Question

A ladder is leaning against a wall. If the ladder is 10 feet long and the base is sliding away from the wall at a rate of __________ feet per minute, how fast is the top of the ladder descending when the base is 6 feet from the wall?

A) 1 ft/min
B) 2 ft/min
C) 3 ft/min
D) 4 ft/min

Answer 1

The rate at which the top of the ladder descends can be derived using related rates in calculus.

Step-by-step explanation:

The student asks how fast the top of the ladder is descending when the base is 6 feet from the wall. This is a related rates problem that can be solved using Pythagoras' theorem and differentiation.

Since the ladder's length is constant at 10 feet, we can write the relationship between the height y of the ladder against the wall and the distance x of the base from the wall as y² + x² = 10².

Differentiating both sides of the equation with respect to time t, we get 2y(dy/dt) + 2x(dx/dt) = 0. We can solve for dy/dt (the rate at which the top of the ladder descends) when x is 6 feet. Assuming the base is moving away from the wall at a constant rate, we plug in dx/dt as the unknown rate z and solve for dy/dt using the given values.

When x is 6 feet, using the Pythagorean theorem, y = √(10² - 6²) = √(100 - 36) = √64 = 8 feet. Therefore, the differentiation gives us 2(8)(dy/dt) + 2(6)(z) = 0, which simplifies to 16(dy/dt) = -12z. Dividing both sides by 16 gives us dy/dt = -¾z. If z = 1 ft/min, then dy/dt = -¾ ft/min; hence, the correct answer is not listed in the options provided.