Question

A train, starting at 11 am, travels east at 45 mph while another, starting at noon from the same point, travels south at 60 mph. How fast are they separating at 3 pm? I need help solving this related rate. Thanks.

Answer 1

Okay, let's take a look.
For related rates, the first step is to try to establish the relationship that's changing. Let's let t=0 be at noon. We know that the position of the first train is going to be directly east of the origin at position 45*(t+1). Similarly, the position of the second train will be directly south of the origin at position 60*t. Then the distance between them can be calculated using the Pythagorean theorem:


d = [(60t)^2 + (45(t+1))^2]^(1/2) = (2025 + 4050t + 5625t^2)^(1/2) =
15*(25t^2 + 18t + 9)^(1/2)

Now, our goal is to find the rate of change of distance wite respect to time, so we want to find dd/dt:
\[\frac{ d }{ dt }(15*\sqrt(25t^2 + 18t + 9)) =\]\[15*\frac{ d }{ dt }(\sqrt(25t^2 + 18t + 9)) =\]\[15*(1/2* [1/\sqrt(25t^2 + 18t + 9)]*(50t+18)) =\]\[[15*(25t+9)]/\sqrt(25t^2 + 18t + 9)\]
Now we solve for the change in distance at 3 o'clock, t=3.
d = [15*(75+9)]/sqrt(225 + 54 + 9) = 1260/sqrt(288) = 1260/(12*sqrt(2)) = 105/sqrt(2) ~ 74.2462 mph

Answer 2

That helped me so much thank you!

Answer 3

I am terrible at related rates haha thanks