The distance dd a car travels varies directly as the time tt. If the car travels 180km⁡180\operatorname{\mathrm{km}} in 33 hours, find how far it will travel in 55 hours.

300km⁡300\operatorname{\mathrm{km}}

60km⁡60\operatorname{\mathrm{km}}

180km⁡180\operatorname{\mathrm{km}}

900km⁡900\operatorname{\mathrm{km}}

10. Relations and Functions

Direct & Inverse Variation

10. Relations and Functions

Direct & Inverse Variation: Videos & Practice Problems

Video Lessons

Topic summary

Direct variation and inverse variation describe how two quantities are related. In direct variation, both quantities change in the same direction, and the relationship is written as $y=kx$ . In inverse variation, the quantities change in opposite directions, and the relationship is written as $y=\\frac{k}{x}$ . You may also see the phrases directly proportional and inversely proportional.

A key skill in Direct & Inverse Variation is finding the constant of variation, then writing the equation that relates the variables. For direct variation, substitute known values into $y=kx$ . For inverse variation, substitute into $y=\\frac{k}{x}$ , noting that $x\\neq 0$ . These models are used to describe relationships such as price and quantity, distance and time, or time and speed.

Concept

Direct Variation

Video duration:

Direct Variation Video Summary

Direct variation describes a relationship between two variables where one variable increases or decreases in direct proportion to the other. This means if one quantity goes up, the other also goes up; if one goes down, the other follows suit. Mathematically, this relationship is expressed as y = kx, where y and x are the variables, and k is the constant of variation. This constant k represents the rate at which y changes with respect to x, making the equation a linear function.

Direct variation is often described as y varies directly as x or y is directly proportional to x. These phrases indicate the same concept and are commonly used in problems involving proportional relationships.

To find the specific equation for a direct variation problem, you start with the general formula:

\[y = kx\]

Given a pair of values for x and y, you substitute them into the equation to solve for the constant k. For example, if y = 10 when x = 2, substitute these values:

\[10 = k \times 2\]

Solving for k involves dividing both sides by 2:

\[k = \frac{10}{2} = 5\]

With k found, the direct variation equation becomes:

\[y = 5x\]

Once the equation is established, it can be used to find y for any given x. For instance, to find y when x = 6, substitute 6 into the equation:

\[y = 5 \times 6 = 30\]

This approach highlights the importance of identifying the constant of variation to model linear relationships accurately. Direct variation is a foundational concept that extends linear equations to real-world applications, where quantities change proportionally. Understanding this concept enables solving problems involving proportional relationships efficiently and accurately.

Problem

If $y$ varies directly as $x$ , and $y equals 18$ when $x equals 6$ , find $y$ when $x equals 10$ .

$3$

$30$

$60$

$9$

Problem

The distance $d$ a car travels varies directly as the time $t$ . If the car travels $180 k m$ in $3$ hours, find how far it will travel in $5$ hours.

$60 k m$

$180 k m$

$300 k m$

$900 k m$

Example

Direct Variation Example 1

Video duration:

Direct Variation Example 1 Video Summary

In problems involving direct variation, the relationship between two variables can be expressed as y = kx, where k is the constant of variation. When applying this to real-world scenarios, such as pricing, it is crucial to carefully define the variables to match the context. For example, if a bakery sells eggs at \$3 per dozen, the price (p) varies directly with the quantity (q) measured in dozens. This can be modeled as p = kq.

To find the constant of variation k, use the given price for one unit of quantity. Since the price is \(3 per dozen, and the quantity is measured in dozens, set q = 1 (representing one dozen). Substituting these values gives 3 = k × 1, so k = 3. Thus, the direct variation model becomes:

\[p = 3q\]

This equation means the price is three times the number of dozens purchased. To calculate the cost of five dozen eggs, substitute q = 5:

\[p = 3 \times 5 = 15\]

Therefore, five dozen eggs cost \)15.

To determine how many dozens were purchased if the total cost is \$39, set p = 39 and solve for q:

\[39 = 3q \implies q = \frac{39}{3} = 13\]

This means 13 dozen eggs were bought.

Understanding direct variation in practical contexts requires careful attention to the units and quantities involved. By defining variables clearly and applying the formula p = kq, where k is the constant rate per unit quantity, you can solve for unknown prices or quantities efficiently. This approach is essential for mastering direct variation problems, especially when quantities are expressed in grouped units like dozens.

Concept

Inverse Variation

Video duration:

Inverse Variation Video Summary

Inverse variation describes a relationship between two quantities where one increases as the other decreases, which is the opposite of direct variation. In direct variation, the equation relating the variables is expressed as $y = kx$, where $k$ is the constant of variation. However, in inverse variation, the relationship is modeled by the equation $y = \frac{k}{x}$, indicating that $y$ varies inversely as $x$. Here, $k$ remains the constant of variation, but instead of multiplying $x$, it is divided by $x$.

To find the constant $k$ in an inverse variation problem, you substitute known values of $x$ and $y$ into the equation and solve for $k$. For example, if $y = 8$ when $x = 4$, substituting these values gives $8 = \frac{k}{4}$. Multiplying both sides by 4 to isolate $k$ results in $k = 32$. This constant allows you to write the specific inverse variation equation as $y = \frac{32}{x}$.

