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Surface Area The change in the surface area \(S=4 \pi r^{2}\) of a sphere when the radius changes from \(a\) to \(a+d r\)

Short Answer

Expert verified
The change in the surface area of a sphere when the radius changes from \(a\) to \(a + dr\) is \(dS = 8 \pi a \cdot dr\).

Step by step solution

01

Define the Expression for the Surface Area of the Sphere

The surface area of a sphere with radius \(r\) is given by \(S=4 \pi r^{2}\). The task is to compute the change in surface area of the sphere when the radius is changed from \(a\) to \(a + dr\). Here, \(S\) is a function of \(r\), i.e., \(S(r) = 4 \pi r^{2}\).
02

Compute the Derivative of the Surface Area

The derivative of the surface area \(S(r)\) with respect to the radius \(r\), denoted as \(\frac{dS}{dr}\), gives the rate of change of surface area for an infinitesimal change in radius. The derivative of \(S(r) = 4 \pi r^{2}\) with respect to \(r\) is \(\frac{dS}{dr} = 8 \pi r\).
03

Compute the Change in Surface Area

The change in surface area (\(dS\)) when the radius changes from \(a\) to \(a + dr\) is given by \(dS = \frac{dS}{dr} \cdot dr = 8 \pi a \cdot dr\).

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Key Concepts

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

Calculus
Calculus is an extensive branch of mathematics that deals with the calculation of properties and behaviors of functions. It has two major areas: differential calculus, which concerns itself with rates of change and slopes of curves, and integral calculus, which focuses on accumulating quantities and the areas under and between curves. In working with the surface area of a sphere, calculus lets us understand how the surface area changes as the radius of the sphere changes. This kind of problem is approached using differential calculus, as we're interested in how a small change in the radius affects the total surface area.
Derivatives
A derivative represents the rate at which a function changes at any point and is a fundamental tool in calculus. Derivatives measure how a function's output alters as its input changes by a tiny amount. In the context of the surface area of a sphere, the derivative quantifies how sensitive the surface area is to a change in the sphere's radius. Mathematically, the derivative of the surface area with respect to the radius, denoted as \(\frac{dS}{dr}\), is the rate of change of the surface area relative to the radius. When we calculate this derivative for the sphere's surface area, which is a function of the radius squared (\(S(r) = 4 \( \pi \) r^2\)), we can see how surface area would change in response to a slight alteration of the radius.
Rate of Change
The rate of change is a value that describes how one quantity changes in relation to another. In our exercise, we are focused on the rate at which the surface area (S) of a sphere changes with respect to its radius (r). This relationship can be quantified by finding the derivative \( \frac{dS}{dr} \), which tells us that for each unit of change in the radius, the surface area changes by the value of the derivative at that particular radius. This concept is not restricted to spherical objects; it's a universal principle in calculus used to describe any dynamic situation where one variable influences another.
Infinitesimal Change
Infinitesimal change involves adjustments that are extremely small, approaching zero, but are still nonzero. In calculus, infinitesimals are the bread and butter of differential calculus. By studying these minuscule changes, we can predict and understand the behavior of a mathematical function. When we look at the sphere's surface area formula, and we compute the derivative, we are essentially considering the effect of an infinitesimally small change in the radius (denoted as \(dr\)) on the surface area (\(dS\)). Calculus offers a framework for making sense of how slight variations in input can systematically affect the output of a function.

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Most popular questions from this chapter

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 \((x-a)\) .

\(f\) is an even function, continuous on \([-3,3],\) and satisfies the following. (d) What can you conclude about \(f(3)\) and \(f(-3) ?\)

Vertical Motion Two masses hanging side by side from springs have positions \(s_{1}=2 \sin t\) and \(s_{2}=\sin 2 t\) respectively, with \(s_{1}\) and \(s_{2}\) in meters and \(t\) in seconds. (a) At what times in the interval \(t>0\) do the masses pass each other? [Hint: \(\sin 2 t=2 \sin t \cos t ]\) (b) When in the interval \(0 \leq t \leq 2 \pi\) is the vertical distance between the masses the greatest? What is this distance? (Hint: \(\cos 2 t=2 \cos ^{2} t-1 . )\)

Geometric Mean The geometric mean of two positive numbers \(a\) and \(b\) is \(\sqrt{a b}\) . Show that for \(f(x)=1 / x\) on any interval \([a, b]\) of positive numbers, the value of \(c\) in the conclusion of the Mean Value Theorem is \(c=\sqrt{a b} .\)

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 \(r=(2 / 3) r_{0}\) .

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