MTH120 College Algebra Chapter 5.8

 5.8 Modeling Using Variation:

Modeling using variation involves describing and analyzing relationships between variables in the context of different types of variations, such as direct variation, inverse variation, joint variation, and combined variation. Here are some key concepts and approaches related to modeling using variation:

1. Direct Variation: In a direct variation, two variables are related in such a way that when one increases, the other also increases, and when one decreases, the other decreases. The relationship is typically expressed as $�=��$, where $�$ is the constant of variation.

2. Inverse Variation: In an inverse variation, two variables are related in such a way that when one increases, the other decreases, and vice versa. The relationship is typically expressed as $�=\frac{�}{�}$, where $�$ is the constant of variation.

3. Joint Variation: Joint variation combines both direct and inverse variation. It occurs when a variable is directly proportional to one variable and inversely proportional to another. The relationship is expressed as $�=���$, where $�$ is the constant of variation.

4. Combined Variation: Combined variation is used when a variable depends on two or more different types of variation. It's often expressed as $�=���$, where $�$ is the constant of variation.

5. Constant of Variation (k): The constant of variation represents the specific relationship between variables in a given type of variation. It helps in quantifying the relationship.

6. Modeling and Applications: You can use these types of variation to model and solve various real-world problems. For example, direct variation can describe scenarios like cost and quantity, where as one increases, the other increases proportionally. Inverse variation can be used to describe situations like the relationship between speed and time.

7. Graphical Representation: Graphs, including linear, hyperbolic, or other curves, can help visualize and understand the variation between variables. The slope of the graph can be used to determine the constant of variation.

8. Solving Problems: When solving problems involving variation, it's essential to set up the proper equations based on the type of variation involved and the given conditions. Understanding the relationships and interpreting the constants involved is crucial.

9. Unit Analysis: Analyzing the units of variables can help in understanding the relationships between variables in different types of variation.

To model using variation, it's important to recognize the type of variation present in a problem and apply the appropriate mathematical model to describe the relationship between the variables involved. This is commonly used in fields such as physics, engineering, economics, and many other areas where quantities are interrelated.

Solving direct variation problems involves finding the equation that describes the direct relationship between two variables and using that equation to solve specific problems. In direct variation, as one variable increases, the other also increases in a consistent manner. The equation for direct variation is usually expressed as $�=��$, where $�$ represents the constant of variation.

Here are the general steps to solve direct variation problems:

1. Identify the Variables: Recognize the two variables involved and which one depends directly on the other. Typically, one variable (usually $�$) is directly proportional to the other (usually $�$).

2. Write the Direct Variation Equation: Set up the equation $�=��$ to represent the direct variation. The constant $�$ is determined by the specific problem and represents the rate of change.

3. Find the Constant of Variation $�$: To find $�$, use the given information from the problem. You can often do this by substituting values of $�$ and $�$ from a data point in the problem.

4. Use the Equation to Solve Problems: Once you have found the value of $�$, you can use the direct variation equation to solve problems related to the situation. Plug in the given values and find the unknown variable.

Let's work through an example:

Example: Solving a Direct Variation Problem

Suppose you have a situation where the number of hours worked ($�$) is directly proportional to the amount earned ($�$). You know that if someone works 5 hours, they earn $40. Find the equation that relates the number of hours worked to the amount earned and use it to find the amount earned for 8 hours of work.

Step 1: Identify the Variables: The number of hours worked ($�$) depends directly on the amount earned ($�$).

Step 2: Write the Direct Variation Equation: The direct variation equation is $�=��$.

Step 3: Find the Constant of Variation $�$: Use the given information to find $�$. You know that when $�=5$, $�=40$. Plug these values into the equation:

$40=�\cdot 5$

Solve for $�$:

$�=\frac{40}{5}=8$

Step 4: Use the Equation to Solve Problems:

Now that you have found $�$, you can use the equation $�=8�$ to find the amount earned for 8 hours of work:

$�=8\cdot 8=64$

So, if someone works 8 hours, they will earn $64.

