Problem 95 - College Algebra with Modeling and Visualization

Question

Solve the variation problem. Suppose \(T\) varies directly as the \(\frac{3}{2}\) power of \(x .\) When \(x=4, T=20 .\) Find \(T\) when \(x=16\)

Step-by-Step Solution

Answer

When \(x = 16\), \(T = 160\).

Step 1: Understand the Variation Statement

The problem states that \( T \) varies directly as the \( \frac{3}{2} \)-power of \( x \). This means we can express this relationship with the equation \( T = kx^{\frac{3}{2}} \), where \( k \) is the constant of variation.

Step 2: Determine the Constant of Variation

Given that when \( x = 4 \), \( T = 20 \), substitute these values into the equation to find \( k \): \[ 20 = k \cdot (4)^{\frac{3}{2}} \] Calculate \( (4)^{\frac{3}{2}} \): \( (4)^{\frac{3}{2}} = (4^{\frac{1}{2}})^3 = (2)^3 = 8 \). Now substitute: \[ 20 = k \cdot 8 \] \[ k = \frac{20}{8} = 2.5 \]. The constant \( k \) is \( 2.5 \).

Step 3: Use the Constant to Find New T

Now that \( k = 2.5 \), use it to find the new \( T \) when \( x = 16 \). Substitute \( x = 16 \) into the equation: \[ T = 2.5 \times (16)^{\frac{3}{2}} \]. Calculate \( (16)^{\frac{3}{2}} \): \( (16)^{\frac{3}{2}} = (16^{\frac{1}{2}})^3 = (4)^3 = 64 \). Substitute back: \[ T = 2.5 \times 64 = 160 \].

Key Concepts

Constant of Variation

In variation problems, the constant of variation is a crucial component. It acts like a "glue" that holds the variables together in a mathematical relationship. Basically, it's a constant number that shows how one variable changes with respect to another. In direct variation, this relationship can be represented as \( y = kx \), where \( k \) is the constant of variation.

In our problem, we expressed the relationship as \( T = kx^{\frac{3}{2}} \). This equation tells us how \( T \) (the dependent variable) changes when \( x \) (the independent variable) is raised to the power of \( \frac{3}{2} \).

To find \( k \), we used the initial values provided: when \( x = 4 \), \( T = 20 \). Substituting them into the equation, we calculated \( k = 2.5 \). With this constant, you can determine \( T \) for different values of \( x \).

Remember, \( k \) doesn't change for this specific relationship. It remains constant, making it a fundamental part of solving variation problems.

Direct Variation

Direct variation is a beautiful, simple concept in mathematics. It's when two variables change in such a way that their ratio always stays the same. If one variable increases, the other one does too, in a predictable manner.

In our specific scenario, \( T \) varies directly with \( x^{\frac{3}{2}} \). This means as \( x \) changes, \( T \) changes according to the cube root squared of \( x \).

The key feature of direct variation is that the relationship between the two variables is linear with respect to the variations in their powers and expressions. The equation \( T = kx^{\frac{3}{2}} \) shows how \( T \) is directly proportional to \( x^{\frac{3}{2}} \) with a constant ratio \( k = 2.5 \).

So, if you know one variable and the constant of variation, you can always find the other variable using this direct relationship. It's a straightforward but powerful tool for understanding how variables interact.

Exponential Powers

Exponential powers are numbers that show how many times a number (the base) is multiplied by itself. They're an effective way to express large numbers or show recurring patterns. In variation problems, they serve as an essential tool for capturing these relationships.

In our exercise, we dealt with \( x^{\frac{3}{2}} \). This power represents a combination of two operations: first, taking the square root, and then cubing the result. For example, when \( x = 16 \), the calculation goes as follows:

• Find the square root of 16, which is 4.
• Then raise 4 to the 3rd power, resulting in 64.

Multiply by the constant \( k \) to find \( T \). Exponential expressions like \( x^{\frac{3}{2}} \) help us capture the essence of how \( T \) varies with \( x \) beyond simple multiplication or addition.

These expressions enrich the way we understand relationships and enable us to model real-world scenarios with more precision. They highlight the nuances and intricacies that linear calculations might miss.