Difference between revisions of "Modeling using Variation"
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<div id="eip-id1165135440091" class="equation unnumbered" style="text-align: center;" data-type="equation" data-label=""><math>y=\frac{25}{8}{x}^{3}</math></div> | <div id="eip-id1165135440091" class="equation unnumbered" style="text-align: center;" data-type="equation" data-label=""><math>y=\frac{25}{8}{x}^{3}</math></div> | ||
<p id="fs-id1165135432964">Substitute <em>x</em> = 6 and solve for <em>y</em>.</p> | <p id="fs-id1165135432964">Substitute <em>x</em> = 6 and solve for <em>y</em>.</p> | ||
− | <div id="eip-id1165135207297" class="equation unnumbered" style="text-align: center;" data-type="equation" data-label=""><math>\begin{cases}y=\frac{25}{8}{\left(6\right)}^{3} \\ \text{ }=675\end{cases}</math></div> | + | <div id="eip-id1165135207297" class="equation unnumbered" style="text-align: center;" data-type="equation" data-label=""><math>\begin{cases}y=\frac{25}{8}{\left(6\right)}^{3} \\ \text{ }=675 \end{cases}</math></div> |
</div> | </div> | ||
<div id="fs-id1165135533140" class="commentary" data-type="commentary"> | <div id="fs-id1165135533140" class="commentary" data-type="commentary"> | ||
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<h3>Solution</h3> | <h3>Solution</h3> | ||
<p id="fs-id1165137827766">Recall that multiplying speed by time gives distance. If we let <em>t</em> represent the drive time in hours, and <em>v</em> represent the velocity (speed or rate) at which the tourist drives, then <em>vt </em>= distance. Because the distance is fixed at 100 miles, <em>vt </em>= 100. Solving this relationship for the time gives us our function.</p> | <p id="fs-id1165137827766">Recall that multiplying speed by time gives distance. If we let <em>t</em> represent the drive time in hours, and <em>v</em> represent the velocity (speed or rate) at which the tourist drives, then <em>vt </em>= distance. Because the distance is fixed at 100 miles, <em>vt </em>= 100. Solving this relationship for the time gives us our function.</p> | ||
− | <div id="eip-id1165134094568" class="equation unnumbered" style="text-align: center;" data-type="equation" data-label=""><math>\begin{cases}t\left(v\right)=\frac{100}{v} | + | <div id="eip-id1165134094568" class="equation unnumbered" style="text-align: center;" data-type="equation" data-label=""><math>\begin{cases}t\left(v\right)=\frac{100}{v} \\ \text{ }=100{v}^{-1} \end{cases}</math></div> |
<p id="fs-id1165137748472">We can see that the constant of variation is 100 and, although we can write the relationship using the negative exponent, it is more common to see it written as a fraction.</p> | <p id="fs-id1165137748472">We can see that the constant of variation is 100 and, although we can write the relationship using the negative exponent, it is more common to see it written as a fraction.</p> | ||
</div> | </div> | ||
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<h3>Solution</h3> | <h3>Solution</h3> | ||
<p id="fs-id1165137627457">The general formula for inverse variation with a cube is <math>y=\frac{k}{{x}^{3}}</math>. The constant can be found by multiplying <em>y</em> by the cube of <em>x</em>.</p> | <p id="fs-id1165137627457">The general formula for inverse variation with a cube is <math>y=\frac{k}{{x}^{3}}</math>. The constant can be found by multiplying <em>y</em> by the cube of <em>x</em>.</p> | ||
− | <div id="eip-id1165132213474" class="equation unnumbered" style="text-align: center;" data-type="equation" data-label=""><math>\begin{cases}k={x}^{3}y | + | <div id="eip-id1165132213474" class="equation unnumbered" style="text-align: center;" data-type="equation" data-label=""><math>\begin{cases}k={x}^{3}y \\ \text{ }={2}^{3}\cdot 25 \\ \text{ }=200 \end{cases}</math></div> |
