12.4: Inverse Variation Problems
What if you had a circuit with a current of 2 amps and two resistances in parallel of 10 and 20 ohms each? How could you find the circuit's voltage? After completing this Concept, you'll be able to solve real-world applications like this one using rational functions.
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CK-12 Foundation: 1204S Solve Applications Using Rational Functions
Guidance
We are going to investigate some problems that are described by rational functions. Our first example is one which is modeled by inverse variation, which are a special type of rational function. Many formulas in physics are described by inverse variation.
Example A
The frequency, \begin{align*}f\end{align*}
Solution
\begin{align*}\text{The inverse variation relationship is:} \qquad \qquad \ f =\frac {k}{\lambda}\!\\
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\text{Plug in the values:} \ \lambda = 34 \ \text{and} \ f = 10: \qquad \quad 10=\frac {k}{34}\!\\
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\text{Multiply both sides by} \ 34: \qquad \quad \qquad \qquad \quad \ k=340\!\\
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\text{Thus, the relationship is given by:} \qquad \quad \qquad \ f=\frac {340}{\lambda}\!\\
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\text{Plug in} \ \lambda = 120 \ \text{meters:} \qquad \qquad \qquad \qquad \qquad f=\frac {340}{120} \Rightarrow f=2.83 \ \text{Hertz}\end{align*}
Electrical circuits are commonplace is everyday life—for example, they’re in all the electrical appliances in your home. The figure below shows an example of a simple electrical circuit. It consists of a battery which provides a voltage (\begin{align*}V\end{align*}
Ohm’s Law gives a relationship between current, voltage and resistance. It states that
\begin{align*}I = \frac{V}{R}\end{align*}
Your light bulbs, toaster and hairdryer are all basically simple resistors. In addition, resistors are used in an electrical circuit to control the amount of current flowing through a circuit and to regulate voltage levels. One important reason to do this is to prevent sensitive electrical components from burning out due to too much current or too high a voltage level. Resistors can be arranged in series or in parallel.
For resistors placed in a series:
the total resistance is just the sum of the resistances of the individual resistors:
\begin{align*}R_{tot} = R_1 + R_2\end{align*}
For resistors placed in parallel:
the reciprocal of the total resistance is the sum of the reciprocals of the resistances of the individual resistors:
\begin{align*}\frac{1}{R_c} = \frac{1}{R_1} + \frac{1}{R_2}\end{align*}
Example B
Find the quantity labeled \begin{align*}x\end{align*}
Solution
\begin{align*}\text{We use the formula} \ I = \frac{V}{R}.\!\\ \\ \text{Plug in the known values:} I = 2 \ A, V = 12 \ V: \qquad \qquad 2 = \frac{12}{R}\!\\ \\ \text{Multiply both sides by} \ R: \qquad \qquad \qquad \qquad \qquad \qquad 2R = 12\!\\ \\ \text{Divide both sides by}\ 2: \qquad \qquad \qquad \qquad \qquad \qquad \quad R = 6 \Omega \quad \mathbf{Answer}\end{align*}
Example C
Find the quantity labeled \begin{align*}x\end{align*} in the following circuit.
Solution
\begin{align*}\text{Ohm's Law also tells us that} \ I_{total} = \frac{V_{total}}{R_{total}}\!\\ \\ \text{Plug in the values we know}, I = 2.5 \ A \ \text{and}\ E = 9 \ V: \quad \qquad \qquad \qquad \ 2.5 = \frac{9}{R_{tot}}\!\\ \\ \text{Multiply both sides by}\ R: \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \quad 2.5R_{tot} = 9\!\\ \\ \text{Divide both sides by}\ 2.5: \ \ \quad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad R_{tot} = 3.6 \Omega\!\\ \\ \text{Since the resistors are placed in parallel, the total resistance is given by:}\ \frac{1}{R_{tot}} = \frac{1}{X} + \frac{1}{20}\!\\ \\ {\;} \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \Rightarrow \frac{1}{3.6} = \frac{1}{X} + \frac{1}{20}\!\\ \\ \text{Multiply all terms by} \ 72X: \qquad \qquad \qquad \qquad \qquad \qquad \frac{1}{3.6}(72X) = \frac{1}{X}(72X) + \frac{1}{20}(72X)\!\\ \\ \text{Cancel common factors:} \qquad \qquad \qquad \qquad \qquad \qquad \quad \ 20X = 72 + 3.6X\!\\ \\ \text{Solve:} \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \quad \ 16.4X = 72\!\\ \\ \text{Divide both sides by}\ 16.4: \ \quad \qquad \qquad \qquad \qquad \qquad \quad X = 4.39 \Omega \quad \mathbf{Answer}\end{align*}
Watch this video for help with the Examples above.
