is 4 inches. As the circle gets bigger and bigger, the radius will expand at a slower and slower rate.
Example 2: A 25-foot long ladder is leaning against a wall and sliding toward the floor. If the foot of the
ladder is sliding away from the base of the wall at a rate of 15 feet/sec, how fast is the top of the ladder
sliding down the wall when the top of the ladder is 7 feet from the ground?
Here’s another classic related rates problem. As always, a picture is worth 1,000 words.
You can see that the ladder forms a right triangle with the wall. Let x stand for the distance from the foot
of the ladder to the base of the wall, and let y represent the distance from the top of the ladder to the
ground. What’s our favorite theorem that deals with right triangles? The Pythagorean Theorem tells us
here that x^2 + y^2 = 25^2 . Now we have an equation that relates the variables to each other.
Now take the derivative of the equation with respect to t.
2 x + 2y = 0Just plug in what you know and solve. Because we’re looking for the rate at which the vertical distance is
changing, we’re going to solve for .
Let’s see what we know. We’re given the rate at which the ladder is sliding away from the wall: = 15.
The distance from the ladder to the top of the wall is 7 feet (y = 7). To find x, use the Pythagorean
Theorem. If we plug in y = 7 to the equation x^2 + y^2 = 25^2 , x = 24.
Now plug all this information into the derivative equation.