In this example, you've used the direct variation equation to establish the relationship between the variables and then solve a specific problem using that equation. This method can be applied to various direct variation situations, such as those involving cost, distance, or any other quantity that directly depends on another.

Direct variation describes a relationship between two variables in which one variable is directly proportional to the other. This means that as one variable increases, the other also increases in a consistent manner. The equation that represents direct variation is typically in the form $�=��$, where $�$ is the dependent variable, $�$ is the independent variable, and $�$ is the constant of variation.

Let's work through an example to illustrate direct variation:

Example: Direct Variation

Suppose you have a situation where the cost of renting a car is directly proportional to the number of days you rent the car. You know that renting a car for 2 days costs $120. Find the equation that relates the cost of renting a car ($�$) to the number of days ($�$) and use it to calculate the cost for renting the car for 5 days.

Step 1: Identify the Variables

In this problem, the cost of renting a car ($�$) is directly proportional to the number of days ($�$). We are looking for the relationship between $�$ and $�$.

Step 2: Write the Direct Variation Equation

The direct variation equation is $�=��$, where $�$ is the constant of variation.

Step 3: Find the Constant of Variation $�$

To find $�$, you can use the information that renting a car for 2 days costs $120. Plug these values into the equation:

$120=�\cdot 2$

Now, solve for $�$:

$�=\frac{120}{2}=60$

So, the constant of variation is $�=60$.

Step 4: Use the Equation to Calculate the Cost for 5 Days

Now that you've found $�$, you can use the direct variation equation to calculate the cost ($�$) for renting the car for 5 days:

$�=60\cdot 5=300$

So, if you rent the car for 5 days, it will cost $300.

In this example, you've applied direct variation to describe the relationship between the cost of renting a car and the number of days it is rented. You used the equation $�=��$ and found the constant of variation $�$ to calculate the cost for a different number of days. This is a simple illustration of how direct variation works in real-world situations, where one variable is directly proportional to another.

To solve a direct variation problem, you can follow these steps:

1. Identify the Variables: Determine the two variables involved in the problem and identify which one is directly proportional to the other.

2. Write the Direct Variation Equation: Set up the direct variation equation, which is typically in the form $�=��$, where $�$ is the dependent variable, $�$ is the independent variable, and $�$ is the constant of variation.

3. Find the Constant of Variation ($�$): To find $�$, use the given information from the problem. This information typically includes a specific data point where both variables are known. Plug these values into the equation and solve for $�$.

4. Use the Equation to Solve for the Unknown: Once you have found the value of $�$, you can use the direct variation equation to solve for the unknown variable by plugging in the given values of the other variable.

Here's an example:

Example: Solving for an Unknown in a Direct Variation Problem

Suppose you are given that the total cost ($�$) of buying $�$ pounds of apples at a certain store varies directly with the number of pounds of apples purchased. If 5 pounds of apples cost $10, find the cost of buying 8 pounds of apples.

Step 1: Identify the Variables

In this problem, the cost of apples ($�$) is directly proportional to the number of pounds of apples ($�$) purchased. We want to find the cost ($�$) for 8 pounds of apples.

Step 2: Write the Direct Variation Equation

The direct variation equation is $�=��$, where $�$ is the constant of variation.

Step 3: Find the Constant of Variation ($�$)

Use the information that 5 pounds of apples cost $10 to find $�$. Plug these values into the equation:

$10=�\cdot 5$

Solve for $�$:

$�=\frac{10}{5}=2$

So, the constant of variation is $�=2$.

Step 4: Use the Equation to Find the Cost for 8 Pounds

Now that you've found $�$, you can use the direct variation equation to calculate the cost ($�$) for 8 pounds of apples:

$�=2\cdot 8=16$

So, if you buy 8 pounds of apples, it will cost $16.

In this example, you used direct variation to establish the relationship between the cost of apples and the number of pounds purchased and then used that equation to find the cost for a different number of pounds. This process can be applied to various direct variation problems in which one variable is directly proportional to another.

