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Chapter 3: Problem 18

Solve each variation problem.Suppose \(p\) varies directly as the square of \(z,\) and inversely as \(r\). If \(p=\frac{32}{5}\) when \(z=4\) and \(r=10,\) find \(p\) when \(z=3\) and \(r=32\)?

Short Answer

Expert verified
The value of 饾憹 is \( \frac{9}{8} \).

Step by step solution

01

- Define Variation Formula

The problem states that 饾憹 varies directly as the square of 饾懅 and inversely as 饾憻. This means the relationship can be written as: \[ p = k \frac{z^2}{r} \] where 饾憳 is the constant of proportionality.
02

- Substitute Known Values to Find 饾憳

Substitute the known values of 饾憹, 饾懅, and 饾憻 into the equation to find 饾憳. Given 饾憹 = \frac{32}{5}, 饾懅 = 4, and 饾憻 = 10, the equation becomes: \[ \frac{32}{5} = k \frac{4^2}{10} \] Simplify this to solve for 饾憳: \[ \frac{32}{5} = k \frac{16}{10} \] \[ \frac{32}{5} = k \frac{8}{5} \] Thus, \[ k = 4 \]
03

- Use 饾憳 to Find the New Value of 饾憹

Now substitute 饾憳 = 4, 饾懅 = 3, and 饾憻 = 32 back into the variation formula to find the new value of 饾憹: \[ p = 4 \frac{3^2}{32} \] \[ p = 4 \frac{9}{32} \] \[ p = \frac{36}{32} \] Simplify the fraction to get the final 饾憹: \[ p = \frac{9}{8} \]

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Direct Variation
Direct variation is a relationship between two variables where one variable is a constant multiple of the other. If variable 饾懄 varies directly as variable 饾懃, we can express this as 饾懄 = 饾憳饾懃, where 饾憳 is the proportionality constant. In this problem, 饾憹 varies directly as the square of 饾懅, so we can rewrite this as 饾憹 = 饾憳饾懅^2.
This means if you increase 饾懅, 饾憹 increases at a rate proportional to the square of 饾懅. Conversely, if you decrease 饾懅, 饾憹 decreases similarly.
Inverse Variation
Inverse variation describes a situation where one variable increases while another variable decreases, in such a way that their product is constant. If variable 饾懄 varies inversely as variable 饾懃, it can be expressed as 饾懄 = 饾憳/饾懃, where 饾憳 is a constant. In this context, 饾憹 varies inversely as 饾憻, so it can be written as 饾憹 = 饾憳/饾憻.
This means if you increase 饾憻, then 饾憹 decreases, and vice versa. In our problem, 饾憹 is given by the combined effect of direct variation with 饾懅^2 and inverse variation with 饾憻.
Proportionality Constant
The proportionality constant, denoted as 饾憳, is a crucial factor in variation problems. It links the variables together. For direct variation, 饾憳 tells you how much one variable changes when the other changes. In our specific problem, the equation involves both direct and inverse variations:
\[ p = k \frac{z^2}{r} \]
To find 饾憳, we substitute the known values (饾憹 = \frac{32}{5}, 饾懅 = 4, and 饾憻 = 10):
\[ \frac{32}{5} = k \frac{4^2}{10} \]
Solving this gives us 饾憳 = 4.
Once we know 饾憳, we can determine the new value of 饾憹 with different 饾懅 and 饾憻 values.
Algebraic Manipulation
Algebraic manipulation is the process of rearranging equations to solve for an unknown variable. In our problem, we used algebraic steps to find the proportionality constant and then used it to find the unknown value of 饾憹. Let's look at this step by step:
1. Starting with the variation formula:
\[ p = k \frac{z^2}{r} \]
2. Substitute the known values to find 饾憳:
\[ \frac{32}{5} = k \frac{4^2}{10} \]
3. Simplify and solve for 饾憳.
4. With 饾憳 found, substitute new values to find 饾憹.
5. Perform arithmetic operations:
\[ p = 4 \frac{3^2}{32} \]
Therefore:
\[ p = \frac{9}{8} \].
This step-by-step manipulation is essential for solving variation problems.

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