I need help approaching this related rates problem.

A 15 ft ladder leans against a wall. The bottom of the ladder is 3 ft from the wall at time t=0 and slides away from the wall at a rate of 2ft/sec.
Find the velocity of the top of the ladder at time t=3.

Do you use the pythagorean theorem?
What do I do from there?

t = 3

x =
3 + 2 * t =
3 + 6 =
9

y =
sqrt(15^2 - x^2) =
sqrt(225 - (3 + 2t)^2) =
sqrt(225 - (9 + 12t + 4t^2) =
sqrt(225 - 9 - 12t - 4t^2) =
sqrt(216 - 12t - 4t^2) =
sqrt(4) * sqrt(54 - 3t - t^2) =
2 * sqrt(54 - 3t - t^2) =
2 * sqrt(54 - 3 * 3 - 3^2) =
2 * sqrt(54 - 9 - 9) =
2 * sqrt(36) =
2 * 6 =
12

x = 3 + 2t
dx/dt = 2

y = 2 * sqrt(54 - 3t - t^2)
dy/dt = 2 * (1/2) * (-3 - 2t) / sqrt(54 - 3t - t^2) = -(3 + 2t) / sqrt(54 - 3t - t^2)

D^2 = x^2 + y^2
2D * dD/dt = 2x * dx/dt + 2y * dy/dt
D * dD/dt = x * dx/dt + y * dy/dt

D = 15
dD/dt = 0
x = 9
dx/dt = 2
y = 12
dy/dt = ?

15 * 0 = 9 * 2 + 12 * dy/dt
0 = 18 + 12 * dy/dt
-12 * dy/dt = 18
dy/dt = -18/12
dy/dt = -3/2

Do you use the pythagorean theorem? yes. Look at the picture.

z= hypotenuse (ladder)
x = base (distance between bottom of the ladder and the wall)
y = height

z^2 = x^2+y^2
At t=3, x=6
15^2 = 6^2 + y^2
225 = 36 + y^2
y^2 = 225-36 = 189
y = sqrt(189) = 13.7477 ft (top of the ladder at t=3) and sliding down

dx/dt = 2 feet / sec
At t=3. x=(3)(2) = 6

z^2 = x^2+y^2
differentiate both sides with respect to time t
2z dz/dt = 2x dx/dt + 2y dy/dt

z dz/dt =x dx/dt + y dy/dt
dz/dt = 0 since z, height of the ladder is a constant

(15) 0 = (6)(2) + (13.7477) dy/dt
dy/dt = -12/13.7477 = -0.8729 ft/sec

http://www.flickr.com/photos/61664301@N0…

If the question is how fast the top of the ladder is moving, the answer is angular velocity because the ladder top is moving in an arc.

The formula for angular velocity uses radians so you would have to convert 2 ft/sec into radians.