Sure! Here's a related rates problem for AP Calculus AB:

Question: A ladder 10 meters long is leaning against a wall. If the bottom of the ladder is sliding away from the wall at a rate of 2 meters per second, how fast is the top of the ladder sliding down the wall when the bottom of the ladder is 6 meters from the wall?

Answer: Given:

• Length of the ladder (l) = 10 meters
• Rate at which the bottom of the ladder is sliding away from the wall (dx/dt) = 2 m/s
• We need to find the rate at which the top of the ladder is sliding down the wall (dy/dt) when the bottom of the ladder is 6 meters from the wall.

This problem involves a right-angled triangle, with the ladder representing the hypotenuse, and the distance from the base of the wall to the bottom of the ladder representing one of the legs.

Using the Pythagorean theorem, we have: (ladder length)^2 = (distance from the wall)^2 + (height of the ladder)^2 (10)^2 = (x)^2 + (y)^2 Differentiating both sides with respect to time (t), we get: 0 = 2x * (dx/dt) + 2y * (dy/dt)

We want to find dy/dt when x = 6. First, we can find the value of y using the Pythagorean theorem: 10^2 = 6^2 + y^2 y = √(10^2 - 6^2) y = 8

Now, differentiate the initial condition equation: 0 = 2(6)(2) + 2(8)(dy/dt) 0 = 12 + 16(dy/dt) -12 = 16(dy/dt) dy/dt = -12/16 dy/dt = -3/4 m/s

So, when the bottom of the ladder is 6 meters from the wall, the top of the ladder is sliding down the wall at a rate of -3/4 m/s.

I hope this helps! Let me know if you need further assistance.