<p id="fs-id1165135188786">Now we use the constant to write an equation that represents this relationship.</p> | <p id="fs-id1165135188786">Now we use the constant to write an equation that represents this relationship.</p> | ||
− | <div id="eip-id1165133333885" class="equation unnumbered" style="text-align: center;" data-type="equation" data-label=""><math>\begin{cases}y=\frac{k}{{x}^{3}},k=200 | + | <div id="eip-id1165133333885" class="equation unnumbered" style="text-align: center;" data-type="equation" data-label=""><math>\begin{cases}y=\frac{k}{{x}^{3}},k=200 \\ y=\frac{200}{{x}^{3}} \end{cases}</math></div> |
<p id="fs-id1165137653904">Substitute <em>x </em>= 6 and solve for <i>y</i>.</p> | <p id="fs-id1165137653904">Substitute <em>x </em>= 6 and solve for <i>y</i>.</p> | ||
− | <div id="eip-id1165131878567" class="equation unnumbered" style="text-align: center;" data-type="equation" data-label=""><math>\begin{cases}y=\frac{200}{{6}^{3}} | + | <div id="eip-id1165131878567" class="equation unnumbered" style="text-align: center;" data-type="equation" data-label=""><math>\begin{cases}y=\frac{200}{{6}^{3}} \\ \text{ }=\frac{25}{27} \end{cases}</math></div> |
</div> | </div> | ||
<div id="fs-id1165137573081" class="commentary" data-type="commentary"> | <div id="fs-id1165137573081" class="commentary" data-type="commentary"> | ||
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<div id="eip-id1165133305360" class="equation unnumbered" style="text-align: center;" data-type="equation" data-label=""><math>x=\frac{k{y}^{2}}{\sqrt[3]{z}}</math></div> | <div id="eip-id1165133305360" class="equation unnumbered" style="text-align: center;" data-type="equation" data-label=""><math>x=\frac{k{y}^{2}}{\sqrt[3]{z}}</math></div> | ||
<p id="fs-id1165135190190">Substitute <em>x </em>= 6, <em>y </em>= 2, and <em>z </em>= 8 to find the value of the constant <em>k</em>.</p> | <p id="fs-id1165135190190">Substitute <em>x </em>= 6, <em>y </em>= 2, and <em>z </em>= 8 to find the value of the constant <em>k</em>.</p> | ||
− | <div id="eip-id1165133420152" class="equation unnumbered" style="text-align: center;" data-type="equation" data-label=""><math>\begin{cases}6=\frac{k{2}^{2}}{\sqrt[3]{8}} | + | <div id="eip-id1165133420152" class="equation unnumbered" style="text-align: center;" data-type="equation" data-label=""><math>\begin{cases}6=\frac{k{2}^{2}}{\sqrt[3]{8}} \\ 6=\frac{4k}{2} \\ 3=k \end{cases}</math></div> |
<p id="fs-id1165137863719">Now we can substitute the value of the constant into the equation for the relationship.</p> | <p id="fs-id1165137863719">Now we can substitute the value of the constant into the equation for the relationship.</p> | ||
<div id="eip-id1165134503062" class="equation unnumbered" style="text-align: center;" data-type="equation" data-label=""><math>x=\frac{3{y}^{2}}{\sqrt[3]{z}}</math></div> | <div id="eip-id1165134503062" class="equation unnumbered" style="text-align: center;" data-type="equation" data-label=""><math>x=\frac{3{y}^{2}}{\sqrt[3]{z}}</math></div> | ||
<p id="fs-id1165137742401">To find <em>x</em> when <em>y </em>= 1 and <em>z </em>= 27, we will substitute values for <em>y</em> and <em>z</em> into our equation.</p> | <p id="fs-id1165137742401">To find <em>x</em> when <em>y </em>= 1 and <em>z </em>= 27, we will substitute values for <em>y</em> and <em>z</em> into our equation.</p> | ||
− | <div id="eip-id1165131924522" class="equation unnumbered" style="text-align: center;" data-type="equation" data-label=""><math>\begin{cases}x=\frac{3{\left(1\right)}^{2}}{\sqrt[3]{27}} | + | <div id="eip-id1165131924522" class="equation unnumbered" style="text-align: center;" data-type="equation" data-label=""><math>\begin{cases}x=\frac{3{\left(1\right)}^{2}}{\sqrt[3]{27}} \\ \text{ }=1 \end{cases}</math></div> |
</div> | </div> | ||
</div> | </div> |
Revision as of 10:59, 25 October 2021
Contents
- 1 A General Note: Direct Variation
- 2 How To: Given a description of a direct variation problem, solve for an unknown.
