a ladder 10 ft long leans against a vertical wall. if the lower end is being moved away from the wall at the…

a ladder 10 ft long leans against a vertical wall. if the lower end is being moved away from the wall at the rate of 2 ft/sec, how fast is the height of the top changing (this will be a negative rate) when the lower end is 6 feet from the wall? the height of the top is changing at a rate of when the lower end is 6 feet from the wall. (simplify your answer.)

a ladder 10 ft long leans against a vertical wall. if the lower end is being moved away from the wall at the rate of 2 ft/sec, how fast is the height of the top changing (this will be a negative rate) when the lower end is 6 feet from the wall? the height of the top is changing at a rate of when the lower end is 6 feet from the wall. (simplify your answer.)

Answer

Explanation:

Step1: Establish the relationship

By the Pythagorean theorem, (x^{2}+y^{2}=10^{2}), so (y=\sqrt{100 - x^{2}}).

Step2: Differentiate with respect to time (t)

Differentiate (x^{2}+y^{2}=100) with respect to (t): (2x\frac{dx}{dt}+2y\frac{dy}{dt}=0), then (\frac{dy}{dt}=-\frac{x}{y}\cdot\frac{dx}{dt}).

Step3: Find (y) when (x = 6)

When (x = 6), (y=\sqrt{100 - 6^{2}}=\sqrt{100 - 36}=\sqrt{64}=8).

Step4: Substitute values

Given (\frac{dx}{dt}=2), (x = 6), (y = 8) into (\frac{dy}{dt}=-\frac{x}{y}\cdot\frac{dx}{dt}), we get (\frac{dy}{dt}=-\frac{6}{8}\times2).

Answer:

(-\frac{3}{2}\text{ ft/sec})