Let's work through a direct variation problem with an example to demonstrate the process:

Example: Solving a Direct Variation Problem

Suppose you have a situation where the time taken to complete a task is directly proportional to the number of workers involved. When 4 workers are employed, the task is completed in 6 hours. Find the equation that relates the number of workers ($�$) to the time ($�$) and use it to calculate the time needed if 10 workers are employed.

Step 1: Identify the Variables

In this problem, the time taken to complete the task ($�$) is directly proportional to the number of workers ($�$). We are looking for the relationship between $�$ and $�$.

Step 2: Write the Direct Variation Equation

The direct variation equation is $�=��$, where $�$ is the constant of variation.

Step 3: Find the Constant of Variation ($�$)

To find $�$, use the given information that when 4 workers are employed, the task is completed in 6 hours. Plug these values into the equation:

$6=�\cdot 4$

Solve for $�$:

$�=\frac{6}{4}=1.5$

So, the constant of variation is $�=1.5$.

Step 4: Use the Equation to Calculate the Time for 10 Workers

Now that you've found $�$, you can use the direct variation equation to calculate the time ($�$) needed if 10 workers are employed:

$�=1.5\cdot 10=15$

So, if 10 workers are employed, the task will be completed in 15 hours.

In this example, we applied direct variation to describe the relationship between the number of workers and the time needed to complete a task. We used the equation $�=��$ and found the constant of variation $�$ to calculate the time for a different number of workers. This illustrates how direct variation works in real-world situations where one variable is directly proportional to another.

Solving inverse variation problems involves finding the equation that describes the inverse relationship between two variables and using that equation to solve specific problems. In inverse variation, as one variable increases, the other decreases, and vice versa. The equation for inverse variation is typically expressed as $�=\frac{�}{�}$, where $�$ is the dependent variable, $�$ is the independent variable, and $�$ is the constant of variation.

Here are the general steps to solve inverse variation problems:

1. Identify the Variables: Recognize the two variables involved in the problem and which one depends inversely on the other. Typically, one variable (usually $�$) is inversely proportional to the other (usually $�$).

2. Write the Inverse Variation Equation: Set up the inverse variation equation, which is usually in the form $�=\frac{�}{�}$, where $�$ is the constant of variation.

3. Find the Constant of Variation ($�$): To find $�$, use the given information from the problem. You can often do this by substituting values of $�$ and $�$ from a data point in the problem.

4. Use the Equation to Solve for the Unknown: Once you have found the value of $�$, you can use the inverse variation equation to solve problems related to the situation. Plug in the given values and find the unknown variable.

Let's work through an example:

Example: Solving an Inverse Variation Problem

Suppose you have a situation where the speed of a car ($�$) is inversely proportional to the time it takes to travel a certain distance ($�$). You know that when the car travels at 60 miles per hour, it takes 3 hours to complete the journey. Find the equation that relates the speed of the car ($�$) to the time it takes to travel ($�$) and use it to calculate the time it takes to travel at 45 miles per hour.

Step 1: Identify the Variables

In this problem, the speed of the car ($�$) is inversely proportional to the time it takes to travel ($�$). We want to find the relationship between $�$ and $�$.

Step 2: Write the Inverse Variation Equation

The inverse variation equation is $�=\frac{�}{�}$, where $�$ is the constant of variation.

Step 3: Find the Constant of Variation ($�$)

To find $�$, use the information that when the car travels at 60 miles per hour, it takes 3 hours to complete the journey. Plug these values into the equation:

$60=\frac{�}{3}$

Solve for $�$:

$�=60\cdot 3=180$

So, the constant of variation is $�=180$.

Step 4: Use the Equation to Calculate the Time for 45 Miles Per Hour

Now that you've found $�$, you can use the inverse variation equation to calculate the time ($�$) it takes to travel at 45 miles per hour:

$45=\frac{180}{�}$

Solve for $�$:

$�=\frac{180}{45}=4$

So, if the car travels at 45 miles per hour, it will take 4 hours to complete the journey.

In this example, you've applied inverse variation to describe the relationship between the speed of the car and the time it takes to travel. You used the equation $�=\frac{�}{�}$ and found the constant of variation $�$ to calculate the time for a different speed. This is how inverse variation works in real-world situations, where one variable is inversely proportional to another.

