Problem 39 - Essential Calculus Early Transcendentals

Question

A ladder 10 ft long leans against a vertical wall. If the bottom of the ladder slides away from the base of the wall at a speed of 2 \(\mathrm{ft} / \mathrm{s}\) , how fast is the angle between the ladder and the wall changing when the bottom of the ladder is 6 \(\mathrm{ft}\) from the base of the wall?

Step-by-Step Solution

Answer

The angle changes at \(-0.2\) radians per second.

Step 1: Identify Known Quantities

We know the ladder is 10 feet long and the bottom of the ladder is moving away from the wall at 2 feet per second. Let \( x \) be the distance of the ladder's base from the wall and \( \theta \) be the angle between the ladder and the wall.

Step 2: Formulate Expressing Tan as Function of Variables

The height \( y \) of the ladder against the wall is \( y = \sqrt{10^2 - x^2} \). We have \( \tan(\theta) = \frac{x}{y} \), so \( \tan(\theta) = \frac{x}{\sqrt{10^2 - x^2}} \).

Step 3: Differentiate Tan with Respect to Time

Differentiate \( \tan(\theta) = \frac{x}{\sqrt{100-x^2}} \) with respect to time \( t \). Using chain rule: \[ \frac{d}{dt}[\tan(\theta)] = \frac{1}{\cos^2(\theta)} \frac{d\theta}{dt} = \frac{d}{dt}\left(\frac{x}{\sqrt{100-x^2}}\right) \]

Step 4: Differentiate Expression for Tan

Using the quotient rule and chain rule, we have: \[ \frac{d}{dt}\left(\frac{x}{\sqrt{100-x^2}}\right) = \frac{(\sqrt{100 - x^2})\cdot 2 - x\cdot\frac{0.5(100 - x^2)^{-0.5}(2x)} }{(100-x^2)} \]

Step 5: Simplify and Solve for \( \frac{d\theta}{dt} \)

Simplify the differentiated expression and solve for \( \frac{d\theta}{dt} \). Substitute \( x = 6 \), \( \frac{dx}{dt} = 2 \) feet/sec. After solving: \( \cos^2(\theta) \Rightarrow 1 - \sin^2(\theta) = \frac{36}{100} \Rightarrow \cos^2(\theta) = \frac{64}{100} \).

Step 6: Find Value of \( \cos^2(\theta) \) and \( \sin^2(\theta) \)

We find \( \sin(\theta) = \frac{6}{10} = 0.6 \). With \( \cos^2(\theta) = \frac{64}{100} = 0.64 \), \( \cos(\theta) = \sqrt{0.64} = 0.8 \).

Step 7: Final Calculation for Rate of Change of Angle

Substituting back into differentiated equation: \[ \frac{2}{(0.8)^2} \cdot \frac{d\theta}{dt} = 0.8 \] Hence \( \frac{d\theta}{dt} = -0.2 \) radians per second.

Key Concepts

Trigonometry

Trigonometry plays a fundamental role when dealing with problems involving angles and distances, especially in scenarios such as a ladder leaning against a wall. In this exercise, trigonometry helps us relate the angle that the ladder makes with the wall to the positions of the ladder’s base and the height at which the top of the ladder touches the wall.

We use the tangent function, which is one of the primary trigonometric functions relating an angle in a right-angled triangle to the ratio of the lengths of the opposite side to the adjacent side. Here, \( \tan(\theta) = \frac{x}{y} \), where \( \theta \) is the angle between the ladder and the wall, \( x \) is the distance of the ladder's base from the wall, and \( y \) is the height at which the top of the ladder touches the wall.

This relationship is foundational, as it sets up the equation we will differentiate to find how fast the angle between the ladder and the wall is changing. Understanding this initial setup is crucial for tackling related rate problems with trigonometric aspects.

Differentiation

Differentiation is a powerful mathematical tool for finding how a quantity changes as other quantities change. In this related rates problem, we differentiate a trigonometric expression to determine the rate at which the angle between the ladder and the wall changes over time.

First, we express \( \tan(\theta) = \frac{x}{\sqrt{100-x^2}} \). Our goal is to understand how \( \theta \) changes over time as \( x \), the position of the ladder’s base, changes. Using the derivative, we handle functions of a function effectively to calculate this change.

Through differentiation, we find the derivative of \( \tan(\theta) \) with respect to time. Since time does not appear explicitly in the expression \( \tan(\theta) = \frac{x}{y} \), this introduces the need for the chain rule, a technique used when differentiating compositions of functions, making it complex yet feasible to track the interrelated rates of change in this problem.

Chain Rule

The chain rule is an essential method used in calculus for differentiating the composition of functions. In this exercise, it is crucial for differentiating \( \tan(\theta) = \frac{x}{\sqrt{100-x^2}} \) with respect to time, especially as we have functions nested within other functions.

When dealing with \( \tan(\theta) \), the chain rule helps us find the derivative \( \frac{d}{dt}[\tan(\theta)] \) by breaking it down into smaller parts. It's vital for tracking how both \( x \) and \( \theta \) change over time. Using the chain rule, we derive

• how the movement of the ladder's base (\