Problem 31 - Calculus Early Transcendentals

Question

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

Step-by-Step Solution

Answer

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

Step 1: Identify the Function

We start with the function given for volume: \( V = \pi r^3 \). This indicates that the volume \( V \) is expressed as a function of the radius \( r \).

Step 2: Apply the Power Rule for Derivatives

The derivative of \( r^n \) with respect to \( r \) is \( n \cdot r^{n-1} \). Here, the function is \( V = \pi r^3 \), so we're looking for the derivative \( V'(r) \).

Step 3: Differentiate the Function

When differentiating \( V = \pi r^3 \), treat \( \pi \) as a constant. The derivative with respect to \( r \) is: \(V'(r) = \pi \cdot 3r^2 = 3\pi r^2 \).

Key Concepts

Power Rule

The power rule is a fundamental tool in calculus used for differentiating functions of the form \( f(x) = x^n \). It states that the derivative of \( x^n \) with respect to \( x \) is \( n \cdot x^{n-1} \). This rule simplifies the process of finding derivatives because you only need to multiply the exponent by the base and reduce the exponent by one.

In the context of the original exercise, we have the function \( V = \pi r^3 \). Applying the power rule here involves:

• Identifying \( r^3 \), where "3" is the exponent.
• Taking the exponent "3" in \( r^3 \) and multiplying it by the constant \( \pi \). This gives \( 3\pi \).
• Decreasing the power by one: \( r^{3-1} = r^2 \).

Therefore, the derivative of \( V = \pi r^3 \) with respect to \( r \) becomes \( V'(r) = 3\pi r^2 \). Using the power rule turns what could be a complex differentiating process into straightforward arithmetic.

Derivative

A derivative represents the rate at which a function is changing at any given point. In terms of graphs, it's the slope of the tangent line to the curve at a particular point.

When we look at the function \( V = \pi r^3 \), the derivative \( V'(r) \) tells us how the volume changes as the radius \( r \) changes. In practical terms for this problem, it gives the rate of change of volume relative to the radius of a sphere. This is particularly useful when modeling real-world scenarios where it's crucial to know how small changes in one variable affect another.

To find a derivative, especially when using simple rules like the power rule, follow these steps:

• Identify the function and which variable it is differentiating with respect to.
• Apply known differentiation rules like the power rule.
• Simplify the expression to arrive at a clean derivative.

Each derivative carries rich information about how the function behaves and responds to changes in its variables.

Function of a Variable

A function of a variable is a basic notion in mathematics where each input into the function (the variable) corresponds to exactly one output. In the problem context, \( V = \pi r^3 \), the volume, \( V \), is described as a function of the radius \( r \). This implies that for any given value of \( r \), you will get a specific value of volume \( V \).

Understanding functions is vital in calculus and mathematics as they model relationships and eventual changes between varying quantities:

• Each function expresses a relationship. For \( V = \pi r^3 \), it's the relationship between the volume of a sphere and its radius.
• They can be visual, for instance, through graphs depicting how one variable affects another.
• Functions often depict real-world scenarios, predicting outcomes based on changing conditions.

In differentiating, focusing on functions allows us to understand and anticipate how alterations in the input, like the radius in this problem, affect the output, here the volume. Recognizing \( V(r) \) as a function alerts us to the broader relationship modeled by the equation.