- 3 Example 1: Solving a Direct Variation Problem
- 4 Solution
- 5 Analysis of the Solution
- 6 Q & A
- 7 Try It 1
- 8 Solve inverse variation problems
- 8.1 A General Note: Inverse Variation
- 8.2 Example 2: Writing a Formula for an Inversely Proportional Relationship
- 8.3 Solution
- 8.4 How To: Given a description of an indirect variation problem, solve for an unknown.
- 8.5 Example 3: Solving an Inverse Variation Problem
- 8.6 Solution
- 8.7 Analysis of the Solution
- 8.8 Try It 2
- 9 Solve problems involving joint variation
- 10 Key Equations
- 11 Key Concepts
A General Note: Direct Variation
If x and y are related by an equation of the form
then we say that the relationship is direct variation and y varies directly with the nth power of x. In direct variation relationships, there is a nonzero constant ratio , where k is called the constant of variation, which help defines the relationship between the variables.
How To: Given a description of a direct variation problem, solve for an unknown.
- Identify the input, x, and the output, y.
- Determine the constant of variation. You may need to divide y by the specified power of x to determine the constant of variation.
- Use the constant of variation to write an equation for the relationship.
- Substitute known values into the equation to find the unknown.
Example 1: Solving a Direct Variation Problem
The quantity y varies directly with the cube of x. If y = 25 when x = 2, find y when x is 6.
Solution
The general formula for direct variation with a cube is . The constant can be found by dividing y by the cube of x.
Now use the constant to write an equation that represents this relationship.
Substitute x = 6 and solve for y.
Q & A
Do the graphs of all direct variation equations look like Example 1?
No. Direct variation equations are power functions—they may be linear, quadratic, cubic, quartic, radical, etc. But all of the graphs pass through (0, 0).
Try It 1
The quantity y varies directly with the square of x. If y = 24 when x = 3, find y when x is 4.
<a href="https://courses.lumenlearning.com/precalcone/chapter/solutions-18/" target="_blank" rel="noopener">Solution</a>
Solve inverse variation problems
Water temperature in an ocean varies inversely to the water’s depth. Between the depths of 250 feet and 500 feet, the formula gives us the temperature in degrees Fahrenheit at a depth in feet below Earth’s surface. Consider the Atlantic Ocean, which covers 22% of Earth’s surface. At a certain location, at the depth of 500 feet, the temperature may be 28°F.
If we create a table we observe that, as the depth increases, the water temperature decreases.
<thead> </thead> <tbody> </tbody>d, depth | Interpretation | |
---|---|---|
500 ft | At a depth of 500 ft, the water temperature is 28° F. | |
350 ft | At a depth of 350 ft, the water temperature is 40° F. | |
250 ft | At a depth of 250 ft, the water temperature is 56° F. |
We notice in the relationship between these variables that, as one quantity increases, the other decreases. The two quantities are said to be inversely proportional and each term varies inversely with the other. Inversely proportional relationships are also called inverse variations.
For our example, the graph depicts the inverse variation. We say the water temperature varies inversely with the depth of the water because, as the depth increases, the temperature decreases. The formula for inverse variation in this case uses k = 14,000.
A General Note: Inverse Variation
If x and y are related by an equation of the form
where k is a nonzero constant, then we say that y varies inversely with the nth power of x. In inversely proportional relationships, or inverse variations, there is a constant multiple .
Example 2: Writing a Formula for an Inversely Proportional Relationship
A tourist plans to drive 100 miles. Find a formula for the time the trip will take as a function of the speed the tourist drives.
Solution
Recall that multiplying speed by time gives distance. If we let t represent the drive time in hours, and v represent the velocity (speed or rate) at which the tourist drives, then vt = distance. Because the distance is fixed at 100 miles, vt = 100. Solving this relationship for the time gives us our function.
