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Chapter 3: Problem 40

A building's shadow On a morning of a day when the sun will pass directly overhead, the shadow of an 80-ft building on level ground is 60 \(\mathrm{ft}\) long. At the moment in question, the angle \(\theta\) the sun makes with the ground is increasing at the rate of \(0.27^{\circ} / \mathrm{min}\) . At what rate is the shadow decreasing? (Remember to use radians. Express your answer in inches per minute, to the nearest tenth.)

Short Answer

Expert verified
The shadow is decreasing at approximately 0.3 inches per minute.

Step by step solution

01

Convert Degrees to Radians

First, we need to convert the rate at which the angle \( \theta \) is changing from degrees per minute to radians per minute. Since \( 1^\circ = \frac{\pi}{180} \text{ radians} \), the rate is \( 0.27 \times \frac{\pi}{180} \) radians per minute.
02

Set Up the Relationship Between Angle and Shadow

Using the trigonometric relation \( \tan(\theta) = \frac{h}{s} \), where \( h = 80 \text{ ft} \) is the height of the building and \( s = 60 \text{ ft} \) is the shadow length. We have \( \tan(\theta) = \frac{80}{s} \).
03

Differentiate with Respect to Time

Differentiate both sides with respect to \( t \). Use the chain rule: \( \frac{d}{dt}[\tan(\theta)] = \sec^2(\theta) \cdot \frac{d\theta}{dt} \) and \( \frac{d}{dt}[\frac{80}{s}] = -\frac{80}{s^2} \cdot \frac{ds}{dt} \).
04

Calculate \( \sec^2(\theta) \)

We have \( \tan(\theta) = \frac{4}{3} \) from the given shadow and building height. So, \( \sec^2(\theta) = 1 + \tan^2(\theta) = 1 + \left( \frac{4}{3} \right)^2 = \frac{25}{9} \).
05

Solve for \( \frac{ds}{dt} \)

Substitute \( \sec^2(\theta) = \frac{25}{9} \), \( \frac{d\theta}{dt} = 0.27 \times \frac{\pi}{180} \), and \( \tan(\theta) = \frac{4}{3} \) into the differentiated equation: \[ \frac{25}{9} \, \cdot \, \frac{0.27 \pi}{180} = -\frac{80}{60^2} \cdot \frac{ds}{dt} \]. Simplify and solve for \( \frac{ds}{dt} \).
06

Convert Rate to Inches Per Minute

After solving for \( \frac{ds}{dt} \) in feet per minute, multiply by 12 to convert the rate into inches per minute.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Differentiation as a Tool for Analyzing Change
Differentiation is a fundamental concept in calculus used to understand how functions change. In the context of related rates problems, like the shadow of the building changing as the sun moves, differentiation helps us track how one quantity changes in response to another. Think of differentiation as the tool that lets us zoom into the calculus world.
  • In this exercise, we differentiated both sides of a trigonometric equation involving a tangent function to link the rate of change of the angle (due to the moving sun) with the rate of change of the shadow length.
  • By using the chain rule for differentiation, we can simultaneously deal with how the angles measured in radians and distances in feet change with respect to time, measured in minutes.

In essence, differentiation provides a means to convert the real-world problem into mathematical expressions that can help us find rates of change efficiently.
Understanding Trigonometry in Related Rates
Trigonometry plays an essential role in understanding how different lengths and angles in a triangle relate to one another. In the given problem, trigonometry allows us to establish the connection between the angle of the sun and the building鈥檚 shadow using the tangent function.
  • The key trigonometric identity used here is the tangent: \( \tan(\theta) = \frac{h}{s} \), where \(h\) is the height of the building, and \(s\) is the shadow length.
  • Since the angle \(\theta\) changes with the sun鈥檚 movement, understanding how to manipulate and differentiate this trigonometric equation becomes crucial.

We see that as the angle \(\theta\) increases, the shadow length \(s\) changes dynamically. Trigonometry lets us express these relationships mathematically and analyze them as rates of change: precisely what related rates problems examine. This makes trigonometry an indispensable tool for such exercises.
Converting Degrees to Radians
In calculus and trigonometry, angles must often be expressed in radians rather than degrees. This is because derivatives and integrals of trigonometric functions are more straightforward to handle in radians.
  • The conversion between degrees and radians is foundational. Recall that \(1^\circ = \frac{\pi}{180} \text{ radians}\).
  • For the exercise, we converted the rate at which the angle \(\theta\) was changing from degrees per minute to radians per minute using the multiplication by \(\frac{\pi}{180}\).

