If P Varies Directly As R Square P =3.2 When R=4 Find The Value Of P When R=6.5

If P Varies Directly As R Square P =3.2 When R=4 Find The Value Of P When R=6.5

Understanding the concept of direct variation is fundamental in algebra and helps in solving real-world problems involving proportional relationships. In this article, we will explore the problem statement involving the direct variation of P with respect to R squared, analyze how to determine the constant of variation, and find the value of P when R is 6.5 based on the given conditions. Let's delve into the details step by step.

Understanding Direct Variation

What Is Direct Variation?

Direct variation describes a relationship between two variables where one varies directly as the other. Mathematically, if P varies directly as R squared, then the relationship can be expressed as:

\[ P = k R^2 \]

where:

• \( P \) and \( R \) are the variables,


• \( k \) is the constant of variation.

This means that as R increases, P increases proportionally to R squared, and vice versa.

Real-World Examples of Direct Variation

• The area of a square as a function of its side length, where the area varies directly as the square of the side length.
• The force between two charged particles varies directly with the product of their charges and inversely with the square of the distance (Coulomb's Law).
• The volume of a gas at constant temperature varies directly with pressure and inversely with volume, which relates to other laws in physics.

Formulating the Problem

Given:

• \( P \) varies directly as \( R^2 \),


• \( P = 3.2 \) when \( R = 4 \),


• Find \( P \) when \( R = 6.5 \).

The key steps involve:

1. Establishing the constant of variation \( k \),


2. Applying the relationship to find the new value of \( P \).

Step 1: Finding the Constant of Variation \( k \)

Since \( P \) varies directly as \( R^2 \), the general formula is:

\[ P = k R^2 \]

Using the known values \( P = 3.2 \) when \( R = 4 \):

\[ 3.2 = k \times 4^2 \]
\[ 3.2 = k \times 16 \]

Solving for \( k \):

\[ k = \frac{3.2}{16} \]
\[ k = 0.2 \]

This constant \( k \) represents the proportionality factor in the direct variation relationship.

Step 2: Calculating \( P \) When \( R = 6.5 \)

Now that we know \( k = 0.2 \), we can substitute \( R = 6.5 \) into the formula:

\[ P = 0.2 \times (6.5)^2 \]

Calculating \( (6.5)^2 \):

\[ 6.5^2 = 6.5 \times 6.5 = 42.25 \]

Then,

\[ P = 0.2 \times 42.25 \]
\[ P = 8.45 \]

Therefore, the value of \( P \) when \( R = 6.5 \) is 8.45.

Summary of the Solution

• The relationship between \( P \) and \( R^2 \) is \( P = k R^2 \).
• Using the initial data, the constant \( k \) is calculated as 0.2.
• Applying this constant to the new value of \( R \), the value of \( P \) is found to be 8.45.

Additional Tips for Solving Similar Problems

1. Understand the Relationship

Make sure to identify whether the variables are directly or inversely related. This determines the form of the equation you will use.

2. Find the Constant of Variation

Use known data points to solve for the constant \( k \) by substituting the known values into the general formula.

3. Apply the Formula to New Data

Once \( k \) is known, substitute the new value of the independent variable to find the corresponding dependent variable.

Common Mistakes to Avoid

• Forgetting to square \( R \) in the formula when the variation involves \( R^2 \).
• Mixing up the variables or constants, leading to incorrect calculations.
• Not checking units or context if applying to real-world problems.

Practice Problems

1. If \( Q \) varies directly as \( T^2 \), and \( Q = 5 \) when \( T = 3 \), find \( Q \) when \( T = 7 \).
2. The volume \( V \) of a sphere varies directly as the cube of its radius \( r \). If \( V = 36 \) when \( r = 2 \), what is the volume when \( r = 5 \)?
3. The speed \( S \) of an object varies directly as the square root of time \( t \). If \( S = 10 \) when \( t = 4 \), find \( S \) when \( t = 9 \).

Answers:

1. \( Q = \frac{5}{3^2} \times 7^2 = \frac{5}{9} \times 49 = \frac{245}{9} \approx 27.22 \)
2. Find \( k \): \( 36 = k \times 2^3 \Rightarrow 36 = k \times 8 \Rightarrow k = 4.5 \). Then, for \( r=5 \):

\[ V = 4.5 \times 5^3 = 4.5 \times 125 = 562.5 \]

1. Find \( k \): \( 10 = k \times \sqrt{4} \Rightarrow 10 = k \times 2 \Rightarrow k = 5 \). When \( t=9 \):

\[ S = 5 \times \sqrt{9} = 5 \times 3 = 15 \]

Conclusion

Understanding how to work with direct variation problems is essential in algebra and various scientific fields. By mastering the process of identifying the constant of variation and applying the formula correctly, you can solve a wide range of problems involving proportional relationships. The specific example of \( P \) varying directly as \( R^2 \) illustrates this process clearly, providing a solid foundation for tackling similar questions with confidence.