Solving problems involving joint variation means dealing with situations where a variable is influenced by both direct and inverse variation simultaneously. In joint variation, a variable is directly proportional to one variable and inversely proportional to another. The general equation for joint variation is:

$�=���$

Where:

• $�$ is the dependent variable.
• $�$ and $�$ are the independent variables.
• $�$ is the constant of variation.

Here's how to solve problems involving joint variation:

1. Identify the Variables: Determine the variables involved in the problem and recognize which variable is directly proportional to one variable and inversely proportional to another.

2. Write the Joint Variation Equation: Set up the joint variation equation $�=���$, where $�$ is the constant of variation.

3. Find the Constant of Variation ($�$): To find $�$, use the given information from the problem. This often involves specific data points where all variables are known. Substitute these values into the equation to solve for $�$.

4. Use the Equation to Solve for the Unknown: Once you've determined the value of $�$, you can use the joint variation equation to calculate the unknown variable by substituting the known values.

Let's work through an example:

Example: Solving a Joint Variation Problem

Suppose you have a situation where the total cost ($�$) of manufacturing a certain number of units ($�$) varies jointly with the cost per unit ($�$) and inversely with the number of units produced per day ($�$). When 100 units are manufactured at a cost of $2 per unit and 5 units are produced per day, find the equation that relates the total cost ($�$), the cost per unit ($�$), and the number of units produced per day ($�$), and use it to calculate the total cost for 200 units at a cost of $2.50 per unit and 8 units produced per day.

Step 1: Identify the Variables

In this problem, the total cost of manufacturing ($�$) is influenced jointly by the cost per unit ($�$) and inversely by the number of units produced per day ($�$). We want to find the relationship between $�$, $�$, and $�$.

Step 2: Write the Joint Variation Equation

The joint variation equation is $�=���$, where $�$ is the constant of variation.

Step 3: Find the Constant of Variation ($�$)

To find $�$, use the given information that when 100 units are manufactured at a cost of $2 per unit and 5 units are produced per day, the total cost is $200. Plug these values into the equation:

$200=�\cdot 2\cdot 5$

Solve for $�$:

$�=\frac{200}{10}=20$

So, the constant of variation is $�=20$.

Step 4: Use the Equation to Calculate the Total Cost

Now that you've found $�$, you can use the joint variation equation to calculate the total cost ($�$) for 200 units at a cost of $2.50 per unit and 8 units produced per day:

$�=20\cdot 2.5\cdot 8=400$

So, if 200 units are manufactured at a cost of $2.50 per unit and 8 units are produced per day, the total cost will be $400.

In this example, we applied joint variation to describe the relationship between the total cost, the cost per unit, and the number of units produced per day. We used the equation $�=���$ and found the constant of variation $�$ to calculate the total cost for different values of $�$ and $�$. This demonstrates how joint variation works in real-world scenarios, where a variable is influenced by both direct and inverse variation.

Joint variation, also known as combined variation, involves a relationship where a variable depends directly on one variable and inversely on another. Here are some examples of joint variation problems:

Example 1: Joint Variation with Area

Suppose you're dealing with a rectangular field, and the area ($�$) of the field varies jointly with its length ($�$) and inversely with its width ($�$). If the area of the field is 300 square meters when its length is 15 meters and its width is 10 meters, find the equation that relates the area ($�$), length ($�$), and width ($�$) and use it to calculate the area when the length is 20 meters and the width is 8 meters.

Solution:

• Identify the variables: $�$ depends jointly on $�$ and inversely on $�$.
• Write the joint variation equation: $�=�\cdot \frac{�}{�}$.
• Find the constant of variation ($�$): Use the given values to find $�$: $300=�\cdot \frac{15}{10}$ Solve for $�$: $�=\frac{300\cdot 10}{15}=200$.
• Use the equation to calculate the area for the new values: $�=200\cdot \frac{20}{8}=500$ So, if the length is 20 meters, and the width is 8 meters, the area will be 500 square meters.

