Question

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.)

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:

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