Problem 51 - Elementary and Intermediate Algebra

Question

Solve each problem by writing a variation model. When playing with a Skip It toy, a child swings a weighted ball on the end of a string in a circular motion around one leg while jumping over the revolving string with the other leg. See the illustration. The tension \(T\) in the string is directly proportional to the square of the speed \(s\) of the ball and inversely proportional to the radius \(r\) of the circle. If the tension in the string is 6 pounds when the speed of the ball is 6 feet per second and the radius is 3 feet, find the tension when the speed is 8 feet per second and the radius is 2.5 feet. CAN'T COPY THE IMAGE

Step-by-Step Solution

Answer

The tension when the speed is 8 feet per second and the radius is 2.5 feet is 12.8 pounds.

Step 1: Understanding the Variation Model

The problem describes a situation where the tension \( T \) in the string is directly proportional to the square of the speed \( s \) and inversely proportional to the radius \( r \) of the circle. This can be expressed with the equation \( T = k \frac{s^2}{r} \), where \( k \) is the constant of proportionality.

Step 2: Find the Constant of Proportionality

Using the information that when \( T = 6 \) pounds, \( s = 6 \) feet per second, and \( r = 3 \) feet, we substitute these values into the equation: \( 6 = k \frac{6^2}{3} \). Simplify to find \( 6 = k \cdot 12 \). Thus, \( k = \frac{6}{12} = \frac{1}{2} \).

Step 3: Formulate the General Equation

Now we have the general equation \( T = \frac{1}{2} \frac{s^2}{r} \) using the constant \( k = \frac{1}{2} \).

Step 4: Calculate the New Tension

Substituting \( s = 8 \) feet per second and \( r = 2.5 \) feet into the general equation gives \( T = \frac{1}{2} \frac{8^2}{2.5} \). Simplify to find \( T = \frac{1}{2} \frac{64}{2.5} = \frac{32}{2.5} \). Further simplifying, \( T = \frac{32}{2.5} \times \frac{10}{10} = \frac{320}{25} = 12.8 \) pounds.

Key Concepts

Proportionality

Proportionality is a fundamental concept in mathematics that helps us understand how quantities are related to each other. When we say two variables are proportional, it means that as one variable increases or decreases, the other does so as well at a consistent rate. This consistency can be direct or inverse:

• Constant of Proportionality: This is a fixed value that provides the link between two proportional quantities. In our exercise, this constant is represented by the symbol \( k \).
• Direct Proportionality: If one variable increases, the other increases by a proportionate amount.
• Inverse Proportionality: If one variable increases, the other decreases at a corresponding rate.

In mathematical terms, if quantity \( a \) is proportional to quantity \( b \), we express it as \( a = kb \), where \( k \) is the proportionality constant. Understanding this is crucial for solving variation problems.

Direct Variation

In direct variation, two variables move together in the same direction. The relationship between them is straightforward: if one goes up, so does the other, and vice versa. This is often expressed in the form of an equation like \( y = kx \), where \( k \) is the constant of proportionality.

In the given problem, the tension \( T \) in the string is directly proportional to the square of the speed \( s \). Mathematically, this can be expressed as \( T \propto s^2 \), meaning \( T = ks^2 \). Consequently, as the speed of the ball increases, the tension will increase if the radius remains constant. Direct variation is all about keeping things consistent, and the consistency is determined by the constant \( k \), which acts as a scaling factor.

Inverse Variation

Inverse variation is a bit more complex than direct variation because the variables move in opposite directions. When one variable increases, the other decreases in such a way that their product is constant. The typical form for an inverse variation equation is \( xy = k \) or \( y = \frac{k}{x} \).

In our scenario, the tension \( T \) is inversely proportional to the radius \( r \) of the circle. This translates into the equation \( T = \frac{k}{r} \). Here, as the radius of the circle increases, the tension decreases, assuming the speed remains constant. This relationship helps balance the forces present in the system and illustrates how both types of variation work together.

Algebraic Expressions

Algebraic expressions are combinations of numbers, variables, and operations that represent a mathematical idea or relationship. They are essential tools for solving problems, especially those involving proportional relationships.

In this exercise, the expression \( T = k \frac{s^2}{r} \) is a representation that connects tension with speed and radius. Each element serves a specific purpose:

• \( T \) represents the tension in the string.
• \( s^2 \) indicates that speed impacts tension directly and quadratically.
• \( r \) shows how radius inversely affects tension.
• \( k \) ties everything together, ensuring the proportional relationships hold.

Through manipulation of the expression, we can substitute known values to find unknowns, making algebraic expressions invaluable in understanding and solving proportionality problems.