Problem 5 - Algebra and Trigonometry

Question

Use the model given to answer the questions about the object or process being modeled. The power \(P\) measured in horsepower (hp) needed to drive a certain ship at a speed of \(s\) knots is modeled by $$ P=0.06 s^{3} $$ (a) Find the power needed to drive the ship at 12 knots. (b) At what speed will a 7.5 -hp engine drive the ship?

Step-by-Step Solution

Answer

(a) 103.68 hp; (b) 5 knots.

Step 1: Identify Given Values for Part (a)

In part (a) of the problem, we are given the speed \( s = 12 \) knots. We need to find the power \( P \) using the provided formula \( P = 0.06 s^3 \).

Step 2: Substitute and Calculate for Part (a)

Using the given formula \( P = 0.06 s^3 \), substitute \( s = 12 \) into the equation: \( P = 0.06 \times 12^3 \).

Step 3: Compute Exponent and Multiply

Calculate \( 12^3 = 1728 \). Then, compute \( P = 0.06 \times 1728 = 103.68 \). Thus, the power needed at 12 knots is 103.68 hp.

Step 1: Identify Given Values for Part (b)

In part (b), we need to find the speed \( s \) such that the power \( P = 7.5 \) hp. We'll use the formula \( P = 0.06 s^3 \).

Step 2: Set up the Equation for Part (b)

Given \( P = 7.5 \), use the formula \( P = 0.06 s^3 \) and solve for \( s \): \( 7.5 = 0.06 s^3 \).

Step 3: Solve for s

Rearrange the equation to isolate \( s^3 \): \( s^3 = \frac{7.5}{0.06} \). Compute \( s^3 = 125 \).

Step 4: Calculate the Cube Root

Find the cube root of 125: \( s = \sqrt[3]{125} = 5 \). Hence, a 7.5-hp engine will drive the ship at 5 knots.

Key Concepts

Power Equation

In mathematics, a power equation involves exponents, which are numbers that denote repeated multiplication of a base number. The given model in the exercise is a perfect example: - Power, \( P = 0.06s^3 \), links the speed \( s \) of a ship with the power \( P \) needed. - Here, the variable \( s \) is raised to the power of 3, indicating the cube of \( s \).
This type of equation is used often in physics and engineering to model how changes in one variable affect another through exponential relationships. The exponent (in this case, 3) informs us that the power is directly proportional to the cube of the speed, showing a nonlinear relationship. Understanding how to manipulate and apply power equations is crucial in predicting outcomes when the quantities involved are related by an exponent.

Substitution Method

The substitution method is a straightforward technique to evaluate expressions or solve equations by replacing variables with known values. For instance, in part (a) of the exercise:- We substitute the speed \( s = 12 \) into our equation: \( P = 0.06 \, (12^3) \).- This sets up the expression for direct computation of power.
The process of substitution simplifies problems and allows us to focus on arithmetic operations rather than algebraic manipulations. It is a widely used technique not only to find specific values in models but also to validate solutions once the general form of an equation is understood. Practicing substitution reinforces the understanding of variable roles in an equation.

Cube Root Calculation

Calculating a cube root, represented as \( \sqrt[3]{x} \), is the process of determining what number, when multiplied by itself three times, results in \( x \). It was used in part (b) of the exercise:- To find the speed \( s \) from \( s^3 = 125 \).- Solve it by calculating \( \sqrt[3]{125} = 5 \).
Cube roots are particularly essential when solving cubic equations, where the unknown variable is raised to the third power. Understanding how to calculate cube roots is vital for rearranging equations to isolate variables in real-world applications, such as determining unknown values in physics models or during engineering design computations.

Solve for Variable

Solving for a variable involves rearranging an equation to isolate the variable of interest. In our exercise, particularly in part (b):- Given \( P = 7.5 \), we start with \( 7.5 = 0.06s^3 \).- Rearranging for \( s^3 \), we get \( s^3 = \frac{7.5}{0.06} \).
Solving for a variable is a fundamental skill in algebra. It allows us to find unknown values and make predictions based on equations. It's important to apply correct arithmetic operations, maintain balance within the equation and interpret the physical meaning of the solution in the context of a problem. This skill is not only central to algebra but also forms the basis of problem-solving in various scientific and technical fields.