Related Rates?

Question

Two cars start moving from the same point. One travels south at 60mi/h and the other travels west at 25mi/h. At what rate is the distance between the cars increasing two hours later

Answer 1

Let's measure t in hours.

The EW distance is 25t, and the NS distance is 60t. The total distance is (625t^2 + 3600t^2) ^ 1/2, or 65t. The derivative of that is always 65, so there's your answer: 65mph. It doesn't matter how long they've been driving, assuming they're on a flat surface.

Catbarf, you forgot to square the 25 in your expression. Shell... nice job editing your answer after someone else comes up with the right answer :P

Answer 2

this is going to be hard to answer without using a sketch, but see if this makes sense:

if one car travels south and the other travels west....the distance between them forms a right triangle (sketch that and see if it makes sense).....so we can use pythagorean theorem as our equation we need to derive.....so if you derive pythagorean theorem you get

2a da/dt + 2b db/dt = 2c dc/dt

we can easily figure out a, b and c.....we know the rates the cars are traveling and we know the cars are traveling for 2 hours......so the car going south has gone 60*2 = 120 mi, and the car going west has gone 25*2 = 50 mi...to find c we use pythagorean theorem....120^2 + 50^2 = 16900 and the square root of that is 130, so c = 130

Now what we are looking for is dc/dt (the rate at which the distance between the cars is increasing), we already know da/dt (the rate of the car going south) and we know db/dt (the car going west)....so now we just plug into our derivative and solve for dc/dt

2(120)(60) + 2(50)(25) = 2(130) dc/dt
240(60) + 100(25) = 260dc/dt
14400 + 2500 = 260 dc/dt
16900 / 260 = dc/dt
65 = dc/dt

so the rate that the distance between the cars is increasing is 65 mi/hr

Hope that helps!!

Answer 3

You should draw a picture. At any time in hours, the southern car will be 60 t miles south of his initial position. The western car will be 25 t miles west of his initial poistion. The distance between them is sqrt [(60t)2+(25t)2] and the rate of change of distance is d(sqrt(3625 t^2))/dt or
sqrt(3625) *(2)*(1/2)* [t /sqrt t^2]. At t=2 hours, this is sqrt(3625) mph or slightly over 60 mph.