Before answering this I think we need a bit more information aboutOriginally Posted by TopGur
what techniques you are expected to know/use.
In particular have you been taught the method of Lagrange Multipliers?
I'm trying to check the answer for these 2 questions
1. You are in a rowboat two miles from the nearest point on a straight shoreline. You noticed smoke billowing from your house, which is 6miles along the shoreline from the nearest point. You can row your boat 6 miles per hour and you can run then miles per hour. How should you proceed in order to get to your house in the least time?
2. What if the house is ten miles away from the nearest point and I can use a motorboat that travels 8mph or run at 5mph?
I need the function, derivative, and critical points for it.
"...and you can run then miles per hour."Originally Posted by TopGur
I assume the "then" is a typo, and it was intended for "ten", so your speed on land is 10mph.
The question asks for least time, so we need to equate the first derivative of time T to zero.
distance = speed * time ---------*****
So, time = distance/speed -----(i)
Since you run faster than rowing, you need to land some distance from or before your house.
Assuming you have to row a straight line, (a curved line will complicate things---besides, a straight line is the shortest distance between two points; shostest distance-->shortest time to spend traveling on same speed), your total path to reach your house the least of time is an oblique line relative to the shoreline while rowing, and a line parallel to the shoreline while running.
Let point A be your original position, point B be the point on the shoreline nearest A, point C be your landing point on the shoreline, and point D be the poisition of your house on the shoreline.
----AB = 2 mi.
----AB is perpendicular to BD. -----"nearest point...." signifies the 2mi distance is perpendicular to the shoreline.
----BD = 6 mi.
----speed by rowing on water = 6mph.
----speed by running on shoreline = 10mph.
Let also x = BC
so, CD = (6-x)
all in miles.
distance to row = AC
By Pythagorean Theorem,
(AC)^2 = 2^2 +x^2
AC = sqrt(x^2 +4) ---**
time to spend = distance/rate,
t1 = sqrt(x^2 +4) / 6
t1 = (1/6)sqrt(x^2 +4) ------***
distance to run = CD = (6-x)
time to spend = distance/rate,
t2 = (6-x)/10
t2 = (1/10)(6-x) ------***
Total time to spend to reach house, T = t1 +t2
T = (1/6)sqrt(x^2 +4) +(1/10)(6-x) ---------------the time function, answer.
The first derivative of T with respect to x,
dT/dx = (1/6)[(1/2)(x^2 +4)^(-1/2) *2x] +(1/10)[-1]
dT/dx = (1/6)[x /sqrt(x^2 +4)] -1/10 ---------------------answer.
Set dT/dx to zero to find the critical points,
0 = (1/6)[x /sqrt(x^2 +4)] -1/10
Clear the fractions, multiply both sides by 6*10*sqrt(x^2 +4),
0 = 10*x -6sqrt(x^2 +4)
0 = 5x -3sqrt(x^2 +4)
3sqrt(x^2 +4) = 5x
Clear the radical, square both sides,
9(x^2 +4) = 25x^2
9x^2 +36 = 25x^2
36 = 25x^2 -9x^2
36 = 16x^2
Take the square roots of both sides,
+,-6 = 4x
x = +,-6/4 = +,-1.5 miles.
The negative 1.5mi means you land on the side of point B going away from D, and since you are not crazy to do that for you are supposed to be in a hurry to reach D, then,
x = 1.5 miles from B, or (6-1.5) = 4.5 miles from D.
That means, you should land at a point on the shoreline that is 4.5 miles from your house and run that 4.5 miles to reach your house the shortest time. -------answer.
"2. What if the house is ten miles away from the nearest point and I can use a motorboat that travels 8mph or run at 5mph?"
Hey, now it is you on the boat?
Okay, now my "you" in my answers is "politically" correct.
Without detailing all those again, if you follow the answer above,
T = (1/8)sqrt(x^2 +4) +(1/5)(6-x)
dT/dx = (1/8)[x /sqrt(x^2 +4)] -1/5
0 = (1/8)[x/sqrt(x^2 +4)] -1/5
0 = 5x -8sqrt(x^2 +4)
64(x^2 +4) = 25x^2
64(4) = (25 -64)x^2
256 = -39x^2
x^2 = -(256/39) = negative
That means only that you do not have to land before you reach your house and run. It means just boat straight to your house ---because your boat's speed is a lot faster than your running in this version of the problem.
motorboat distance = sqrt(2^2 +6^2) = 6.3245 mi
T = (6.3245)/8 = 0.7906 hr.
If, say, land first at a point 0.2 mi before your house,
motorboat distance = sqrt(2^2 +5.8^2) = 6.135 mi
T = 6.135/8 +0.2/5 = 0.806 hr
See, the 0.806 hr is longer than the 0.7906 hr.
Therefore, here, speed with your boat straight to your house for the least time. ---answer.
Since ticbol has given a detailed answer I will not work this problem, but
will observe that implicit in ticbols solution is that the two segments of
the path followed satisfy Snell's law, and once you have show that this is
the case solving such problems becomes a doddle.