Problem 58 - Basic Technical Mathematics with Calculus

Question

Solve the given problems. In the theory of the motion of a sphere moving through a fluid, the function \(f(r)=4 r^{3}-3 a r^{2}-a^{3}\) is used. Is \((\text { a) } r=a\) or (b) \(r=2 a\) a zero of \(f(r) ?\)

Step-by-Step Solution

Answer

(a) \(r = a\) is a zero; (b) \(r = 2a\) is not a zero.

Step 1: Substitute for Part (a)

For part (a), we substitute \( r = a \) into the function \( f(r) = 4r^3 - 3ar^2 - a^3 \). So, we have:\[ f(a) = 4a^3 - 3a \times a^2 - a^3 \] Simplifying this gives us \( f(a) = 4a^3 - 3a^3 - a^3 \).

Step 2: Simplify Part (a)

Simplify the expression from Step 1:\[ f(a) = (4a^3 - 3a^3 - a^3) = 0 \] Since the result is zero, \( r = a \) is a zero of \( f(r) \).

Step 3: Substitute for Part (b)

For part (b), substitute \( r = 2a \) into the function \( f(r) = 4r^3 - 3ar^2 - a^3 \). Therefore:\[ f(2a) = 4(2a)^3 - 3a(2a)^2 - a^3 \]Calculate the powers:\[ (2a)^3 = 8a^3 \text{ and } (2a)^2 = 4a^2 \]

Step 4: Simplify Part (b)

Now substitute the calculated values back into the expression:\[ f(2a) = 4 \times 8a^3 - 3a \times 4a^2 - a^3 \] Simplify each term:\[ f(2a) = 32a^3 - 12a^3 - a^3 \]\[ f(2a) = (32a^3 - 13a^3) = 19a^3 \] Since \( f(2a) eq 0 \), \( r = 2a \) is not a zero of \( f(r) \).

Key Concepts

Polynomials

Polynomials are mathematical expressions made of variables and constants combined using addition, subtraction, multiplication, and non-negative integer exponents. Each term in a polynomial is composed of a coefficient (a constant) and one or more variables raised to a whole number power. In the function \( f(r) = 4r^3 - 3ar^2 - a^3 \), this is a cubic polynomial since the highest power of the variable \( r \) is 3.

A polynomial can have one or several terms. These terms are typically organized in decreasing order of their powers. In our example, the terms are \( 4r^3 \), \(-3ar^2 \), and \(-a^3 \). These terms indicate that as \( r \) changes, it influences the value of the polynomial based on the power to which it is raised. This is crucial when investigating where or if the polynomial crosses the x-axis (where \( f(r) = 0 \)).

Zeros of Functions

The zeros of a function, often referred to as the roots or solutions, are the values of \( r \) for which the function equals zero, \( f(r) = 0 \). Finding these zeros is important because they are the points where the graph of the function intersects the x-axis.

In our exercise, we need to determine if \( r = a \) or \( r = 2a \) are zeros of the function \( f(r) = 4r^3 - 3ar^2 - a^3 \). This involves substituting these values into the function and simplifying the expressions. If substituting a certain value results in zero, that value is a zero of the function. This provides insight into the behavior of the function, especially in applications like theoretical physics, where understanding the intersection points (or lack thereof) might signify something crucial about the system being described.

Step-by-Step Solutions

Step-by-step solutions are extremely helpful for understanding how to solve problems, especially in mathematics. They break down the problem-solving process into manageable parts, making it easier to follow and understand.

For the function \( f(r) = 4r^3 - 3ar^2 - a^3 \), the solution steps are:

• Substitute the value: First, plug the suspected zero into the function where \( r \) appears, one at a time (e.g., \( r = a \) or \( r = 2a \)).
• Simplify the expression: After substitution, simplify the algebraic expression completely.
• Test the result: Check if the final simplified value equals zero. If it does, the substituted value is a zero of the function; if not, it isn't.

By dividing the process into these steps, complex algebra becomes clear and organized, reducing mistakes and enhancing understanding. This methodology is not only specific to our problem but applies to any algebraic solving scenario, ensuring students develop strong problem-solving skills.