Related Rates Calculus

Question

A man is running over a bridge at a rate of 20 ft/sec while a boat passes under the bridge and immediately below him at a rate of 10ft/sec/ The boat's course is at right angles to the man's and is 20 ft below it. How fast are the man and the boat seperating one second later?

Answer 1

Let x=man's distance traveled, y=boat's distance traveled, and z=height distance

The formula for the distance from the man to the boat is p²=x²+y²+z².

The derivative of this is 2p dp = 2x dx + 2y dy + 2z dz and we want to know dp. Eight variables and 1 equation mean we need to nail down the other seven.

Find the distance at 1 second. p²=10²+20²+20² = 100+400+400 = 900, so p=30

dx=10, dy=20, and z=20 from the problem, and since 1 second has elapsed x=10 and y=20. This leaves dz, but the vertical distance he is from the boat is not changing, so dz=0

Plug in everything.

2(30) dp = 2(10)(10)+2(20)(20)+2(20)(0)

60 dp = 200+800+0

60 dp = 1000

dp = 50/3 ≈ 16.67 ft/sec²

Answer 2

Hi Amir: Let's do this one!

The diagram here is tricky because it's a 3-dimensional situation.

WLOG, let's have the runner, moving to the right on the positive X-axis in 3D space. After time t, the runner is at the position, (20t, 0, 0)

Similarly, WLOG, the boat is initially on the Y-axis at position 20, moving along the Z axis, at right angles to the runner's path. So, at time t, the boat is at position (0, 20, 10t).

By the usual distance formula, the distance between them is, L = sqrt{(20t - 0)^2 + (0 - 20)^2 + (0 - 10t)^2)}

= sqrt{ 400t^2 + 400 + 100t^2}

= sqrt (500t^2 + 400)

= (500t^2 + 400)^1/2

Now, the rate that L is changing with time, t, is just dL/dt.

By the Chain Rule, dL/dt = 1/2 (500t^2 + 400)^(-1/2) (1000t)

= 500t / (500t^2 + 400)^(1/2)

The question asks for the position after 1 second, so we plug t = 1 into this equation

=> dL/dT (at t = 1) = 500 / sqrt (900)

= 500 / 30

= 50 /3 ft/sec