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Chapter 7: Problem 54

Solve each problem. The weight of a trout varies jointly as its length and the square of its girth. One angler caught a trout that weighed 10.5 lb and measured 26 in. long with an 18 -in. girth. Find the weight of a trout that is 22 in. long with a 15 -in. girth.

Short Answer

Expert verified
The second trout weighs approximately 6.18 lb.

Step by step solution

01

Understand the given relationship

The weight of the trout varies jointly as its length and the square of its girth. This relationship can be expressed mathematically as: \[ W = k \times L \times G^2 \]where:- \(W\) is the weight of the trout,- \(L\) is the length of the trout,- \(G\) is the girth of the trout,- and \(k\) is the constant of variation.
02

Determine the constant of variation

Given the first trout weighs 10.5 lb, is 26 in. long, and has an 18 in. girth, substitute these values into the equation to find \(k\):\[ 10.5 = k \times 26 \times 18^2 \]Solve for \(k\):\[ 10.5 = k \times 26 \times 324 \]\[ 10.5 = 8424k \]\[ k = \frac{10.5}{8424} \approx 0.0012472 \]
03

Use the constant to find the weight of the second trout

Using the determined constant of variation (\(k\)), find the weight of the trout that is 22 in. long with a 15 in. girth. Substitute these values into the equation:\[ W = 0.0012472 \times 22 \times 15^2 \]\[ W = 0.0012472 \times 22 \times 225 \]\[ W = 0.0012472 \times 4950 \]\[ W \approx 6.1774 \]Therefore, the weight of the second trout is approximately 6.18 lb.

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Key Concepts

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

algebraic equations
Lastly, understanding algebraic equations is fundamental in solving joint variation problems. An equation represents the relationship between different mathematical expressions. In our example, \( W = k \times L \times G^2 \) is an algebraic equation where we solve for either the weight (W) or the constant of variation (k) by reorganizing and substituting the given values.
Algebraic equations are the backbone of solving many real-world problems. They translate complex relationships into concise mathematical language that can be manipulated to find solutions.

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Most popular questions from this chapter

Solve each problem. The force with which Earth attracts an object above Earth's surface varies inversely as the square of the distance of the object from the center of Earth. If an object \(4000 \mathrm{mi}\) from the center of Earth is attracted with a force of 160 lb, find the force of attraction if the object were \(6000 \mathrm{mi}\) from the center of Earth.

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