Problem 34 - Algebra and Trigonometry Real Mathematics, Real People

Question

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) \(V\left(\frac{3}{2}\right)\) (c) \(V(2 r)\)

Step-by-Step Solution

Answer

The volume value of the sphere with radius 3 is \(V(3) = 36\pi\), with radius 1.5 is \(V(3/2) = \frac{27}{2}\pi \) and with radius 2r is \(V(2r) = \frac{32}{3} r^{3} \pi \).

Step 1: Evaluate V(3)

To find V(3), substitute r = 3 into the equation. This yields \( V(3)=\frac{4}{3} \pi (3)^{3} = 36\pi \).

Step 2: Evaluate V(3/2)

To find V(3/2), substitute r = 3/2 in the function. This yields \( V(3/2) = \frac{4}{3} \pi (1.5)^{3} = \frac{27}{2}\pi \).

Step 3: Evaluate V(2r)

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

Key Concepts

Function Evaluation

When we talk about function evaluation, we refer to the process of finding the result of a function at given values of its variables. In the context of the exercise provided, the function is the volume of a sphere, given by the formula: \[ V(r)=\frac{4}{3} \pi r^{3} \] Function evaluation involves substituting specific values for the variable \( r \) to calculate the volume. For example, evaluating \( V(3) \) means you replace \( r \) with 3 in the formula, resulting in the calculation \( 36\pi \). The core idea is to understand how input values affect the output of the function. This is essential in mathematics and helps in understanding how functions model real-world situations.

Independent Variable

In mathematical expressions and equations, an independent variable is a variable whose variation does not depend on another variable. In our volume function, \( r \) is the independent variable.

• The independent variable is what you change to see how the function behaves. In the formula \( V(r)=\frac{4}{3} \pi r^{3} \), changing \( r \) will change the volume of the sphere.
• This demonstrates the relationship between the radius \( r \) and the volume \( V \).
• It's important because it allows us to comprehend how variations in \( r \) result in changes to the calculated volume.

Understanding the role of an independent variable is vital for exploring relationships in mathematical models.

Algebraic Expressions

Algebraic expressions are combinations of numbers, variables, and mathematical operators. They are used to represent mathematical relationships in a compact form. In the sphere's volume formula, \( \frac{4}{3} \pi r^{3} \) is an algebraic expression.

• The components \( \frac{4}{3}, \pi, \) and \( r^{3} \) denote multiplication, a constant, and exponentiation, respectively.
• These elements together express how the radius \( r \) relates to the volume \( V \).
• Algebraic expressions like this one allow us to perform substitution and manipulation based on algebraic rules.

Grasping algebraic expressions is crucial for solving and simplifying mathematical problems, as seen in evaluating the volume function.

Mathematical Simplification

Mathematical simplification is the process of altering a mathematical expression into a simpler or more comprehensible form. In the example of our sphere volume function, once you perform the specific substitutions of \( r \), it’s important to simplify your results to get clear and usable answers.

• For instance, when you evaluate \( V(3) \), the expression simplifies to \( 36\pi \) from \( \frac{4}{3} \pi 27 \).
• Similarly, evaluating \( V(\frac{3}{2}) \) simplifies from a more complex fraction to \( \frac{27}{2}\pi \).
• Simplification often involves reducing fractions, combining like terms, or canceling out terms.

Simplification is key in mathematics as it makes expressions easier to work with and understand, especially in developing solutions to problems like the ones in this exercise.