/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 37 Volume The change in the volume ... [FREE SOLUTION] | 91Ó°ÊÓ

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Volume The change in the volume \(V=x^{3}\) of a cube when the edge lengths change from \(a\) to \(a+d x\)

Short Answer

Expert verified
The change in volume \(dV\) when the edge length changes from \(a\) to \(a + dx\) is approximately \(3a^{2}dx\).

Step by step solution

01

Understand the Volume of a Cube

The volume \(V\) of a cube with edge length \(x\) is given by \(V=x^{3}\). Thus, if the edge length changes from \(a\) to \(a + dx\), the new volume is \((a+dx)^{3}\).
02

Find the Change in Volume Expression

The change in volume, denoted as \(dV\), is given by the new volume minus the old volume, \(dV = (a + dx)^{3} - a^{3}\). Expand the binomial expression \((a + dx)^{3}\) to \(a^{3} +3a^{2}dx + 3adx^{2} + dx^{3}\).
03

Simplify the Change in Volume Expression

Simplify the change in volume expression: \(dV = 3a^{2}dx + 3adx^{2} + dx^{3}\), realizing that terms with \(dx^{2}\) and \(dx^{3}\) are insignificant due to \(dx\) being very small. The expression simplifies to: \(dV = 3a^{2}dx\). This tells us that when the edge length of a cube increases by an infinitesimal amount, the volume increases roughly by 3 times the square of the original length times the change.

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Key Concepts

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

Volume of a Cube
The volume of a cube is one of the fundamental concepts in geometry. A cube has equal edge lengths, and its volume is calculated by raising the edge length to the third power. This formula can be expressed as \( V = x^3 \), where \( x \) represents the length of each edge in the cube.

This concept is essential in understanding how changes in dimension affect the cube's overall size. For example, if a cube's edge length changes from a measure \( a \) to \( a + dx \), the volume of the new cube becomes \( (a + dx)^3 \).

This simple relationship powerfully displays how small changes in dimension can lead to significant changes in volume, providing a foundation for further exploration in calculus and physics.
Binomial Expansion
Binomial expansion allows us to expand expressions raised to a power, such as \( (a + dx)^3 \).

The Binomial Theorem gives a formula to expand such expressions into a sum of terms:\[ (a + b)^n = a^n + \binom{n}{1}a^{n-1}b + \binom{n}{2}a^{n-2}b^2 + \ldots + b^n \]When applied to the expression \( (a + dx)^3 \), the expansion is: \( a^3 + 3a^2dx + 3adx^2 + dx^3 \). The coefficients (such as \( \binom{3}{1} = 3 \)) arise from combinations, simplifying the process significantly.

Understanding binomial expansion is crucial for dealing with polynomial expressions in calculus, physics, and engineering. It shows us how terms develop as powers increase and is especially helpful for approximating values when handling infinitesimals.
Infinitesimals
Infinitesimals are incredibly small quantities, and they're a key concept in calculus. When we talk about changes that are infinitesimal, we mean they approach zero but aren't zero.

In calculus, infinitesimals are used to analyze how small changes affect functions, such as the volume of a cube. In our exercise, as the length changes by \( dx \), a small amount, we observe its effect on the volume change \( dV \).

Notice that in \( dV = 3a^2dx + 3adx^2 + dx^3 \), terms involving \( dx^2 \) and \( dx^3 \) are negligible because they are products of even smaller quantities. This simplification helps us understand and predict the behavior of functions and variables in advanced mathematics.

In essence, infinitesimals enable us to make accurate predictions and calculations about real-world phenomena by considering the impacts of minute changes. They are pivotal in differential calculus, allowing a deeper insight into the behavior of mathematical and physical systems.

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Most popular questions from this chapter

In Exercises 62 and \(63,\) feel free to use a CAS (computer algebra system), if you have one, to solve the problem. Logistic Functions Let \(f(x)=c /\left(1+a e^{-h x}\right)\) with \(a>0\) \(a b c \neq 0\) (a) Show that \(f\) is increasing on the interval \((-\infty, \infty)\) if \(a b c>0\) and decreasing if \(a b c<0\) . (b) Show that the point of inflection of \(f\) occurs at \(x=(\ln |a|) / b\)

Multiple Choice The \(x\) -coordinates of the points of inflection of the graph of \(y=x^{5}-5 x^{4}+3 x+7\) are \(\mathrm \) (A) 0 only (B) 1 only (C) 3 only (D) 0 and 3 (E) 0 and 1

Newton's Method Suppose your first guess in using Newton's method is lucky in the sense that \(x_{1}\) is a root of \(f(x)=0 .\) What happens to \(x_{2}\) and later approximations?

Industrial Production (a) Economists often use the expression expression "rate of growth" in relative rather than absolute terms. For example, let \(u=f(t)\) be the number of people in the labor force at time \(t\) in a given industry. (We treat this function as though it were differentiable even though it is an integer-valued step function.) Let \(v=g(t)\) be the average production per person in the labor force at time \(t .\) The total production is then \(y=u v\) . If the labor force is growing at the rate of 4\(\%\) per year year \((d u / d t=\) 0.04\(u\) ) and the production per worker is growing at the rate of 5\(\%\) per year \((d v / d t=0.05 v),\) find the rate of growth of the total production, y. (b) Suppose that the labor force in part (a) is decreasing at the rate of 2\(\%\) per year while the production per person is increasing at the rate of 3\(\%\) per year. Is the total production increasing, or is it decreasing, and at what rate?

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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