related rates question

Question

A man starts walking north at 3 ft/s from a point P. Five minutes later a woman starts walking south at 4 ft/s from a point 500 ft due east of P. At what rate are the people moving apart 15 min after the woman starts walking?

Answer 1

First order of business is to unify our units, notice your velocities are measured in ft/s but the time in question is 5 mins and 15 mins. a quick conversion will have us multiplying our mins by 60 (5)(60) and (15)(60) to give us 300 and 900 seconds respectively

t1 = 300 secs

t2 = 900 secs

Second is to place a cartesian coordinate system over our 4 cardinal directions. Let north and south be in the y direction with north positive and south negative. east and west will be the x axis with east positive and west negative.

this means that the man is moving in the positive y direction at a rate of 3ft/s and the woman is moving in the negative y direction at a rate of 4ft/sec. For the x direction, there is a distance of 500 ft in between them. this distance will always remain constant through out the problem. ultimately between the two of them they create a right triangle.

Right triangles are best described by the pythagreon theorem x(t)² + y(t)² = d(t)²

notice that x y and d(the distance between the man and woman) are functions of time.recall that even tho x is a function of t, that distance will remain constant throughout the problem, 500ft, that distance will never change because we have determined that the mean and woman move only in the y direction

now recall we are looking for the rate of the distance between the man and woman after 900 seconds(15mins). therefore we will take a derivative with respect to time of our Pythagreon function to find the rate of change of the total distance in between them

that is

d/dt[x(t)² + y(t)² ] = d/dt[d(t)²]

remember related rates problems always involve IMPLICIT DIFFERENTIATION

d/dt[x(t)²] = 2x(t)*x'(t) x'(t) is the rate or change in the x direction

d/dt[y(t)² ] = 2y(t)*y'(t) y'(t) is the rate or change in the y direction

d/dt[d(t)²] = 2d(t)*d'(t) d'(t) is the rate at which the people are moving apart

therefore

d/dt[x(t)² + y(t)² ] = d/dt[d(t)²]

becomes

2x(t)*x'(t) + 2y(t)*y'(t) = 2d(t)*d'(t)

using algebra to solve for the rate at which the people are moving apart we have

[2x(t)*x'(t) + 2y(t)*y'(t)]/2d(t) = d'(t)

or

[x(t)*x'(t) + y(t)*y'(t)]/d(t) = d'(t)

ok now we have 2 equations. (1) represents distance and the other (2) the rate of change of the distances need to plug in our numbers into this

x(t)² + y(t)² = d(t)² (1)

[x(t)*x'(t) + y(t)*y'(t)]/d(t) = d'(t) (2)

we now just have to plug in the numbers if the information provided. but we must proceed with caution. equation 2 can be reduced down to a much simpler form if we use the right logic. recall that the x value will always remain constant through out the problem. it stays 500ft. that means that the rate of change in the x direction will be 0.

x(t) = 500 x'(t) = 0

evaluating into equation 2 yields

[(500)*(0) + y(t)*y'(t)]/d(t) = d'(t)

or

[y(t)*y'(t)]/d(t) = d'(t) (3)

now we need to determine y(t) and y'(t) which will take careful examination.

the man is walking in the positive y direction at a rate of 3ft/sec. because the woman is walking in the negative direction, we say she is walking at a rate of - 4ft/sec. now because they are walking away from one another, we will take the difference in their rates to find the to rate in the y direction

y'(t) = 3 - (-4)

y'(t) = 7 ft/sec

for y(t) we need to account for the fact that the man was walking at a rate of 3 ft/sec for five mins (300 seconds) before the woman began walking. this means that the man had traveled a distance of (3)(300) = 900 ft before the woman had even moved. this just means we need an additional 1800 ft to the distance separating them in the y direction.

y(t) = 900 + (man and woman walking apart)

We find by taking the time it took for them to walk in separate directions 15 mins(900 secs) and multiply it to their combined rate in the y direction

(7 ft/sec)(900 sec) = 6300 ft

y(t) = 900 + 6300 = 7200 ft

now we need to find the distance between them at 15 mins(900 secs) and that is found by using equation 1 or the pythagreon theorem

x(t)² + y(t)² = d(t)²

(500)² + (7200)² = d(t)²

using algebra we have approximately d(t) = 7217.34 ft

we finally have all 3 pieces, evaluating them into equation 3 gives

[y(t)*y'(t)]/d(t) = d'(t)

[(7200)*(7)]/7217.34 = d'(t)

which is approximately

d'(t) = 6.98 ft per second

notice that the rate at which the people are moving apart is nearly 7 seconds the exact same rate of change in the y direction. this is because there was no change in the x direction and therefore y'(t) practically equals d'(t)