Problem 39 - Precalculus with Limits

Question

In Exercises 37-52, evaluate the function at each specified value of the independent variable and simplify. \(V(r) = \frac{4}{3}\pi r^3\) (a) \(V(3)\) (b) \(V(\frac{3}{2})\) (c) \(V(2r)\)

Step-by-Step Solution

Answer

\(V(3) = 36\pi\) cube units, \(V(3/2) = \frac{27\pi}{2}\) cube units, and \(V(2r) = \frac{32}{3}r^3\pi\) cube units

Step 1: Evaluate V(3)

To find \(V(3)\), substitute \(r = 3\) in the equation for V(r). \n Therefore it's \(V(3) = \frac{4}{3}\pi (3)^3 = 36\pi\) cubes units.

Step 2: Evaluate V(3/2)

To find \(V(3/2)\), substitute \(r = 3/2\) in the equation for V(r). \n Therefore it's \(V(\frac{3}{2}) = \frac{4}{3}\pi (\frac{3}{2})^3 = \frac{27\pi}{2}\) cube units.

Step 3: Evaluate V(2r)

To find \(V(2r)\), substitute \(r = 2r\) in the function. We get \(V(2r) = \frac{4}{3}\pi (2r)^3 = \frac{32}{3}r^3\pi\) cube units.

Key Concepts

Precalculus Functions

Precalculus functions play a critical role in forming the foundation for calculus. They describe mathematical relationships and processes that will be explored in depth in calculus. In this scenario, we explore the function that calculates the volume of a sphere, a classic example of a precalculus problem. The volume function is given by \[V(r) = \frac{4}{3}\pi r^3\]where \(r\) represents the radius of the sphere. With precalculus, we interpret this function to see how changes in \(r\) affect \(V(r)\). This function exemplifies how altering an input value modifies the output, thus teaching us about direct relationships and dependencies in functions.
This understanding prepares students for more advanced calculus topics such as derivatives and integrals.

Substitution

Substitution is a technique used to evaluate functions by replacing variables with specific values or expressions. In this exercise, the concept of substitution is used to compute the volume for different values of \(r\).

• For \(V(3)\), substitution involves replacing \(r\) with 3, leading to the calculation of the specific volume for a sphere with radius 3.
• When evaluating \(V(\frac{3}{2})\), we substitute \(r\) with \(\frac{3}{2}\), demonstrating how substitution can involve fractions or decimals.
• Lastly, for \(V(2r)\), \(r\) is substituted by \(2r\), showcasing how substitution allows for variable expressions as inputs.

This method is powerful for evaluating the function at particular instances or for transforming expressions into a more workable form.

Simplifying Expressions

Simplifying expressions involves the process of rewriting expressions to their most comprehensible form. Once substitution is done, the expressions frequently need simplification to make sense of the results.

• In \(V(3)\), \((3)^3\) is simplified to 27, leading to \(36\pi\).
• With \(V(\frac{3}{2})\), \((\frac{3}{2})^3\) is simplified to \(\frac{27}{8}\), which further manipulates to \(\frac{27\pi}{2}\).
• For \(V(2r)\), \((2r)^3\) simplifies to \(8r^3\), resulting in \(\frac{32}{3}r^3\pi\).

Simplification ensures that the results are not only correct but also easily interpretable and concise, preparing expressions for further work or analysis.

Mathematical Operations

Mathematical operations form the backbone of solving equations and evaluating functions. This includes basic operations like addition, subtraction, multiplication, and division, as well as exponentiation and working with irrational numbers like \(\pi\).
For instance,

• The operation \(r^3\) involves exponentiation, where the radius is cubed, indicating a three-dimensional volumetric quantity.
• Multiplication is used to scale the cube's value by \(\frac{4}{3}\pi\), providing the sphere's volume after integrating \(\pi\).
• Each step follows a logical sequence where operations align with defined mathematical principles to achieve accurate results and solidify understanding of spatial calculations.

Through mastering these operations, students enhance their skill in tackling a vast array of problems, from basic to advanced mathematical challenges.