A 10-foot ladder leaning against a wall. The top slides down with dy/dt = -2 ft/s when the top is at height y = 6 ft. How fast is the bottom moving away from the wall when y = 6 ft?

A 10-foot ladder leaning against a wall. The top slides down with dy/dt = -2 ft/s when the top is at height y = 6 ft. How fast is the bottom moving away from the wall when y = 6 ft?

• A. 1
• B. 3/2
• C. -3/2
• D. 0

Answer: (B)

Related rates with a constant ladder length. Let the bottom’s distance from the wall be x and the top’s height on the wall be y. The ladder gives a right triangle with x^2 + y^2 = 10^2. Differentiate: 2x dx/dt + 2y dy/dt = 0, so dx/dt = −(y dy/dt)/x. When y = 6, the bottom distance is x = sqrt(10^2 − 6^2) = sqrt(64) = 8. The top is moving down at dy/dt = −2 ft/s, so dx/dt = −(6)(−2)/8 = 12/8 = 3/2 ft/s. So the bottom is moving away from the wall at 3/2 ft/s.

Related rates with a constant ladder length. Let the bottom’s distance from the wall be x and the top’s height on the wall be y. The ladder gives a right triangle with x^2 + y^2 = 10^2.

Differentiate: 2x dx/dt + 2y dy/dt = 0, so dx/dt = −(y dy/dt)/x.

When y = 6, the bottom distance is x = sqrt(10^2 − 6^2) = sqrt(64) = 8. The top is moving down at dy/dt = −2 ft/s, so

dx/dt = −(6)(−2)/8 = 12/8 = 3/2 ft/s.

So the bottom is moving away from the wall at 3/2 ft/s.