Question

question 1 of 14, step 1 of 1
an electrician is working at the top of a ladder that is 9 ft long and leaning against the wall when its bottom starts sliding at a rate of 1 ft/s. fortunately, a fellow worker catches it when the ladders bottom is 3 ft from the wall. at what speed is the top of the ladder (along with the electrician) sliding down the wall at that instant? round any intermediate calculations, if needed, to no less than six decimal places, and round your final answer to two decimal places.
answer
ft/s
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Explanation:

Step1: Use the Pythagorean theorem

Let \(x\) be the distance of the bottom of the ladder from the wall and \(y\) be the distance of the top of the ladder from the ground. The length of the ladder \(L = 9\) ft. By the Pythagorean theorem \(x^{2}+y^{2}=L^{2}\). Substituting \(L = 9\), we have \(x^{2}+y^{2}=81\).

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

Differentiating both sides of \(x^{2}+y^{2}=81\) with respect to \(t\) gives \(2x\frac{dx}{dt}+2y\frac{dy}{dt}=0\). We know that \(\frac{dx}{dt}=1\) ft/s. When \(x = 3\) ft, we can find \(y\) from \(x^{2}+y^{2}=81\). So \(y=\sqrt{81 - x^{2}}=\sqrt{81-9}=\sqrt{72}=6\sqrt{2}\) ft.

Step3: Solve for \(\frac{dy}{dt}\)

Substitute \(x = 3\), \(y = 6\sqrt{2}\), and \(\frac{dx}{dt}=1\) into \(x\frac{dx}{dt}+y\frac{dy}{dt}=0\). We get \(3\times1+6\sqrt{2}\times\frac{dy}{dt}=0\). Then \(\frac{dy}{dt}=-\frac{3}{6\sqrt{2}}=-\frac{1}{2\sqrt{2}}\approx - 0.35\) ft/s.

Answer:

\(-0.35\) ft/s