Help! Hard math problem?

Question

Jane is 2mi offshore in a boat and wishes to reach a coastal village down a straight shoreline from the point nearest the boat. She can row 2mph and can walk 5mph. Where should she land her boat to reach the village in the least amount of time?

Answer 1

She lands 0.7332 miles from the point on the shore which is 2 miles from the boat(call this point A) Note: the point where she should land (call it B) does not depend on the distance to the village but only upon the relative speeds of walking and rowing.
Let x be the angle from the perpendicular she heads towards the shore and d be the distance in miles that the boat will travel at this angle, then 2/d = cosx and so d = 2/cosx miles.The amount of time taken to go this distance is d/2mph =1/cosx hrs.
She will land at point B that is y miles from point A and y/d = sinx so y = d*sinx = 2sinx/cosx miles.
If the distance to the village is D then she will have to walk D-2sins/cosx miles to the village. This will take her (D-2sinx/cosx)/5 hrs
The total time,t , taken is therefore

t = (1/cosx +(D-2sinx/cosx)/5) hrs

dt/dx = (secx*tanx) -(2/5)sec^2(x)

= secx(tanx-(2/5)secx)

Since secx is never 0, the derivative will be 0 when tanx = 2/5*secx, which is true when sinx = 2/5 and cosx=0.917

So y = 2sinx/cosx = 0.7332 miles from point on the shore closest to the boat.

Answer 2

An interesting aspect here is that you do not know how far down the shore the village is. Your solution should show that this doesn't matter because (within limits) it doesn't.

For better understanding, solve this algebraically and then put in the numbers. Call the distance to the shore S, the downshore distance to the village D, the landing point, measured form the nearest point on the shore X, the row speed R and the walking speed W.

The distance you row is: sqrt(S^2 + X^2)
The distance you walk is: D - X

The rowing time (RT) is the rowing distance divided by rowing speed: RT = sqrt(S^2 + X^2)/R

The walking time (WT), similarly is: WT = (D - X)/W

Adding the two times together to get total time T:
T = sqrt(S^2 + X^2)/R + (D - X)/W

Now to find the minimum time, take the derivative with respect to the landing point X:

dT/dX = X/(sqrt(S^2 + X^2) * R) - 1/W

To find the minimum, set this equal to zero:

0 = X/(sqrt(S^2 + X^2) * R) - 1/W

Notice that the distance down the shore (D) is not in this equation. As long as it is greater than X, the answer is the same.

Rearranging the above and simplifying step by step:

1/W = X/(sqrt(S^2 + X^2) * R)

sqrt(S^2 + X^2) = XW/R

Squaring:

S^2 + X^2 = (XW/R)^2

Solving for X^2:

X^2*((W/R)^2 - 1) = S^2

X = sqrt((S^2)/((W/R)^2 - 1))

You might notice that a real solution exists only for W > R, that is you walk faster than you row, which makes sense.

X = sqrt(4/(2.5^2 - 1) = 0.873 miles

Answer 3

I think we also need the distance along the shoreline to the village to answer this question.

I solved this problem setting the distance the town is down the coast as D. She can row her boat to any area along the shore, and that distance (lets call this R for rowed distance) is given by the length of a right triangle with one leg being her distance from the shore (2 miles), and the other being her distance travelled down the shoreling when she reaches it (I called this distance X). This gives me this equation R^2 = X^2 + 2^2

The total time it will take her to reach the village is given by the distance divided by the rate, which gives us this equation:

time travelling = [Square Root of (X^2 + 4) divided by 2 mph] + [(D - X)/5mph]

You just have to find the minimum point along this equation. You can do that by graphing it, or using calculus to find the minimum points. You can't actually solve it until you have a value of D though (that was the distance the village is along the coast).

Editted: lol after I did the calculus I figured out that the distance of the village doesn't matter, because the time function that I came up with is minized at zero, meaning that she should row straight for shore! oy, I suppose that should have been obvious from the start, but I prefer to proove it! :)

here's why in case you need to justify your answer: Using the eqation above, I took the derivate and set it equal to zero (this tells us the critical points of the curve, and possible where it is changing directions (the minimum or maximum, but you have to use the second derivative test to be sure).

dt/dx = x*(x^2+4)^(-1/2) - 1/5
solve for where this equals zero, and it's only when x equals zero, therefore she should not angle her boat at all, heading straight to shore is the fastest way. Uck, I hope I don't do this badly on my GREs next semester, I spent ten minutes on this problem when it should have taken 1. :(

Answer 4

You want to maximize the amount of time walking and minimize the amount of time rowing. We don't know how far away the village is , if it's very close to the straight line to the coast you could do a formula to figure out what the payoff is, rowing and walking, otherwise, get to shore fast and start walking.

Answer 5

She should row straight to shore. Cutting the angle in the slow boat doesn't help.