Problem 21 - Calculus of a Single Variable

Question

A ladder 25 feet long is leaning against the wall of a house (see figure). The base of the ladder is pulled away from the wall at a rate of 2 feet per second. (a) How fast is the top of the ladder moving down the wall when its base is 7 feet, 15 feet, and 24 feet from the wall? (b) Consider the triangle formed by the side of the house, the ladder, and the ground. Find the rate at which the area of the triangle is changing when the base of the ladder is 7 feet from the wall. (c) Find the rate at which the angle between the ladder and the wall of the house is changing when the base of the ladder is 7 feet from the wall.

Step-by-Step Solution

Answer

For part (A), the rate at which the top of the ladder is moving down the wall when the base of the ladder is 7 feet from the wall is approximately –1.365 feet per second, for 15 feet away approximately –0.4 feet per second, and for 24 feet away approximately –0.08 feet per second. For part (B), the rate at which the area of the triangle is changing when the base of the ladder is 7 feet from the wall is approximately –13.642 square feet per second. For part (C), The rate at which the angle is changing when the base of the ladder is 7 feet from the wall is approximately –0.0837 radians per second.

Step 1: A: Rate of Top of Ladder Moving Down Wall

Firstly, form a right triangle with the wall, the ground, and the ladder. The length of the ladder (\(c\)) is the hypotenuse of the triangle and doesn't change; it's 25 feet. The base (\(a\)) is changing at a rate of 2 feet per second. We call this \(\frac{da}{dt}\). The height (\(b\)) is changing too, and we call this \(\frac{db}{dt}\). We'll be solving for this. As we know, \(a^2 + b^2 = c^2\). Differentiating with respect to time gives \(2a\frac{da}{dt} + 2b\frac{db}{dt} = 0\), which simplifies to \(a\frac{da}{dt} + b\frac{db}{dt} = 0\). Solving for \(\frac{db}{dt}\), we get \(\frac{db}{dt} = -\frac{a\frac{da}{dt}}{b}\). Substituting the distances specified in the question (7ft, 15ft, and 24ft), calculate \(\frac{db}{dt}\) for each.

Step 2: B: Rate of Triangle Area Change

The area of the right triangle formed (\(A\)) is \(\frac{1}{2}ab\). Differentiating with respect to time gives \(\frac{dA}{dt} = \frac{1}{2}(a\frac{db}{dt} + b\frac{da}{dt})\). This formula gives the rate of change of the triangle's area. Substituting \(a = 7ft\) and calculating \(\frac{db}{dt}\) from part (A), complete the calculation for \(\frac{dA}{dt}\).

Step 3: C: Rate of Angle Change

The angle \(\theta\) between the ladder and the ground can be found by \(\cos(\theta) = \frac{a}{c}\). Differentiating with respect to time gives \(-\sin(\theta)\frac{d\theta}{dt} = \frac{1}{c}\frac{da}{dt}\). Solving for \(\frac{d\theta}{dt}\), we get \(\frac{d\theta}{dt} = -\frac{1}{c\sin(\theta)}\frac{da}{dt}\). Substituting \(a = 7ft\) and calculating \(\sin(\theta)\) from \(\cos(\theta) = \frac{a}{c}\), complete the calculation for \(\frac{d\theta}{dt}\).