A ladder $51 \mathrm{ft}$ long leans against a vertical wall. If the lower end is being moved away from the wall at the rate of 3 $\mathrm{ft} / \mathrm{sec}$, how fast is the height of the top changing (this will be a negative rate) when the lower end is 24 feet from the wall? The height of the top is changing at a rate of when the lower end is 24 feet from the wall. (Simplify your answer.)

Step 1 ：We have a right triangle where the hypotenuse is the length of the ladder, one leg is the distance of the base of the ladder from the wall, and the other leg is the height of the ladder up the wall.

Step 2 ：We can use the Pythagorean theorem to relate these three quantities: \(x^2 + y^2 = 51^2\), where \(x\) is the distance of the base of the ladder from the wall and \(y\) is the height of the ladder up the wall.

Step 3 ：We are given that \(x = 24\) feet and \(\frac{dx}{dt} = 3\) feet per second. We want to find \(\frac{dy}{dt}\) when \(x = 24\).

Step 4 ：Differentiating both sides of the equation with respect to \(t\), we get \(2x\frac{dx}{dt} + 2y\frac{dy}{dt} = 0\).

Step 5 ：Substituting \(x = 24\), \(\frac{dx}{dt} = 3\), and \(y = \sqrt{51^2 - 24^2}\) into the equation, we can solve for \(\frac{dy}{dt}\).

Step 6 ：Doing so, we find that \(\frac{dy}{dt} = -\frac{x\frac{dx}{dt}}{y} = -\frac{24*3}{\sqrt{51^2 - 24^2}}\).

Step 7 ：Simplifying this expression, we find that \(\frac{dy}{dt} = \boxed{-1.5}\) feet per second.