/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 2 (a) Write the area A of a circle... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

(a) Write the area A of a circle as a function of the circumference C. (b) Find the (instantaneous) rate of change of the area A with respect to the circumference C. (c) Evaluate the rate of change of \(A\) at \(C=\pi\) and \(C=6 \pi\) (d) If \(C\) is measured in inches and \(A\) is measured in square inches, what units would be appropriate for \(d A / d C ?\)

Short Answer

Expert verified
The area as a function of the circumference is \(A(C) = \frac{C^2}{4\pi}\). The rate of change of the area with respect to the circumference is \(dA/dC = \frac{C}{2\pi}\). The rate of change at \(C = \pi\) is \(1/2\) and at \(C = 6\pi\) is \(3\). The units for \(dA/dC\) are inches.

Step by step solution

01

Expressing area as a function of circumference

From the formula for circumference \(C = 2\pi r\), we can solve for \(r\), giving us \(r = \frac{C}{2\pi}\). Substituting \(r\) into the formula for the area \(A = \pi r^2\), we get \(A(C) = \pi * \left(\frac{C}{2\pi}\right)^2 = \frac{C^2}{4\pi}\).
02

Computing the rate of change of area with respect to circumference

To find the rate of change of \(A\) with respect to \(C\), we differentiate \(A(C)\) with respect to \(C\). The derivative of \(\frac{C^2}{4\pi}\) with respect to \(C\) is \(dA/dC = \frac{C}{2\pi}\).
03

Evaluating the rate of change at specific circumferences

Substitute \(C = \pi\) and \(C = 6\pi\) into the derivative to find the rate of change at these specific values. For \(C = \pi\), we have \(dA/dC = \frac{\pi}{2\pi} = \frac{1}{2}\). For \(C = 6\pi\), we have \(dA/dC = \frac{6\pi}{2\pi} = 3\).
04

Determining the appropriate units

The units of the rate of change \(dA/dC\) are given by the units of \(A\) divided by the units of \(C\), which in this case, are square inches per inch, or simply, inches.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

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

Rate of Change
Understanding the rate of change is essential in calculus, especially when dealing with functions that describe real-world scenarios. In simple terms, the rate of change measures how rapidly one quantity changes in relation to another. It's like asking, "How fast is something happening?"
In mathematical terms, when we talk about the rate of change, we are often referring to derivatives. A derivative represents the instantaneous rate of change of a function. It tells us how the function's output changes as we make tiny changes in its input.
In the context of the exercise from the textbook, the rate of change of the area of a circle with respect to its circumference is what we're finding. We are differentiating the area, a function of circumference, to see how a slight increase in the circle's circumference affects its area. This derivative gives us a precise mathematical way to describe this relationship.
Functions in Calculus
Functions are the cornerstone of calculus. A function is a relationship that uniquely associates each input with exactly one output. In calculus, functions are used to describe and understand relationships between changing quantities.
A classic example of a function relationship in geometry is between a circle's circumference and its area. Here, we can express the area as a function of the circumference.
The mathematical description stems from known formulas: Circumference \( C = 2\pi r \) and Area \( A = \pi r^2 \). By manipulating these formulas, we find that the area in terms of circumference is \( A(C) = \frac{C^2}{4\pi}\).
So, calculus functions help clarify how varying the circumference of a circle directly alters its area. This relationship is not merely theoretical. It has practical implications in fields ranging from physics to engineering, where understanding how changes in one quantity affect another is crucial.
Derivatives
Derivatives are a fundamental tool in calculus. They quantify how a function changes as its input changes. A derivative gives the slope of the tangent line to a function's graph at any given point. This slope is synonymous with the rate of change we previously discussed.
To compute a derivative, you apply specific rules to transform the original function into another function that describes its rate of change. In our circle area problem, we differentiated the function \( A(C) = \frac{C^2}{4\pi}\) to obtain \( \frac{dA}{dC} = \frac{C}{2\pi}\), which signifies how the area changes with respect to changes in the circumference.
The derivative itself can also be evaluated at specific points, like \( C = \pi\) and \( C = 6\pi \). These values provide insights into exactly how quickly the circle's area is increasing at those points. Furthermore, dimensionally, this derivative has units of inches, indicating the area change per unit change in circumference, again highlighting the practical nature of these computations.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Local Linearity This is the graph of the function \(y=\sin x\) close to the origin. Since sin \(x\) is differentiable, this graph resembles a line. Find an equation for this line. \(y=x\)

In Exercises \(1-8,\) find the derivative of \(y\) with respect to the appropriate variable. $$y=\frac{1}{\sin ^{-1}(2 x)}$$

Thoroughbred Racing A racehorse is running a 10 -furlong race. (A furlong is 220 yards, although we will use furlongs and seconds as our units in this exercise.) As the horse passes each furlong marker \((F),\) a steward records the time elapsed \((t)\) since the beginning of the race, as shown in the table below: \(\begin{array}{c|ccccccccccc}{\mathbf{F}} & {0} & {1} & {2} & {3} & {4} & {5} & {6} & {7} & {8} & {9} & {10} \\ \hline t & {0} & {20} & {33} & {46} & {59} & {73} & {86} & {100} & {112} & {124} & {135}\end{array}\) (a) How long does it take the horse to finish the race? (b) What is the average speed of the horse over the first 5 furlongs? (c) What is the approximate speed of the horse as it passes the 3 -furlong marker? (d) During which portion of the race is the horse running the fastest? (e) During which portion of the race is the horse accelerating the fastest?

(a) Write the area A of an equilateral triangle as a function of the side length s. (b) Find the (instantaneous) rate of change of the area A with respect to a side s. (c) Evaluate the rate of change of A at s " 2 and s " 10. (d) If \(s\) is measured in inches and \(A\) is measured in square inches, what units would be appropriate for \(d A / d s ?\)

In Exercises \(7-12,\) find the horizontal tangents of the curve. $$y=x^{4}-7 x^{3}+2 x^{2}+15$$

See all solutions

Recommended explanations on Math Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.