a ladder 15 ft long leans against a vertical wall. if the lower end is being moved away from the wall at the…

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}))