It is important to note that $x$ cannot be zero in inverse variation because division by zero is undefined. This restriction is inherent in the nature of the inverse variation formula.

Once the equation is established, you can find the value of $y$ for any given $x$ by substituting the $x$ value into the equation. For instance, if $x = 2$, then $y = \frac{32}{2} = 16$. This demonstrates how inverse variation equations can be used to predict one variable based on the other.

Inverse variation is also commonly described using terms like "y varies inversely as x" or "y is inversely proportional to x," both indicating the same mathematical relationship. Understanding this concept is crucial for solving problems where quantities have an inverse relationship, such as in physics with concepts like pressure and volume or in economics with supply and demand.

In summary, inverse variation contrasts with direct variation by showing an opposite effect between two variables. The key steps involve identifying the relationship, calculating the constant of variation $k$, and then using the equation $y = \frac{k}{x}$ to find unknown values. Mastery of this concept enhances problem-solving skills and prepares learners for real-world applications involving inverse proportionality.

Problem

$p$ varies inversely as $q$ , and $p equals 18$ when $q equals 9$ . If $q equals 12$ , what is the value of $p$ ?

$13.5$

$162$

$24$

$12$

Problem

The time $t$ (in hours) required to travel a fixed distance varies inversely as the speed $s$ (in $m i / h r$ ). If it takes $6$ hours to travel a certain distance at $50 m i divided by h of r$ , how long will it take to travel the same distance at $75 m i divided by h of r$ ?

$6$ hours

$4$ hours

$9$ hours

$8$ hours

Example

Inverse Variation Example 2

Video duration:

Inverse Variation Example 2 Video Summary

In problems involving the inverse variation between volume and pressure of a gas, the volume varies inversely as the pressure applied. This means that as pressure increases, volume decreases proportionally, and vice versa. The general form of the equation representing this relationship is given by:

\[ V = \frac{k}{P} \]

where V is the volume, P is the pressure, and k is the constant of variation. To determine the constant k, use known values of volume and pressure from the problem. For example, if a helium gas has a volume of 370 cubic inches at a pressure of 15 pounds per square inch (PSI), substitute these values into the equation:

\[ 370 = \frac{k}{15} \]

Multiplying both sides by 15 to isolate k yields:

\[ k = 370 \times 15 = 5550 \]

Thus, the equation relating volume and pressure becomes:

\[ V = \frac{5550}{P} \]

This formula allows calculation of the volume at any given pressure. For instance, if the pressure increases to 20 PSI, the volume can be found by substituting P = 20:

\[ V = \frac{5550}{20} = 277.5 \text{ cubic inches} \]

This value can also be expressed as the fraction:

\[ V = \frac{555}{2} \text{ cubic inches} \]

Understanding inverse variation is essential in real-world applications such as gas laws, where pressure and volume are inversely proportional under constant temperature conditions. This relationship is a practical example of how mathematical models describe physical phenomena, enabling predictions and problem-solving in scientific contexts.

Here's what students ask on this topic:

Direct variation and inverse variation describe how two quantities relate to each other. In direct variation, both quantities increase or decrease together, meaning if one goes up, the other also goes up, and if one goes down, the other goes down. The equation for direct variation is $y = k x$ , where $k$ is the constant of variation. In inverse variation, the quantities change in opposite directions: as one increases, the other decreases. The equation for inverse variation is $y = \frac{k}{x}$ , with $x $\neq$ 0$ . Understanding these differences helps in modeling real-world relationships like speed and time or price and quantity.

To find the constant of variation $k$ in direct variation, you use the equation $y = k x$ . Given values of $x$ and $y$ , substitute them into the equation. For example, if $y = 10$ when $x = 2$ , substitute to get $10 = k $\times$ 2$ . Solve for $k$ by dividing both sides by $2$ , resulting in $k = 5$ . This constant $k$ is then used to write the direct variation equation for the problem.

For inverse variation, the general equation is $y = \frac{k}{x}$ . To find $k$ , substitute the known values of $x$ and $y$ . For example, if $y = 8$ when $x = 4$ , substitute to get $8 = \frac{k}{4}$ . Multiply both sides by $4$ to isolate $k$ : $k = 32$ . The inverse variation equation is then $y = \frac{32}{x}$ . Remember, $x$ cannot be zero because division by zero is undefined.

Direct and inverse variation models are useful in many real-world situations. For direct variation, an example is the relationship between distance and time when speed is constant: as time increases, distance increases proportionally. This can be modeled by $d = k t$ . For inverse variation, consider the relationship between speed and travel time for a fixed distance: as speed increases, travel time decreases. This is modeled by $t = \frac{k}{s}$ . These models help in understanding and solving practical problems involving proportional relationships.

Once you have the variation equation, solving for $y$ given $x$ is straightforward. For direct variation, use $y = k x$ . Substitute the given $x$ value and multiply by $k$ . For example, if $k = 5$ and $x = 6$ , then $y = 5 $\times$ 6 = 30$ . For inverse variation, use $y = \frac{k}{x}$ . Substitute $x$ and divide $k$ by $x$ . For example, if $k = 32$ and $x = 2$ , then $y = \frac{32}{2} = 16$ .

Your Beginning & Intermediate Algebra tutors

Direct & Inverse Variation: Videos & Practice Problems