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Chapter 2: Problem 31

Find the indicated derivative. $$V^{\prime}(r), \text { where } V=\pi r^{3}$$

Short Answer

Expert verified
The derivative is \( V'(r) = 3\pi r^2 \).

Step by step solution

01

Identify the Function

The function given is the volume function, where \( V = \pi r^3 \). This indicates that \( V \) is a function of \( r \).
02

Apply the Power Rule for Derivatives

The power rule for derivatives states that if you have a function of the form \( r^n \), its derivative is \( n \times r^{n-1} \). For our function \( V = \pi r^3 \), \( n = 3 \). Thus, the derivative of \( r^3 \) is \( 3r^2 \).
03

Multiply by the Constant Coefficient

Since \( \pi \) is a constant multiplier in the original function \( V = \pi r^3 \), we multiply the derivative \( 3r^2 \) by \( \pi \). Therefore, \( V'(r) = 3\pi r^2 \).
04

State the Derivative

The derivative of the volume function \( V \) with respect to \( r \) is \( V'(r) = 3 \pi r^2 \). This tells us how the volume changes as the radius \( r \) changes.

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Key Concepts

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

Power Rule
The power rule is a fundamental concept in calculus used to find derivatives of functions with exponents. When you have a function like \( r^n \), the power rule helps us determine the derivative efficiently. The rule states: take the exponent \( n \) and multiply it by the base raised to the power of \( n-1 \).
For example:
  • If \( f(r) = r^3 \), then the derivative \( f'(r) \) is \( 3r^2 \).
This operation reduces the exponent by one, which simplifies the process of finding the rate at which the function changes. The power rule is particularly useful as it can be applied to any polynomial function, simplifying each individual term.
Volume Function
In calculus, a volume function represents how the volume of a geometric shape changes with respect to a variable. For a sphere or similar objects, it could depend on the radius. In the exercise given, the volume function is \( V = \pi r^3 \).
Here,
  • \( \pi \) is a constant, which represents the mathematical constant pi.
  • \( r \) is the radius of the sphere.
The function describes how the volume of the sphere changes as the radius \( r \) increases or decreases. Understanding the volume function is crucial in applied physics and engineering, where the size and mass of objects must be modeled accurately.
Constant Multiplier
A constant multiplier in a function is a coefficient that remains unchanged regardless of the variable. When taking the derivative of such a function, this constant simply multiplies the derivative of the variable component. For the given function \( V = \pi r^3 \), \( \pi \) is the constant multiplier.
  • When differentiating, you first apply the power rule to \( r^3 \) to get \( 3r^2 \).
  • Then, multiply this result by the constant \( \pi \), giving us \( 3\pi r^2 \).
This step shows how constants affect the rate of change of a function, retaining their presence while not altering the math operations on the variable components.
Function of a Variable
A function of a variable expresses how one quantity, dependent on another, changes as that variable changes. In our exercise, the volume \( V \) depends on the radius \( r \), which means \( V \) is a function of \( r \). This relationship is essential because it helps to understand how a change in radius impacts the volume.
When dealing with functions of a variable, consider:
  • The independent variable (\( r \)) is the input.
  • The dependent variable (\( V \)) is what changes according to the input.
By differentiating \( V \) with respect to \( r \), we get \( V'(r) \), showing the sensitivity of the volume to changes in the radius. This concept is key in calculus as it allows us to model dynamic systems in science and engineering.

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Most popular questions from this chapter

Find \(d y / d x\) $$y=\left(\frac{1+x^{2}}{1-x^{2}}\right)^{17}$$

Find \(d y / d x\) $$y=x^{3} \sin ^{2}(5 x)$$

Given the following table of values, find the indicated derivatives in parts (a) and (b). $$\begin{array}{|c|c|c|} \hline x & f(x) & f^{\prime}(x) \\ \hline 2 & 1 & 7 \\\ \hline 8 & 5 & -3 \\ \hline \end{array}$$ (a) \(g^{\prime}(2),\) where \(g(x)=[f(x)]^{3}\) (b) \(h^{\prime}(2),\) where \(h(x)=f\left(x^{3}\right)\)

A function \(y=f(x)\) and an \(x\) -value \(x_{0}\) are given. (a) Find a formula for the slope of the tangent line to the graph of \(f\) at a general point \(x=x_{0}\). (b) Use the formula obtained in part (a) to find the slope of the tangent line for the given value of \(x_{0}\). $$f(x)=x^{2}-1 ; x_{0}=-1$$

Find the indicated derivative. $$y=\cot ^{3}(\pi-\theta) ; \text { find } \frac{d y}{d \theta}$$
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