AP Calculus AB Exam Question

A ladder is leaning against a wall at a point $(x, y)$. The bottom of the ladder is sliding away from the wall at a rate of 2 ft/s. At the instant when the bottom of the ladder is 5 ft away from the wall, the bottom is moving horizontally at a rate of 3 ft/s.

1. Find the rate at which the top of the ladder is sliding down the wall when the bottom is 5 ft away from the wall.

Solution:

Let's denote the position of the bottom of the ladder as $x$ feet, the height of the wall as $y$ feet, and the length of the ladder as $L$ feet. We are given that $\frac{dx}{dt} = 2$ ft/s when $x = 5$. We need to find $\frac{dy}{dt}$ when $x = 5$.

We can form a right triangle where the ladder is the hypotenuse, the vertical distance from the bottom of the wall to the top of the ladder is $y$, and the horizontal distance from the bottom of the ladder to the wall is $x$. Using the Pythagorean theorem, we have:

Differentiating both sides of the equation with respect to time $t$, we get:

Since we are interested in finding $\frac{dy}{dt}$, we can rearrange the equation as follows:

We know that $\frac{dx}{dt} = 2$ ft/s, so the equation becomes:

Substituting $x = 5$ ft and $\frac{dx}{dt} = 2$ ft/s, we have:

Simplifying further, we have:

Now, we need to find the value of $\frac{dL}{dt}$ at $x = 5$ ft. Since the bottom of the ladder is sliding away from the wall, the vertical distance $y$ remains constant. Therefore, $\frac{dy}{dt} = 0$ ft/s.

From $\triangle x, y, L$, we can also find that $\frac{dL}{dt} = \frac{dx}{dt} = 2$ ft/s.

Substituting these values into the equation, we have:

Thus, the length of the ladder is 5 ft.

Now we can substitute the values of $x = 5$, $L = 5$, and $\frac{dL}{dt} = 2$ ft/s into the equation to find $\frac{dy}{dt}$:

Therefore, when the bottom of the ladder is 5 ft away from the wall, the top of the ladder is not sliding down the wall ($\frac{dy}{dt} = 0$ ft/s).

(answer: $\frac{dy}{dt} = 0$ ft/s)