This conversion ensures that when we differentiate, the functions involving trigonometry behave predictably and smoothly. Radians are preferred because they often simplify the math involved, notably when calculating derivatives, as most calculus formulas are based on radian measure. Understanding and converting angles correctly into radians is crucial for solving related rates problems efficiently.

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Most popular questions from this chapter

Use a CAS to perform the following steps in Exercises \(53-60 .\) \begin{equation} \begin{array}{l}{\text { a. Plot the equation with the implicit plotter of a CAS. Check to }} \\ {\text { see that the given point } P \text { satisfies the equation. }} \\ {\text { b. Using implicit differentiation, find a formula for the deriva- }} \\ {\text { tive } d y / d x \text { and evaluate it at the given point } P .}\end{array} \end{equation} \begin{equation} \begin{array}{l}{\text { c. Use the slope found in part (b) to find an equation for the tan- }} \\ {\text { gent line to the curve at } P \text { . Then plot the implicit curve and }} \\ {\text { tangent line together on a single graph. }}\end{array} \end{equation} \begin{equation} x^{3}-x y+y^{3}=7, \quad P(2,1) \end{equation}

Changing dimensions in a rectangle The length \(l\) of a rectangle is decreasing at the rate of 2 \(\mathrm{cm} / \mathrm{sec}\) while the width \(w\) is increasing at the rate of 2 \(\mathrm{cm} / \mathrm{sec} .\) When \(l=12 \mathrm{cm}\) and \(w=5 \mathrm{cm},\) find the rates of change of (a) the area, (b) the perimeter, and (c) the lengths of the diagonals of the rectangle. Which of these quantities are decreasing, and which are increasing?

Heating a plate When a circular plate of metal is heated in an oven, its radius increases at the rate of 0.01 \(\mathrm{cm} / \mathrm{min.}\) . At what rate is the plate's area increasing when the radius is 50 \(\mathrm{cm} ?\).

Diagonals If \(x, y,\) and \(z\) are lengths of the edges of a rectangular box, the common length of the box's diagonals is \(s=\) \(\sqrt{x^{2}+y^{2}+z^{2}}\) a. Assuming that \(x, y,\) and \(z\) are differentiable functions of \(t\) how is \(d s / d t\) related to \(d x / d t, d y / d t,\) and \(d z / d t\) ? b. How is \(d s / d t\) related to \(d y / d t\) and \(d z / d t\) if \(x\) is constant? c. How are \(d x / d t, d y / d t,\) and \(d z / d t\) related if \(s\) is constant?

Quadratic approximations $$ \begin{array}{l}{\text { a. Let } Q(x)=b_{0}+b_{1}(x-a)+b_{2}(x-a)^{2} \text { be a quadratic }} \\ {\text { approximation to } f(x) \text { at } x=a \text { with the properties: }}\end{array} $$ $$ \begin{aligned} \text { i) } Q(a) &=f(a) \\ \text { ii) } Q^{\prime}(a) &=f^{\prime}(a) \\ \text { iii) } Q^{\prime \prime}(a) &=f^{\prime \prime}(a) \end{aligned} $$ $$ \text{Determine the coefficients}b_{0}, b_{1}, \text { and } b_{2} $$ $$ \begin{array}{l}{\text { b. Find the quadratic approximation to } f(x)=1 /(1-x) \text { at }} \\ {\quad x=0 .} \\ {\text { c. } \operatorname{Graph} f(x)=1 /(1-x) \text { and its quadratic approximation at }} \\ {x=0 . \text { Then zoom in on the two graphs at the point }(0,1) \text { . }} \\ {\text { Comment on what you see. }}\end{array} $$ $$ \begin{array}{l}{\text { d. Find the quadratic approximation to } g(x)=1 / x \text { at } x=1} \\ {\text { Graph } g \text { and its quadratic approximation together. Comment }} \\ {\text { on what you see. }}\end{array} $$ $$ \begin{array}{l}{\text { e. Find the quadratic approximation to } h(x)=\sqrt{1+x} \text { at }} \\ {x=0 . \text { Graph } h \text { and its quadratic approximation together. }} \\ {\text { Comment on what you see. }} \\\ {\text { f. What are the linearizations of } f, g, \text { and } h \text { at the respective }} \\ {\text { points in parts (b), (d), and (e)? }}\end{array} $$
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