = 3/2 or 1.5) (All this occurs because light too follows a "stationary-time-path" between any two points. This is known as Fermat's Principle.) The critical angle made with the normal to the shoreline/interface is tan^(-1) [2/sqrt(5)] = 41.81... degs. At this angle, the emergent ray just slides along the shoreline/interface. For steeper approaches, you have complete internal reflection.

(It's amusing to swim to the bottom of a swimming pool, look up, and see the relatively narrowly confined if wave-disturbed cone into which the distorted view of the upper world is squeezed. That's the fish-eye view, with squashed-looking anglers near the outer edge of that dancing cone. Beyond that, the surface behaves much as a shifting and dancing silvered mirror. I find it quite mesmerizing.)

Live long and prosper.

Answer 3

The swimmer has got to get to shore first, so let her swim
to a point that is x metres from the shore's closest point
towards the cottage. This makes a right-angled triangle in
which she swims the hypotenuse, the distance being equal to
sqrt(x^2 + 500^2) metres.

At 4m/s, the time she takes to swim this is :
T1 = sqrt(x^2 + 500^2) / 4 seconds

Her walking distance from this point is now (1800 - x) metres.
At 6 m/s, the time taken to walk this is :
T2 = (1800 - x) / 6 seconds

Adding these gives the total time as :
T = sqrt(x^2 + 500^2) / 4 + (1800 - x) / 6 seconds

Take the derivative and set it equal to zero to find the
shortest time required.

dT/dx = (1/4)(1/2)(x^2 + 500^2)^(-1/2) * 2x - 1/6 = 0

Therefore, x / sqrt(x^2 + 500^2) = 2/3

Squaring both sides gives :
x^2 / (x^2 + 500^2) = 4/9

Cross-multiplying gives :
9x^2 = 4x^2 + 4 * 500^2
5x^2 = 4 * 500^2
x^2 = 4 * 500^2 / 5
x = 2 * 500 / sqrt(5)
= 200 * sqrt(5), or approximately 447.2 metres.

Thus, 1800 - x = 1800 - 200 * sqrt(5) = 200[9 - sqrt(5)],
which is approximately 1352.8 metres.

Therefore, she should come ashore 1352.8 metres from
the cottage to arrive in the shortest time, which comes to
(25/3)[36 + 5 * sqrt(5)] seconds,
or approximately 6 minutes, 33 seconds.

Answer 4

If the simmer comes straight to shore (500m) and then walks the 1800m to the cottage, it will take 125 seconds to swim and 300 seconds of walking for a total of 425.

Swimming at an angle the entire distance will shave considerable distance, but actually takes longer (468.75 seconds) because the distance of 1875 meters is traversed at 4m per second.

Therefore, the fastest method is a combination of the two, and is the one which yields the most beneficial point of arrival on the shore where the extra 2m per second on foot can overtake the drag of swimming on the lake.

This is calculated as a 45 degree angle from where the swimmer begins, to the shoreline. Therefore this yields a swim time of 156 seconds (625 meters) and a walk time of 229 seconds (1375 meters) for a total of 385 seconds.

Answer 5

This problem needs calculus to optimize her time. Put your origin at the point on the shore closest to the swimmer. Call the point x the distance from the origin where she lands. Call the distance from her to shore at the start (500m) D and the distance to the cottage from the origin (1800 m) C. Call her swim speed S and her walk speed W.

By Pythagorus, the distance to point x from the start is sqrt(D^2 + x^2). Since she is swimming, the time to get to x is the distance divided by swim speed: sqrt(D^2 + x^2)/S

The distance from x to the cottage is C-x, the time to walk from x to the cottage is (C-x)/W

So the total time is the sum of the two times:

t = sqrt(D^2 + x^2)/S + (C-x)/W

To optimize, take the derivative with respect to the variable (x) and set it to zero:

0 = (x/sqrt(D^2 + x^2))/S - 1/W

Rearrange to get an expression in x:

x/sqrt(D^2 + x^2) = S/W

x^2 = (S/W)^2 * (D^2 + x^2)

x^2 = ((SD/W)^2)/(1 - (S/W)^2)

x = (SD/W)/sqrt(1 - (S/W)^2)

x = D/sqrt((W/S)^2 - 1)

Putting in the numbers:x = 447.2 m

This is the distance from the closest point on shore so the distance from the cottage is C-x = 1353 m

Answer 6

she should swim as far as she needs to but if shecan walk then she needs to that instead of swimming because she walks quicker than she swims

Answer 7

1352.8 m

Answer 8

first thing is to tell her to get a boat!!!!!NOW!!!!!! NEXT QUESTION???????