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In Exercises \(1-41,\) assume all variables are differentiable functions of \(t\) . Area The radius \(r\) and area \(A\) of a circle are related by the equation \(A=\pi r^{2} .\) Write an equation that relates \(d A / d t\) to $d r / d t .

Short Answer

Expert verified
The rate at which the area of the circle is changing with respect to time \(\frac{dA}{dt}\) is related to the rate at which the radius of the circle is changing with respect to time \(\frac{dr}{dt}\) by the formula \(\frac{dA}{dt} = 2\pi r \frac{dr}{dt}\).

Step by step solution

01

Identify the given equation

We are given the equation for the area of the circle, which is \(A = \pi r^{2}\).
02

Differentiate both sides with respect to time

We take the derivative on both sides of the equation with respect to time \(t\). Doing so, we have to consider that both \(A\) and \(r\) depend on time, which implicates a product rule: \(\frac{dA}{dt} = 2\pi r \frac{dr}{dt}\). Here, the derivative of \(r^{2}\) comes out as \(2r\) due to the application of the power rule, but because \(r\) itself depends on \(t\), we multiply by \(\frac{dr}{dt}\), that is the chain rule.
03

Write the final formula

Thus, the relationship between \(\frac{dA}{dt}\) and \(\frac{dr}{dt}\) comes out as: \(\frac{dA}{dt} = 2\pi r \frac{dr}{dt}\)

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Key Concepts

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

Differentiation
Differentiation is a concept in calculus that concentrates on the rate at which quantities change. It involves finding the derivative, which represents how a function changes as its input changes. In the context of differentiating an area of a circle, we are interested in how the area changes as the radius changes. This helps to interpret the 'speed' aspect of various physical changes in real-world applications.
For instance, if we think of a circle slowly expanding, differentiation allows us to measure how quickly the area grows concerning time. In this exercise, the equation for the area of a circle is given by \(A = \pi r^2\). However, because both area (\(A\)) and the radius (\(r\)) are functions of time (\(t\)), we need to use differentiation to express how they change with time.
Chain Rule
The chain rule is an essential differentiation rule used when dealing with composite functions, where one function depends on another. In simple terms, if a variable \(y\) is a function of \(u\), and \(u\) is a function of \(x\), then the derivative of \(y\) with respect to \(x\) can be found by multiplying the derivative of \(y\) with respect to \(u\) by the derivative of \(u\) with respect to \(x\).
In our exercise, the chain rule is crucial because the variables \(A\) (area) and \(r\) (radius) depend on \(t\) (time). When we differentiate the area \(A = \pi r^2\) with respect to time, we discover how both the area and radius change over time.
  • The derivative of \(r^2\) is \(2r\), due to the power rule.
  • However, since \(r\) is a function of time, we further multiply by \(\frac{dr}{dt}\), emphasizing the chain rule.
This leads us to the final derivative which describes how the area changes: \(\frac{dA}{dt} = 2\pi r \frac{dr}{dt}\).
Related Rates
Related rates problems focus on finding a relationship between the rates at which two or more related variables change over time. These scenarios often appear in problems involving quantities that change cohesively, such as geometry in motion or growth over time.
The exercise given is a perfect illustration of related rates. As the radius of a circle expands, its area grows concurrently. Here, the objective is to determine how the change in radius impacts the change in area over time. Using the related rate equation:
\[\frac{dA}{dt} = 2\pi r \frac{dr}{dt}\]- \(\frac{dA}{dt}\) represents how quickly the area of the circle is changing. - \(\frac{dr}{dt}\) signifies the rate at which the radius is changing.The equation revealed from differentiation showcases the direct relationship between these two rates, primarily governed by the factor of \(2\pi r\), indicating their instantaneous connection.
Area of a Circle
The area of a circle is a fundamental concept in geometry, and its formula \(A = \pi r^2\) is foundational in many applications involving circular shapes. This formula calculates the total space contained within the borders of a circle, where \(r\) is the radius, a line segment from the center of the circle to its perimeter.
Understanding the area of a circle is vital for solving problems involving both static and dynamic circles, especially in physics and engineering contexts.
In the problem at hand, we dig deeper into how this area changes as the radius varies over time. The analysis leads directly to the relation between the rates: as the circle grows or shrinks, we're able to quantify exactly how the growing circle's area shifts moment-by-moment. This powerful insight is fundamental for applications where precise changes in spatial measurements are critical.

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Most popular questions from this chapter

Writing to Learn The function $$ f(x)=\left\\{\begin{array}{ll}{x,} & {0 \leq x<1} \\ {0,} & {x=1}\end{array}\right. $$ is zero at \(x=0\) and at \(x=1 .\) Its derivative is equal to 1 at every point between 0 and \(1,\) so \(f^{\prime}\) is never zero between 0 and 1 and the graph of \(f\) has no tangent parallel to the chord from \((0,0)\) to \((1,0) .\) Explain why this does not contradict the Mean Value Theorem.

How We Cough When we cough, the trachea (windpipe) contracts to increase the velocity of the air going out. This raises the question of how much it should contract to maximize the velocity and whether it really contracts that much when we cough. Under reasonable assumptions about the elasticity of the tracheal wall and about how the air near the wall is slowed by friction, the average flow velocity \(v(\) in \(\mathrm{cm} / \mathrm{sec})\) can be modeled by the equation $$v=c\left(r_{0}-r\right) r^{2}, \quad \frac{r_{0}}{2} \leq r \leq r_{0}$$ where \(r_{0}\) is the rest radius of the trachea in \(\mathrm{cm}\) and \(c\) is a positive constant whose value depends in part on the length of the trachea. (a) Show that \(v\) is greatest when \(r=(2 / 3) r_{0},\) that is, when the trachea is about 33\(\%\) contracted. The remarkable fact is that \(X\) -ray photographs confirm that the trachea contracts about this much during a cough. (b) Take \(r_{0}\) to be 0.5 and \(c\) to be \(1,\) and graph \(v\) over the interval \(0 \leq r \leq 0.5 .\) Compare what you see to the claim that \(v\) is a maximum when \(r=(2 / 3) r_{0}\) .

Formulas for Differentials Verify the following formulas. (a) \(d(c)=0(c\) a constant \()\) (b) \(d(c u)=c d u(c\) a constant \()\) (c) \(d(u+v)=d u+d v\) (d) \(d(u \cdot v)=u d v+v d u\) (e) \(d\left(\frac{u}{v}\right)=\frac{v d u-u d v}{v^{2}}\) (f) \(d\left(u^{n}\right)=n u^{n-1} d u\)

The Linearization is the Best Linear Approximation Suppose that \(y=f(x)\) is differentiable at \(x=a\) and that \(g(x)=m(x-a)+c(m\) and \(c\) constants). If the error \(E(x)=f(x)-g(x)\) were small enough near \(x=a,\) we might think of using \(g\) as a linear approximation of \(f\) instead of the linearization \(L(x)=f(a)+f^{\prime}(a)(x-a) .\) Show that if we impose on \(g\) the conditions i. \(E(a)=0\) ii. \(\lim _{x \rightarrow a} \frac{E(x)}{x-a}=0\) then \(g(x)=f(a)+f^{\prime}(a)(x-a) .\) Thus, the linearization gives the only linear approximation whose error is both zero at \(x=a\) and negligible in comparison with \((x-a)\) .

Calculus and Geometry How close does the curve \(y=\sqrt{x}\) come to the point \((3 / 2,0) ?[\)Hint: If you minimize the square of the distance, you can avoid square roots.

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