Problem 20 - Algebra and Trigonometry

Question

13–22 ? Express the statement as an equation. Use the given information to find the constant of proportionality. \(H\) is jointly proportional to the squares of \(l\) and \(w .\) If \(l=2\) and \(w=\frac{1}{3},\) then \(H=36.\)

Step-by-Step Solution

Answer

The constant of proportionality is 81.

Step 1: Understanding Joint Variation

In joint variation, a variable is proportional to the product of other variables. Here, the problem says that \( H \) is jointly proportional to the squares of \( l \) and \( w \). This means \( H = k \cdot l^2 \cdot w^2 \), where \( k \) is the constant of proportionality.

Step 2: Formulate the Equation

Given the relationship that \( H \) is jointly proportional to \( l^2 \) and \( w^2 \), we can write the equation as:\[ H = k imes l^2 imes w^2 \]

Step 3: Substitute Given Values

We are given that when \( l=2 \), \( w=\frac{1}{3} \), then \( H=36 \). Substituting these values into the equation, we get:\[ 36 = k imes (2)^2 imes \left(\frac{1}{3}\right)^2 \]

Step 4: Simplify the Expression

First, calculate the squares: \( l^2 = (2)^2 = 4 \) and \( w^2 = \left(\frac{1}{3}\right)^2 = \frac{1}{9} \).Now substitute these into the equation:\[ 36 = k imes 4 imes \frac{1}{9} \]

Step 5: Solve for the Constant of Proportionality \( k \)

Simplify the equation:\( 36 = k imes \frac{4}{9} \)To isolate \( k \), multiply both sides by \( \frac{9}{4} \):\[ k = 36 imes \frac{9}{4} \]Calculate \( k \):\[ k = 81 \]

Step 6: Write the Full Equation with the Constant

Now that we have \( k = 81 \), the equation expressing the relationship between \( H \), \( l \), and \( w \) is:\[ H = 81 imes l^2 imes w^2 \]

Key Concepts

Constant of Proportionality

The constant of proportionality is a fundamental concept in understanding how variables relate to each other in joint variation. It represents the ratio that remains constant as conditions change. When we say that a variable, say \( H \), is jointly proportional to the squares of \( l \) and \( w \), we formulate an equation like \( H = k \cdot l^2 \cdot w^2 \), where \( k \) is our constant of proportionality. This constant helps link the variables together, showing how changes in \( l \) and \( w \) can affect \( H \).
By substituting actual values for \( l \), \( w \), and \( H \), we can solve for \( k \). In the given problem, when \( l = 2 \), \( w = \frac{1}{3} \), and \( H = 36 \), solving the equation \( 36 = k \times (2)^2 \times \left(\frac{1}{3}\right)^2 \), we find that \( k = 81 \). This means every time we double \( l \) and triple \( w \), \( H \) isn't just doubled or tripled; it's scaled by \( 81 \) times the product of the squares of these new values.

Squares of Variables

The term "squares of variables" refers to a mathematical operation where a number or variable is multiplied by itself. Understanding the concept of squaring is crucial when dealing with joint variation problems, like the one given here. In our problem, both \( l \) and \( w \) are squared. This means we calculate \( l^2 \) and \( w^2 \) before further operations.
Squaring can dramatically change the values: if \( l = 2 \), then \( l^2 = 4 \); similarly, if \( w = \frac{1}{3} \), then \( w^2 = \frac{1}{9} \). These squared values are then used in the joint variation equation. It's important to work out these squares correctly, as mistakes can lead to incorrect solutions. The increased value that results from squaring shows how the effect of \( l \) and \( w \) on \( H \) is multiplied and not merely additive.

Algebraic Equations

Algebraic equations are expressions that use algebraic symbols and numbers to describe relationships between variables. In this exercise, the algebraic equation is used to formalize the joint variation relationship: \( H = k \cdot l^2 \cdot w^2 \). This is a perfect example of using algebra to express complex relations simply and concisely.
Solving algebraic equations involves isolation of variables, substitution of known values, and simplification to find unknowns. Here, we substituted \( l = 2 \), \( w = \frac{1}{3} \), and \( H = 36 \) into the equation to solve for \( k \), the constant of proportionality. The process highlighted the importance of careful manipulation and organization of terms to arrive at the correct solution. Mastery of such techniques is essential for solving both simple and complex algebraic equations.