Question

Kevin and Austin are trying to get into an upstairs window. They lean a 15 ft. ladder against the wall. But, the ladder is slipping so that its base moves away from the wall at a rate of 3 ft/sec. How fast is the angle formed by the ladder and the ground decreasing when the ladder's base is 9 ft. from the wall?

Answer 1

To find the rate at which the angle formed by the ladder and the ground is decreasing, we can use similar triangles. Let x be the distance of the base of the ladder from the wall. We are given that dx/dt = -3 ft/sec. Let θ be the angle formed by the ladder and the ground. Using the properties of similar triangles, we can differentiate to find dθ/dt.

Step-by-step explanation:

To find the rate at which the angle formed by the ladder and the ground is decreasing, we can use similar triangles. Let x be the distance of the base of the ladder from the wall. We are given that dx/dt = -3 ft/sec. Let θ be the angle formed by the ladder and the ground. Using the properties of similar triangles, we have tan(θ) = (15 ft - x) / x. Differentiating both sides of this equation with respect to time, we get sec^2(θ) * dθ/dt = (x * -1 - (15 ft - x)) / x^2 * dx/dt. Plugging in the given values dx/dt = -3 ft/sec and x = 9 ft, we can solve for dθ/dt, which represents the rate at which the angle is decreasing.