Example 2: Joint Variation with Work

Suppose the amount of work ($�$) done by a team varies jointly with the number of team members ($�$) and inversely with the time ($�$) it takes. If 6 team members can complete a task in 4 hours, find the equation that relates the work ($�$), the number of team members ($�$), and the time ($�$) and use it to calculate the work done by 8 team members in 3 hours.

Solution:

• Identify the variables: $�$ depends jointly on $�$ and inversely on $�$.
• Write the joint variation equation: $�=�\cdot \frac{�}{�}$.
• Find the constant of variation ($�$): Use the given values to find $�$: $�=�\cdot \frac{6}{4}$ Solve for $�$: $�=\frac{�\cdot 4}{6}=\frac{2�}{3}$.
• Use the equation to calculate the work for the new values: $�=\frac{2�}{3}\cdot \frac{8}{3}=\frac{16�}{9}$ Solving for $�$: $�=\frac{9}{16}\cdot \frac{16�}{9}=�$ So, if 8 team members work for 3 hours, they will complete the same amount of work as 8 team members in 3 hours.

These are two examples of joint variation where a variable depends both directly and inversely on other variables. The key is to set up the equation, find the constant of variation, and then use the equation to solve problems involving changes in the variables.

For the following exercises, write an equation describing the relationship of the given variables. 1. y varies directly as x and when x=5,y=11. 2. y varies directly as the square of x and when x=2,y=60.  3. y varies directly as the square root of x and when x=38,y=26. 4. y varies directly as the cube of x and when x=34,y=21.

To write equations describing the relationships between the given variables, we need to identify the type of variation (direct, inverse, or joint variation) and find the constant of variation ($�$). Here, we have direct variations. The general form for direct variation is $�=��$, where $�$ is the constant of variation.

Let's write the equations for the given scenarios:

1. $�$ varies directly as $�$, and when $�=5$, $�=11$.

   The equation for this direct variation is $�=��$. We can find $�$ using the provided values:

   $11=�\cdot 5$

   Solving for $�$:

   $�=\frac{11}{5}$

   So, the equation is:

   $�=\frac{11}{5}�$

2. $�$ varies directly as the square of $�$, and when $�=2$, $�=60$.

   The equation for this direct square variation is $�=�{�}^2$. Find $�$:

   $60=�\cdot 2^2$

   Solve for (k:

   $�=\frac{60}{4}=15$

   So, the equation is:

   $�=15{�}^2$

3. $�$ varies directly as the square root of $�$, and when $�=38$, $�=26$.

   The equation for this direct square root variation is $�=�\sqrt{�}$. Find $�$:

   $26=�\sqrt{38}$

   Solve for $�$:

   $�=\frac{26}{\sqrt{38}}$

   The equation is:

   $�=\frac{26}{\sqrt{38}}\sqrt{�}$

4. $�$ varies directly as the cube of $�$, and when $�=34$, $�=21$.

   The equation for this direct cube variation is $�=�{�}^3$. Find $�$:

   $21=�\cdot 34^3$

   Solve for $�$:

   $�=\frac{21}{34^3}$

   The equation is:

   $�=\frac{21}{34^3}{�}^3$

These equations describe the relationships between the given variables in terms of direct variation.

MTH120 Algebra Chapter 5 quiz

Give the degree and leading coefficient of the following polynomial function.

1.  f(x)=x3(3−3x−6x2)

Determine the end behavior of the polynomial function.

2.  f(x)=4x3−7x2+1x−4

3.  f(x)=−2x2(6−3x−6x2)

Write the quadratic function in standard form. Determine the vertex and axes intercepts and graph the function.

4.  f(x)=x2+2x−6

Given information about the graph of a quadratic function, find its equation.

5. Vertex  (2,0)

  and point on graph  (2,14).

Solve the following application problem.

6. A rectangular field is to be enclosed by fencing. In addition to the enclosing fence, another fence is to divide the field into two parts, running parallel to two sides. If 1,200 feet of fencing is available, find the maximum area that can be enclosed.

Find all zeros of the following polynomial functions, noting multiplicities.

