Question

a ladder 15 ft long leans against a vertical wall. if the lower end is being moved away from the wall at the rate of 5 ft/sec, how fast is the height of the top changing (this will be a negative rate) when the lower end is 9 feet from the wall? the height of the top is changing at a rate of when the lower end is 9 feet from the wall (simplify your answer)

Explanation:

Step1: Establish the relationship

By the Pythagorean theorem, \(x^{2}+y^{2}=15^{2}\), where \(x\) is the distance of the lower - end from the wall and \(y\) is the height of the top of the ladder on the wall.

Step2: Differentiate with respect to time \(t\)

Differentiating \(x^{2}+y^{2}=225\) with respect to \(t\) gives \(2x\frac{dx}{dt}+2y\frac{dy}{dt}=0\). Then simplify to \(x\frac{dx}{dt}+y\frac{dy}{dt}=0\), and solve for \(\frac{dy}{dt}\): \(\frac{dy}{dt}=-\frac{x}{y}\cdot\frac{dx}{dt}\).

Step3: Find \(y\) when \(x = 9\)

When \(x = 9\), from \(x^{2}+y^{2}=225\), we have \(y=\sqrt{225 - x^{2}}=\sqrt{225-81}=\sqrt{144} = 12\).

Step4: Substitute values

We know that \(\frac{dx}{dt}=5\) ft/sec, \(x = 9\), and \(y = 12\). Substitute into \(\frac{dy}{dt}=-\frac{x}{y}\cdot\frac{dx}{dt}\), so \(\frac{dy}{dt}=-\frac{9}{12}\times5\).

Step5: Simplify

\(\frac{dy}{dt}=-\frac{15}{4}=- 3.75\) ft/sec.

Answer:

\(-\frac{15}{4}\text{ ft/sec}\) (or \(-3.75\text{ ft/sec}\))