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Chapter 3: Problem 7

Find the rate of change of the volume of a sphere with respect to the radius \(r\).

Short Answer

Expert verified
The rate of change of the volume of a sphere with respect to the radius \(r\) is \(4\pi r^2\).

Step by step solution

01

Identify the formula for the volume of a sphere

The formula for the volume of a sphere with radius \(r\) is \(V = \frac{4}{3} \pi r^3\)
02

Derive the volume formula with respect to \(r\)

We take the derivative of \(V\) with respect to \(r\) using the power rule for derivatives, which states that the derivative of \(x^n\) with respect to \(x\) is \(nx^{n-1}\). Applying this rule, we get \(\frac{dV}{dr} = 4\pi r^2\).
03

Interpret the result

The rate of change of the volume of a sphere with respect to the radius \(r\) is \(\frac{dV}{dr} = 4\pi r^2\). This means that for each unit increase in the radius, the volume of the sphere increases by \(4\pi r^2\).

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Key Concepts

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

Sphere Volume
The volume of a sphere is a measure of the space contained within its surface. For a sphere with radius \( r \), the volume \( V \) can be calculated using the formula:
  • \( V = \frac{4}{3} \pi r^3 \)
This formula shows us that the volume of a sphere not only depends on its radius but increases dramatically as the radius increases. This is because the radius is cubed in the formula, highlighting how even a small increase in radius leads to a large increase in volume.
It is important to know that \( \pi \) (Pi) is a constant, approximately equal to 3.14159, and it represents the ratio of the circumference of a circle to its diameter.
Derivative
A derivative is a fundamental tool in calculus that describes how a function changes as its input changes. In simple terms, the derivative of a function gives us the rate at which one quantity changes with respect to another.
When we calculate the derivative, we are looking to understand how one variable influences another variable's rate of change. For the volume of a sphere, we calculate the derivative \( \frac{dV}{dr} \), which represents the rate of change of volume with respect to the radius.
Derivatives are crucial in understanding behaviors of mathematical functions, and they are widely used in various scientific fields to model real-world phenomena.
Power Rule
The power rule is a basic derivative rule used in calculus to simplify the differentiation process of functions that are powers of a variable. The power rule states:
  • If \( f(x) = x^n \), then the derivative \( f'(x) = nx^{n-1} \).
Essentially, to find the derivative, you multiply by the original power and subtract one from the exponent.
This rule makes finding derivatives straightforward, especially for polynomials. In the context of our sphere volume example, we apply the power rule to \( r^3 \) when determining \( \frac{dV}{dr} \). Hence, \( 3r^2 \) turns into \( 4\pi r^2 \) with \( 4\pi \) as a constant factor.
Radius
The radius of a sphere is the distance from its center to any point on its surface. It is a crucial measurement, as it directly affects not just the volume, but other properties of the sphere as well.
In our formula for the volume of a sphere, the radius is cubed \(( r^3 )\), pointing to its significant influence on the calculation. Changes in the radius will lead to substantial changes in the sphere's volume due to this cubed relationship.
Understanding changes in radius is vital for applications involving spherical objects, whether it's in physics, engineering, or geometry.

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Most popular questions from this chapter

Find a function \(f\) with the given derivative. Check your answer by differentiation. $$f^{\prime}(x)=2 \cos x-3 \sin x$$

Let \(f(x)=1 / x, x>0 .\) Show that the triangle that is formed by each line tangent to the graph of \(f\) and the coordinate axes has an area of 2 square units.

Use a graphing utility to determine where (a) \(f^{\prime}(x)=0 ; \quad\) (b) \(f^{\prime}(x)=0 ; \quad\) (c) \(f^{\prime}(x)<0\). \(f(x)=\frac{x^{2}+1}{\sqrt{x}}\).

Find equations for all the lines tangent to the graph of \(f(x)=x^{3}-x\) that pass through the point (-2,2)

Find two lines through the point (2,8) that are tangent to the graph of \(f(x)=x^{3}\).
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