7.  f(x)=(x−5)3(1x−1)(x−1)2

8.  f(x)=2x3−11x5+14x4

1. For the polynomial function (f(x) = x^3(3 - 3x - 6x^2):

• Degree: The degree of a polynomial is the highest power of $�$ in the expression. In this case, the highest power of $�$ is ${�}^3$, so the degree is 3.
• Leading Coefficient: The leading coefficient is the coefficient of the term with the highest power of $�$. In this case, the leading coefficient is 1 (the coefficient of ${�}^3$).

End Behavior:

For the end behavior of a polynomial, consider the highest-degree term.

• As $�$ approaches positive or negative infinity, the term with the highest power dominates. In this case, ${�}^3$ is the leading term.
• Since the leading coefficient is positive, the polynomial will behave similarly to $�={�}^3$ for large values of $�$.
• As $�$ becomes very large, the function will increase or decrease without bound, depending on the sign of the leading coefficient.

2. For the polynomial function (f(x) = 4x^3 - 7x^2 + x - 4):

• Degree: The degree is 3, as the highest power of $�$ is ${�}^3$.
• Leading Coefficient: The leading coefficient is 4, which is the coefficient of the term with the highest power.

3. For the polynomial function (f(x) = -2x^2(6 - 3x - 6x^2):

• Degree: The degree is 3, as the highest power of $�$ is ${�}^3$ (from multiplying ${�}^2$ by $-6{�}^2$).
• Leading Coefficient: The leading coefficient is -12, which is the result of multiplying -2 and -6 (the coefficients of ${�}^2$ and $-6{�}^2$).

4. For the quadratic function (f(x) = x^2 + 2x - 6):

• To write it in standard form, you should express it as $�{�}^2+��+�$. In this case, you already have it in standard form.
• Vertex: The vertex of a quadratic function in standard form is given by $(ℎ,�)$, where $ℎ$ and $�$ are the values that minimize or maximize the function.
  • The vertex in this case is $(-1,-7)$.
• Axis Intercepts: To find the $�$-intercepts, set $�(�)=0$ and solve for $�$.
  • ${�}^2+2�-6=0$ can be factored as $(�+3)(�-1)=0$.
  • The $�$-intercepts are $�=-3$ and $�=1$.
  • To find the $�$-intercept, set $�=0$ and solve for $�(0)$.
    • $�(0)=0^2+2(0)-6=-6$.
  • The $�$-intercept is $(0,-6)$.
• Graph: To graph the function, you can plot the vertex and the intercepts, and then draw the parabola. The vertex is at (-1, -7), and the intercepts are (-3, 0), (1, 0), and (0, -6).

5. Given the vertex (2, 0) and a point on the graph (2, 14), we can use the vertex form of a quadratic equation:

$�=�(�-ℎ{)}^2+�$, where (h, k) is the vertex.

• Plug in the values:
  • $14=�(2-2{)}^2+0$
  • $14=�\cdot 0+0$
  • This means that $�$ must be 14.
• So, the equation of the quadratic function is $�=14(�-2{)}^2$.

6. Application Problem:

In this problem, you have 1,200 feet of fencing to enclose a rectangular field and divide it into two parts.

Let the length of the rectangular field be $�$ feet and the width be $�$ feet. You need to use the fencing to enclose three sides of the rectangular field and one side to divide it into two parts. This means:

• The fencing used to enclose the field is $2�+�$ feet.
• The fencing used to divide the field is $�$ feet.

The total amount of fencing you have is 1,200 feet, so you can write the equation:

$2�+�+�=1,200$

Simplify the equation:

$3�+�=1,200$

Now, you want to maximize the area of the rectangular field, which is given by $�=��$. You need to express one of the variables in terms of the other and substitute it into the area formula.

From the equation $3�+�=1,200$, solve for $�$:

$�=1,200-3�$

Now, substitute this expression for $�$ into the area formula:

$�=�(1,200-3�)$

To find the maximum area, take the derivative of $�$ with respect to $�$, set it equal to zero, and solve for $�$. This will give you the value of $�$ that maximizes the area.