We can see that the constant of variation is 100 and, although we can write the relationship using the negative exponent, it is more common to see it written as a fraction.
How To: Given a description of an indirect variation problem, solve for an unknown.
- Identify the input, x, and the output, y.
- Determine the constant of variation. You may need to multiply y by the specified power of x to determine the constant of variation.
- Use the constant of variation to write an equation for the relationship.
- Substitute known values into the equation to find the unknown.
Example 3: Solving an Inverse Variation Problem
A quantity y varies inversely with the cube of x. If y = 25 when x = 2, find y when x is 6.
Solution
The general formula for inverse variation with a cube is . The constant can be found by multiplying y by the cube of x.
Now we use the constant to write an equation that represents this relationship.
Substitute x = 6 and solve for y.
Analysis of the Solution
The graph of this equation is a rational function.
Try It 2
A quantity y varies inversely with the square of x. If y = 8 when x = 3, find y when x is 4.
<a href="https://courses.lumenlearning.com/precalcone/chapter/solutions-18/" target="_blank" rel="noopener">Solution</a>
Solve problems involving joint variation
Many situations are more complicated than a basic direct variation or inverse variation model. One variable often depends on multiple other variables. When a variable is dependent on the product or quotient of two or more variables, this is called joint variation. For example, the cost of busing students for each school trip varies with the number of students attending and the distance from the school. The variable c, cost, varies jointly with the number of students, n, and the distance, d.
A General Note: Joint Variation
Joint variation occurs when a variable varies directly or inversely with multiple variables.
For instance, if x varies directly with both y and z, we have x = kyz. If x varies directly with y and inversely with z, we have . Notice that we only use one constant in a joint variation equation.
Example 4: Solving Problems Involving Joint Variation
A quantity x varies directly with the square of y and inversely with the cube root of z. If x = 6 when y = 2 and z = 8, find x when y = 1 and z = 27.
Solution
Begin by writing an equation to show the relationship between the variables.
Substitute x = 6, y = 2, and z = 8 to find the value of the constant k.
Now we can substitute the value of the constant into the equation for the relationship.
To find x when y = 1 and z = 27, we will substitute values for y and z into our equation.
Try It 3
x varies directly with the square of y and inversely with z. If x = 40 when y = 4 and z = 2, find x when y = 10 and z = 25.
<a href="https://courses.lumenlearning.com/precalcone/chapter/solutions-18/" target="_blank" rel="noopener">Solution</a>
<section id="fs-id1165137898092" class="key-equations" data-depth="1">
Key Equations
<tbody> </tbody>Direct variation | . |
Inverse variation | . |
</section> <section id="fs-id1165137419773" class="key-concepts" data-depth="1">
Key Concepts
- A relationship where one quantity is a constant multiplied by another quantity is called direct variation.
- Two variables that are directly proportional to one another will have a constant ratio.
- A relationship where one quantity is a constant divided by another quantity is called inverse variation.
- Two variables that are inversely proportional to one another will have a constant multiple.
- In many problems, a variable varies directly or inversely with multiple variables. We call this type of relationship joint variation.
Glossary
- constant of variation
- the non-zero value k that helps define the relationship between variables in direct or inverse variation
- direct variation
- the relationship between two variables that are a constant multiple of each other; as one quantity increases, so does the other
- inverse variation
- the relationship between two variables in which the product of the variables is a constant
- inversely proportional
- a relationship where one quantity is a constant divided by the other quantity; as one quantity increases, the other decreases
- joint variation
- a relationship where a variable varies directly or inversely with multiple variables
- varies directly
- a relationship where one quantity is a constant multiplied by the other quantity
- varies inversely
- a relationship where one quantity is a constant divided by the other quantity
Resources
- Modeling Using Variation, Lumen Learning
- Variation Word Problems, Open Text Intermediate Algebra
- Intro to Direct & Inverse Variation, Khan Academy
- Direct Inverse and Joint Variation Word Problems, The Organic Chemistry Tutor
Analysis of the Solution
The graph of this equation is a simple cubic, as shown below.
Figure 2