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Chapter 3: Problem 31

Circle's changing area What is the rate of change of the area of a circle \(\left(A=\pi r^{2}\right)\) with respect to the radius when the radius is \(r=3 ?\)

Short Answer

Expert verified
The rate of change of the area is \(6\pi\) when the radius is 3.

Step by step solution

01

Understand the Problem

We need to find how fast the area of a circle changes as the radius changes, specifically when the radius is 3. In other words, we want the derivative of the area with respect to the radius, evaluated at this given radius.
02

Write the Formula for Circle's Area

The formula for the area of a circle is given as \( A = \pi r^2 \). This equation tells us how the area depends on the radius \( r \).
03

Differentiate the Area with Respect to Radius

To find the rate of change of the area with respect to the radius, we differentiate \( A \) with respect to \( r \). The derivative \( \frac{dA}{dr} \) is found by differentiating \( \pi r^2 \) with respect to \( r \), which gives \( \frac{dA}{dr} = 2\pi r \).
04

Evaluate the Derivative at r = 3

Now that we have \( \frac{dA}{dr} = 2\pi r \), substitute \( r = 3 \) into this derivative to find the rate of change at that specific radius. Thus, \( \frac{dA}{dr} = 2\pi \times 3 = 6\pi \).

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Key Concepts

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

Rate of Change
The rate of change is a fundamental concept in calculus. It measures how fast a certain quantity changes with respect to another quantity. In the context of our circle problem, we are interested in how quickly the area of the circle changes as its radius increases.
For example, as you stretch a balloon, the area expands. The speed of this expansion, at any given instant, is the rate of change of the area with respect to the radius.
  • Formula: The rate of change can be expressed as a derivative. In mathematical terms, if you have a function, then its derivative shows the rate of change.
  • Practical use: Knowing the rate of change helps in predicting how changes in one dimension affect another, invaluable for fields like physics and engineering.
In calculus, the derivative of a function at a point gives the instantaneous rate of change at that point.
Circle Area
The area of a circle is determined using a straightforward formula:
\[A = \pi r^2\]
This formula essentially tells you how much space your circle occupies, measured in square units, depending on its radius.
  • Understanding \( \pi \): The symbol \( \pi \) represents a constant approximately equal to 3.14159. It's crucial in circular calculations, as it relates the circumference of a circle to its diameter.
  • Dependence on Radius: Notice that the area is directly proportional to the square of the radius. This relationship highlights how significantly the area increases when you stretch out the radius.
This quadratic relationship means that even small increases in the radius can result in large increases in the area.
Differentiation
Differentiation is a mathematical process used to find the rate at which a function is changing at any point. It plays a crucial role in understanding how variable change impacts an entire system.
In our circle problem, differentiation helps us determine how the area changes as the radius changes. Specifically, we find the derivative of \( A = \pi r^2 \), which gives us \( \frac{dA}{dr} = 2\pi r \).
  • Product Rule: When differentiating, we apply rules such as the product rule, quotient rule, or chain rule, depending on the function's composition.
  • Importance: In practical terms, differentiation helps in optimizing and analyzing real-world problems by providing critical information on rates of change and slope of lines.
In essence, differentiation allows us to "zoom in" to see precisely how the area changes with each infinitesimally small increase in the radius.
Geometry
Geometry is the branch of mathematics concerned with the properties and relations of points, lines, surfaces, and solids. It provides the foundational understanding necessary for solving problems involving shapes and space.
The concept of circle area stems directly from geometry. The way circles are defined and calculated鈥攗sing properties like radius and 蟺鈥攃omes from geometric principles.
  • Visual Understanding: Geometry provides a visual framework which helps in comprehending various spatial relationships.
  • Applications: This knowledge is pivotal in numerous fields such as architecture, engineering, and astronomy, where spatial understanding is a must.
In our rate of change problem, geometric principles guide us on how changes in one dimension (radius) translate to changes in area.

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Most popular questions from this chapter

Measuring acceleration of gravity When the length \(L\) of a clock pendulum is held constant by controlling its temperature, the pendulum's period \(T\) depends on the acceleration of gravity \(g\). The period will therefore vary slightly as the clock is moved from place to place on Earth's surface, depending on the change in \(g\). By keeping track of \(\Delta T\), we can estimate the variation in \(g\) from the equation \(T=2 \pi(L / g)^{1 / 2}\) that relates \(T, g,\) and \(L\) a. With \(L\) held constant and \(g\) as the independent variable, calculate \(d T\) and use it to answer parts (b) and (c). b. If \(g\) increases, will \(T\) increase or decrease? Will a pendulum clock speed up or slow down? Explain. c. A clock with a \(100-\mathrm{cm}\) pendulum is moved from a location where \(g=980 \mathrm{cm} / \mathrm{sec}^{2}\) to a new location. This increases the period by \(d T=0.001\) sec. Find \(d g\) and estimate the value of g at the new location.

In Exercises \(69-74,\) use a CAS to estimate the magnitude of the error in using the linearization in place of the function over a specified interval I. Perform the following steps: a. Plot the function \(f\) over \(I\) b. Find the linearization \(L\) of the function at the point \(a\). c. Plot \(f\) and \(L\) together on a single graph. d. Plot the absolute error \(|f(x)-L(x)|\) over \(I\) and find its maximum value. e. From your graph in part (d), estimate as large a \(\delta>0\) as you can that satisfies \(|x-a|<\delta \Rightarrow|f(x)-L(x)|<\varepsilon\) for \(\varepsilon=0.5,0.1,\) and \(0.01 .\) Then check graphically to see whether your \(\delta\) -estimate holds true. $$f(x)=x 2^{x}, \quad[0,2], \quad a=1$$

Find the values of a. \(\sec ^{-1}(-3)\) b. \(\csc ^{-1} 1.7\) c. \(\cot ^{-1}(-2)\)

Find \(d y / d x\) $$x^{y}=y^{x}$$

You will explore some functions and their inverses together with their derivatives and tangent line approximations at specified points. Perform the following steps using your CAS: a. Plot the function \(y=f(x)\) together with its derivative over the given interval. Explain why you know that \(f\) is one-to-one over the interval. b. Solve the equation \(y=f(x)\) for \(x\) as a function of \(y,\) and name the resulting inverse function \(g\). c. Find an equation for the tangent line to \(f\) at the specified $$ \text { point }\left(x_{0}, f\left(x_{0}\right)\right) $$ d. Find an equation for the tangent line to \(g\) at the point \(\left(f\left(x_{0}\right), x_{0}\right)\) located symmetrically across the \(45^{\circ}\) line \(y=x\) (which is the graph of the identity function). Use Theorem 3 to find the slope of this tangent line. e. Plot the functions \(f\) and \(g\), the identity, the two tangent lines, and the line segment joining the points \(\left(x_{0}, f\left(x_{0}\right)\right)\) and \(\left(f\left(x_{0}\right), x_{0}\right) .\) Discuss the symmetries you see across the main diagonal (the line \(y=x\) ). $$y=2-x-x^{3}, \quad-2 \leq x \leq 2, \quad x_{0}=\frac{3}{2}$$
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