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Chapter 4: Problem 20

Assume that the radius \(r\) and the area \(A=\pi r^{2}\) of a circle are differentiable functions of \(t\). Express \(d A / d t\) in terms of \(d r / d t\).

Short Answer

Expert verified
\(\frac{dA}{dt} = 2\pi r \frac{dr}{dt}\).

Step by step solution

01

Identify Given Information

We know that the area of a circle is given by the formula \(A = \pi r^2\), and both the radius \(r\) and the area \(A\) are functions of time \(t\). We need to find \(\frac{dA}{dt}\) in terms of \(\frac{dr}{dt}\).
02

Apply the Chain Rule

To find \(\frac{dA}{dt}\), we apply the chain rule for differentiation. Since \(A = \pi r^2\), we differentiate both sides with respect to \(t\), obtaining:\[ \frac{dA}{dt} = \frac{d}{dt}(\pi r^2). \]
03

Differentiate the Expression

Using the chain rule on \(\pi r^2\), we have:\[\frac{d}{dt}(\pi r^2) = 2\pi r \cdot \frac{dr}{dt}\]Here, \(2\pi r\) is the derivative of \(\pi r^2\) with respect to \(r\), and we multiply by \(\frac{dr}{dt}\) since \(r\) is a function of \(t\).
04

Write the Final Expression

The expression for \(\frac{dA}{dt}\) is:\[\frac{dA}{dt} = 2\pi r \frac{dr}{dt}.\]This formula shows the rate of change of the area \(A\) in terms of the rate of change of the radius \(r\).

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Key Concepts

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

Understanding the Chain Rule
Differentiation is a powerful tool in calculus that helps us understand how quantities change. The chain rule is particularly useful when we have a function composed of another function. It allows us to differentiate complex relationships. Suppose we have a function of a function, for example, the area of a circle as a function of its radius, where the radius itself is a function of time. Here, we can apply the chain rule to find how the area changes with time.
  • Consider a composite function: if we have a function \( A(r(t)) = \pi r^2 \), the chain rule helps us differentiate \( A \) with respect to \( t \).
  • The chain rule states: \( \frac{dA}{dt} = \frac{dA}{dr} \cdot \frac{dr}{dt} \).
  • In the context of the circle's area: since \( A = \pi r^2 \), we find \( \frac{dA}{dr} = 2 \pi r \), and multiply by \( \frac{dr}{dt} \).
Applying the chain rule helps us find \( \frac{dA}{dt} = 2\pi r \frac{dr}{dt} \). This tells us that the rate of change of the area is dependent on both the radius and the rate of change of the radius over time.
The Role of Radius in Differentiation
The radius is a fundamental part of circle geometry and plays a crucial role in differentiation, especially when relating to areas and rates of change. Understanding how slight changes in the radius affect the overall circle can help us model various real-world scenarios.
  • The formula for the area of a circle is \( A = \pi r^2 \), where \( r \) is the radius.
  • When \( r \) changes, so does the area, and this is described through differentiation.
  • If \( r \) is changing over time, it makes sense to look at how quickly these changes affect the area鈥攖his is where \( \frac{dr}{dt} \) comes in.
By performing differentiation, especially using the chain rule, we label \( 2\pi r \) as the rate at which the area changes per unit radius, multiplied by the rate at which the radius is changing. This concept is not just theoretical but applicable in real life, for example, in fields like meteorology, physics, and engineering.
Exploring the Rate of Change
Understanding the rate of change is central to the study of calculus and applicable across various scientific domains. It tells us how a quantity evolves over a period of time. In the context of the circle, when the radius changes, it directly impacts the rate at which the area grows.
  • The rate of change of any function with respect to time is given by its derivative with respect to that time period.
  • For a circle's area: \( \frac{dA}{dt} = 2\pi r \frac{dr}{dt} \) shows how the change in radius controls the rate at which the area increases.
  • This formula is significant because it connects the growth of the circle鈥檚 surface area with how fast its radius expands or contracts.
Finding the rate of change helps us answer questions like: How fast is an area expanding? Or shrinking? Such insights are crucial, for instance, in determining the impact of changes in physical phenomena, such as the thawing of polar ice caps, investing in expanding sectors, or predicting demographic growth.

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Most popular questions from this chapter

Find the equation of the normal line to the curve \(y=\sqrt{x}-1\) at the point \((4,1)\).

(a) Use the formal definition to find the derivative of \(y=\frac{1}{x}\) at \(x=2\). (b) Show that the point \(\left(2, \frac{1}{2}\right)\) is on the graph of \(y=\frac{1}{x}\), and find the equation of the normal line at the point \(\left(2, \frac{1}{2}\right)\). (c) Graph \(y=\frac{1}{x}\) and the tangent line at the point \(\left(2, \frac{1}{2}\right)\) in the same coordinate system.

Find the derivatives of the following functions: $$ f(x)=\cos x-\sin 2 x $$

Calculate the linear approximation for \(f(x)\) : $$f(x) \approx f(a)+f^{\prime}(a)(x-a)$$ . \(f(x)=e^{2 x+1}\) at \(a=-1 / 2\)

Use the quotient rule to show that $$ \frac{d}{d x} \sec x=\sec x \tan x $$
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