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Chapter 2: Problem 53

A spherical balloon is being inflated. Find the rate of increase of the surface area \(\left(S-4 \pi r^{2}\right)\) with respect to the radius \(r\) when \(r\) is (a) 1 \(\mathrm{ft}\) , (b) 2 \(\mathrm{ft}\) , and (c) 3 ft. What conclusion can you make?

Short Answer

Expert verified
As the radius increases, the rate at which the surface area increases also grows linearly, being directly proportional to the radius.

Step by step solution

01

Understand the Problem

The problem asks for the derivative of the surface area of a spherical balloon with respect to its radius. This means we need to find how fast the surface area increases as the radius changes.
02

Recall the Formula for Surface Area

The surface area of a sphere is given by the formula \(S = 4\pi r^2\). We need to find the rate of change of this surface area (\(\frac{dS}{dr}\)) with respect to the radius \(r\).
03

Differentiate the Surface Area Formula

Differentiate \(S = 4\pi r^2\) with respect to \(r\), using the power rule. \[ \frac{dS}{dr} = \frac{d}{dr}(4\pi r^2) = 8\pi r \]
04

Substitute the Given Radii

Substitute each given radius into the derivative to find the rate of increase of the surface area: For (a) \(r = 1\): \[ \frac{dS}{dr} = 8\pi(1) = 8\pi \] For (b) \(r = 2\): \[ \frac{dS}{dr} = 8\pi(2) = 16\pi \] For (c) \(r = 3\): \[ \frac{dS}{dr} = 8\pi(3) = 24\pi \]
05

Analyze the Results

The rate of increase of the surface area is proportional to the radius. Specifically, \(\frac{dS}{dr} = 8\pi r\), which means the rate of increase grows linearly with \(r\).

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Key Concepts

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

surface area of a sphere
The surface area of a sphere is an important concept in geometry. Imagine blowing up a balloon; this shape is a perfect example of a sphere. As it gets bigger, its surface doesn't stretch to a bigger square or rectangle but to more of the balloon's curved surface. The surface area of a sphere can be calculated using the formula:
  • \( S = 4\pi r^2 \)
Essentially, \( S \) represents the total outer area of the sphere, and \( r \) is the radius, a straight line from the center to any point on the sphere's surface. This formula is derived from calculus and geometry, combining them to measure that curvy outer shell of a sphere. The formula shows the relationship between the surface area and the radius, highlighting that as the radius increases, the surface area grows quite quickly, specifically, with the radius squared.
rate of change
The rate of change in mathematics is a way of showing how one quantity changes in relation to another. In our balloon example, we are interested in understanding how fast the surface area of a spherical balloon increases as we blow it up鈥攍iterally, how the surface grows as the radius grows. We refer to this as the rate of change of the surface area with respect to the radius, commonly noted as \( \frac{dS}{dr} \).
  • This notation is from calculus, where \( \frac{dS}{dr} \) indicates the derivative of \( S \) with respect to \( r \).
This tells us how much the surface area changes for every tiny (infinitesimal) change in the radius. Understanding this rate helps in predicting how much material might be needed to cover a sphere as it increases in size鈥攍ike knowing how much rubber is needed for bigger and bigger balloons!
derivative
In calculus, the derivative is a fundamental tool that helps us understand the "instantaneous rate of change" of a quantity with respect to another. Specifically, for a function like the surface area of a sphere \( S = 4\pi r^2 \), the derivative with respect to \( r \), written as \( \frac{dS}{dr} \), is crucial.The derivative, \( \frac{dS}{dr} = 8\pi r \), is derived using the power rule of differentiation: If \( f(x) = x^n \), then \( f'(x) = nx^{n-1} \). Applying this to our surface area function shows us how fast the surface area increases at any point along the sphere's size.
  • This derivative indicates that for each unit increase in the radius, the surface area increases by \( 8\pi \) times the radius.
It's like having a speedometer for growth鈥攕howing the exact speed (rate) at which the surface expands as the sphere gets larger.

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Most popular questions from this chapter

If a ball is thrown vertically upward with a velocity of \(80 \mathrm{ft} / \mathrm{s},\) then its height after \(t\) seconds is \(s=80 t-16 t^{2}\) . (a) What is the maximum height reached by the ball? (b) What is the velocity of the ball when it is 96 \(\mathrm{ft}\) above the ground on its way up? On its way down?

A tangent line is drawn to the hyperbola \(x y=c\) at a point \(P .\) (a) Show that the midpoint of the line segment cut from this tangent line by the coordinate axes is \(P\) . (b) Show that the triangle formed by the tangent line and the coordinate axes always has the same area, no matter where \(P\) is located on the hyperbola.

Find a parabola with equation \(y=a x^{2}+b x+c\) that has slope 4 at \(x=1,\) slope \(-8\) at \(x=-1,\) and passes through the point \((2,15) .\)

(a) The van der Waals equation for moles of a gas is $$\left(P+\frac{n^{2} a}{V^{2}}\right)(V-n b)=n R T$$ where \(P\) is the pressure, \(V\) is the volume, and \(T\) is the temperature of the gas. The constant \(R\) is the universal gas constant and \(a\) and \(b\) are positive constants that are characteristic of a particular gas. If \(T\) remains constant, use implicit differentiation to find \(d V / d P .\) (b) Find the rate of change of volume with respect to pressure of 1 mole of carbon dioxide at a volume of \(V=10 \mathrm{L}\) and a pressure of \(P=2.5\) atm. Use \(a=3.592 \mathrm{L}^{2}-\mathrm{atm} / \mathrm{mole}^{2}\) and \(b=0.04267 \mathrm{L} / \mathrm{mole}\)

A lighthouse is located on a small island 3 km away from the nearest point \(P\) on a straight shoreline and its light makes four revolutions per minute. How fast is the beam of light moving along the shoreline when it is 1 \(\mathrm{km}\) from \(P ?\)
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