A ladder 5 m long is leaning against a wall. The bottom of the ladder is pulled along the ground, away from the wall, at the rate of 2 cm/s. How fast is its height on the wall decreasing when the foot of the ladder is 4 m away from the wall?
The height on the wall is decreasing at the rate of 8/3 cm/s.
Explanation
This is a related rates problem involving a right triangle formed by the ladder, wall, and ground. The ladder length remains constant at 5 m. Using Pythagoras theorem and implicit differentiation with respect to time, we can find the relationship between the rates of change of the horizontal distance and vertical height. The key insight is that when one variable increases (distance from wall), the other must decrease (height on wall) to maintain the constant hypotenuse.
Solution Steps
Step 1: Let x be the distance of the foot of the ladder from the wall, and y be the height on the wall. By Pythagoras theorem: x² + y² = 5² = 25
Step 2: Differentiate both sides with respect to time t: 2x(dx/dt) + 2y(dy/dt) = 0, which gives x(dx/dt) + y(dy/dt) = 0
Step 3: Find y when x = 4 m: y² = 25 - 16 = 9, so y = 3 m
Step 4: Given dx/dt = 2 cm/s = 0.02 m/s. Substitute in the differentiated equation: 4(0.02) + 3(dy/dt) = 0
Step 5: Solve for dy/dt: dy/dt = -0.08/3 m/s = -8/3 cm/s. The negative sign indicates the height is decreasing at 8/3 cm/s.