Questions: A 25-foot ladder is placed against a vertical wall. Suppose the bottom of the ladder slides away from the wall at a constant rate of 2 feet per second. How fast is the top of the ladder sliding down the wall when the bottom is 20 feet from the wall? The ladder is sliding down the wall at a rate of ft/sec. (Type an integer or a simplified fraction.)

Solution Steps

Let $x$ be the distance from the bottom of the ladder to the wall, and $y$ be the distance from the top of the ladder to the ground. The ladder has a length of 25 feet. We are given that $\frac{dx}{dt} = 2$ ft/s, and we want to find $\frac{dy}{dt}$ when $x = 20$ feet.

By the Pythagorean theorem, we have $x^2 + y^2 = 25^2 = 625$.

Differentiating both sides with respect to $t$, we get $2x\frac{dx}{dt} + 2y\frac{dy}{dt} = 0$.

When $x = 20$, we have $20^2 + y^2 = 625$, so $y^2 = 625 - 400 = 225$, and $y = 15$. Substituting $x = 20$, $y = 15$, and $\frac{dx}{dt} = 2$ into the differentiated equation, we get: $2(20)(2) + 2(15)\frac{dy}{dt} = 0$ $80 + 30\frac{dy}{dt} = 0$ $30\frac{dy}{dt} = -80$ $\frac{dy}{dt} = -\frac{80}{30} = -\frac{8}{3}$

Final Answer