Question 11
$5 \mathrm{pts}$

A ladder is slipping down a vertical wall. If the ladder is 20 feet long and the top of the ladder is slipping down at the constant rate of $4 \frac{\mathrm{ft}}{\mathrm{s}}$, how fast is the bottom of the ladder moving along the ground when the bottom is 16 feet from the wall?
3 feet per second
0.25 feet per second
5 feet per second
0.8 feet per second

Step 1 ：Given that the length of the ladder \(L = 20\) feet, the rate at which the top of the ladder is slipping down \(\frac{dy}{dt} = -4\) ft/s, and we want to find \(\frac{dx}{dt}\) when \(x = 16\) feet.

Step 2 ：We start by using the Pythagorean theorem to relate the lengths of the sides of the right triangle formed by the wall, the ground, and the ladder: \(x^2 + y^2 = L^2\).

Step 3 ：Differentiating both sides with respect to time \(t\), we get: \(2x\frac{dx}{dt} + 2y\frac{dy}{dt} = 0\).

Step 4 ：First, we need to find \(y\) when \(x = 16\) feet. We can use the Pythagorean theorem: \(16^2 + y^2 = 20^2\) which simplifies to \(y^2 = 20^2 - 16^2\) and further simplifies to \(y = \sqrt{400 - 256}\) which gives us \(y = \sqrt{144}\) and finally \(y = 12\) feet.

Step 5 ：Now, we can substitute \(x = 16\) feet, \(y = 12\) feet, \(\frac{dy}{dt} = -4\) ft/s into the differentiated equation: \(2*16*\frac{dx}{dt} + 2*12*(-4) = 0\) which simplifies to \(32*\frac{dx}{dt} - 96 = 0\) and further simplifies to \(32*\frac{dx}{dt} = 96\).

Step 6 ：Solving for \(\frac{dx}{dt}\) we get \(\frac{dx}{dt} = \frac{96}{32}\) which simplifies to \(\frac{dx}{dt} = 3\) ft/s.

Step 7 ：So, the bottom of the ladder is moving along the ground at a rate of \(\boxed{3}\) feet per second when the bottom is 16 feet from the wall.