CK-12 Foundation: Solve Applications Using Rational Functions
Vocabulary
- Ohm’s Law gives a relationship between current, voltage and resistance. It states that
\begin{align*}I = \frac{V}{R}\end{align*}
- For resistors placed in a series:
the total resistance is just the sum of the resistances of the individual resistors:
\begin{align*}R_{tot} = R_1 + R_2\end{align*}
- For resistors placed in parallel:
the reciprocal of the total resistance is the sum of the reciprocals of the resistances of the individual resistors:
\begin{align*}\frac{1}{R_c} = \frac{1}{R_1} + \frac{1}{R_2}\end{align*}
Guided Practice
Electrostatic force is the force of attraction or repulsion between two charges. The electrostatic force is given by the formula \begin{align*}F = \frac{Kq_1 q_2}{d^2},\end{align*} where \begin{align*}q_1\end{align*} and \begin{align*}q_2\end{align*} are the charges of the charged particles, \begin{align*}d\end{align*} is the distance between the charges and \begin{align*}k\end{align*} is a proportionality constant. The charges do not change, so they too are constants; that means we can combine them with the other constant \begin{align*}k\end{align*} to form a new constant \begin{align*}K\end{align*}, so we can rewrite the equation as \begin{align*}F = \frac{K}{d^2}\end{align*}.
If the electrostatic force is \begin{align*}F = 740\end{align*} Newtons when the distance between charges is \begin{align*}5.3 \times 10^{-11}\end{align*} meters, what is \begin{align*}F\end{align*} when \begin{align*}d = 2.0 \times 10^{-10}\end{align*} meters?
Solution
\begin{align*}\text{The inverse variation relationship is:} \qquad \qquad \qquad \qquad F=\frac {K}{d^2}\!\\ \\ \text{Plug in the values} \ F = 740 \ \text{and} \ d = 5.3\times10^{-11}: \qquad \quad 740=\frac {K}{\left ( {5.3\ \times \ 10^{-11}} \right )^2}\!\\ \\ \text{Multiply both sides by} \ (5.3\times10^{-11})^2: \qquad \qquad \qquad \quad \ K=740 \left ( {5.3\times 10^{-11}} \right )^2\!\\ \\ {\;} \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad K = 2.08\times 10^{-18}\!\\ \\ \text{The electrostatic force is given by:} \qquad \qquad \qquad \quad \qquad F=\frac {2.08\ \times \ 10^{-18}}{d^2}\!\\ \\ \text{When} \ d = 2.0 \times 10^{-10}: \qquad \qquad \qquad \qquad \qquad \qquad \quad \ \ \ F=\frac {2.08\ \times \ 10^{-18}}{\left ( 2.0 \ \times \ 10^{-10} \right )^2}\!\\ \\ \text{Use scientific notation to simplify:} \qquad \qquad \qquad \qquad \quad F=52 \ \text{Newtons}\end{align*}
Practice
For 1-4, find the quantity labeled \begin{align*}x\end{align*} in each of the following circuits.
For 5-7, the intensity of light is inversely proportional to the square of the distance between the light source and the object being illuminated.
- A light meter that is 10 meters from a light source registers 35 lux. What intensity would it register 25 meters from the light source?
- A light meter that is registering 40 lux is moved twice as far away from the light source illuminating it. What intensity does it now register? (Hint: let \begin{align*}x\end{align*} be the original distance from the light source.)
- The same light meter is moved twice as far away again (so it is now four times as far from the light source as it started out). What intensity does it register now?
- Ohm’s Law states that current flowing in a wire is inversely proportional to the resistance of the wire. If the current is 2.5 Amperes when the resistance is 20 ohms, find the resistance when the current is 5 Amperes.
For 9-10, the volume of a gas varies directly with its temperature and inversely with its pressure. At 273 degrees Kelvin and pressure of 2 atmospheres, the volume of a certain gas is 24 liters.
- Find the volume of the gas when the temperature is 220 Kelvin and the pressure is 1.2 atmospheres.
- Find the temperature when the volume is 24 liters and the pressure is 3 atmospheres.
For 11-13, the volume of a square pyramid varies jointly with the height and the square of the side length of the base. A pyramid whose height is 4 inches and whose base has a side length of 3 inches has a volume of \begin{align*}12 \ in^3\end{align*}.
- Find the volume of a square pyramid that has a height of 9 inches and whose base has a side length of 5 inches.
- Find the height of a square pyramid that has a volume of \begin{align*}49 \ in^3\end{align*} and whose base has a side length of 7 inches.
- A square pyramid has a volume of \begin{align*}72 \ in^3\end{align*} and its base has a side length equal to its height. Find the height of the pyramid.
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Function
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Here you'l learn how to solve circuitry and other real-wold applications that involve inverse variations and rational functions.