3.4 Modeling With Equations
- Page ID
- 152975
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\(\newcommand{\avec}{\mathbf a}\) \(\newcommand{\bvec}{\mathbf b}\) \(\newcommand{\cvec}{\mathbf c}\) \(\newcommand{\dvec}{\mathbf d}\) \(\newcommand{\dtil}{\widetilde{\mathbf d}}\) \(\newcommand{\evec}{\mathbf e}\) \(\newcommand{\fvec}{\mathbf f}\) \(\newcommand{\nvec}{\mathbf n}\) \(\newcommand{\pvec}{\mathbf p}\) \(\newcommand{\qvec}{\mathbf q}\) \(\newcommand{\svec}{\mathbf s}\) \(\newcommand{\tvec}{\mathbf t}\) \(\newcommand{\uvec}{\mathbf u}\) \(\newcommand{\vvec}{\mathbf v}\) \(\newcommand{\wvec}{\mathbf w}\) \(\newcommand{\xvec}{\mathbf x}\) \(\newcommand{\yvec}{\mathbf y}\) \(\newcommand{\zvec}{\mathbf z}\) \(\newcommand{\rvec}{\mathbf r}\) \(\newcommand{\mvec}{\mathbf m}\) \(\newcommand{\zerovec}{\mathbf 0}\) \(\newcommand{\onevec}{\mathbf 1}\) \(\newcommand{\real}{\mathbb R}\) \(\newcommand{\twovec}[2]{\left[\begin{array}{r}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\ctwovec}[2]{\left[\begin{array}{c}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\threevec}[3]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\cthreevec}[3]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\fourvec}[4]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\cfourvec}[4]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\fivevec}[5]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\cfivevec}[5]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\mattwo}[4]{\left[\begin{array}{rr}#1 \amp #2 \\ #3 \amp #4 \\ \end{array}\right]}\) \(\newcommand{\laspan}[1]{\text{Span}\{#1\}}\) \(\newcommand{\bcal}{\cal B}\) \(\newcommand{\ccal}{\cal C}\) \(\newcommand{\scal}{\cal S}\) \(\newcommand{\wcal}{\cal W}\) \(\newcommand{\ecal}{\cal E}\) \(\newcommand{\coords}[2]{\left\{#1\right\}_{#2}}\) \(\newcommand{\gray}[1]{\color{gray}{#1}}\) \(\newcommand{\lgray}[1]{\color{lightgray}{#1}}\) \(\newcommand{\rank}{\operatorname{rank}}\) \(\newcommand{\row}{\text{Row}}\) \(\newcommand{\col}{\text{Col}}\) \(\renewcommand{\row}{\text{Row}}\) \(\newcommand{\nul}{\text{Nul}}\) \(\newcommand{\var}{\text{Var}}\) \(\newcommand{\corr}{\text{corr}}\) \(\newcommand{\len}[1]{\left|#1\right|}\) \(\newcommand{\bbar}{\overline{\bvec}}\) \(\newcommand{\bhat}{\widehat{\bvec}}\) \(\newcommand{\bperp}{\bvec^\perp}\) \(\newcommand{\xhat}{\widehat{\xvec}}\) \(\newcommand{\vhat}{\widehat{\vvec}}\) \(\newcommand{\uhat}{\widehat{\uvec}}\) \(\newcommand{\what}{\widehat{\wvec}}\) \(\newcommand{\Sighat}{\widehat{\Sigma}}\) \(\newcommand{\lt}{<}\) \(\newcommand{\gt}{>}\) \(\newcommand{\amp}{&}\) \(\definecolor{fillinmathshade}{gray}{0.9}\)By the end of this section, you will be able to:
- recognize and describe direct and inverse variation using equations containing variables
- combine different types of variation into a single equation of multiple variables
- set up equations describing the relationships between real life quantities
- solve an equation for a variable in terms of others
A mathematical model is a way of describing a real-world situation or phenomenon using math symbols in an equation or formula. This involves representing quantities with variables and then writing down how they relate to each other. Since they depend upon each other, changing one variable induces a change in the other(s). Sometimes the relationship between them is very simple.
Direct Variation
If the quantities \(x\) and \(y\) are related by an equation of the form \( y = kx \), for some constant \(k \neq 0\), then we say \(y\) varies directly as \(x\), or \(y\) is directly proportional to \(x\). The constant (plain ol' number) \(k\) is the constant of proportionality.
Translation: one variable is a constant multiple of the other. This is also just called being proportional, like the default type of proportionality is "direct." Notice that the equation \(y = kx\) looks like the equation of a straight line with slope \(k\) and \(y\)-intercept \( (0,0)\). Looking at the graph below, we see that when \(k\) is positive, increasing \(x\) causes an increase in \(y\) (as I go to the right, I also go up). This shows how \(y\) is directly proportional to \(x\).
Lots of real-world physical situations can be modeled by direct variation.
1. As I produce more and more washing machines, my cost goes up, because I'm paying for more materials and more labor. Each washing machine costs me $225.37 to make. Find a model relating cost, denoted \(C\), and the number of machines I make, denoted \(x\). Then use it to figure out what it will cost to make 100 washing machines.
2. I'm driving down I-70 in my 2005 Scion xA at a constant speed of 100mph. (Well, I don't have cruise control because it's a manual, but I put a brick on the gas pedal. (Do not do this.)) As time in hours, denoted \(t\), goes on, I cover more distance in miles, denoted \(d\). Find a model relating distance and time, and use it to figure out how long it will take to travel 150 miles.
Solution
1. Every washing machine costs $225.37 to make, so the total cost can be calculated with multiplication: \(C = 225.37 x \). This model shows that \(C\) is directly proportional to the number of machines I make, \(x\), with the constant of proportionality \(k = 225.37\). If I want to make 100 machines, I can calculate the associated cost by plugging in for \(x\): \(C = 225.37(100) = $22,537\).
2. Distance covered and time are directly proportional, with equation \(d = 100t \). If I want to figure out how long it will take to cover 150 miles, I just plug that quantity in for \(d\) and solve for \(t\). I get
\[ 150 = 100 t \quad \rightarrow \quad t = \frac{150}{100} = 1.5 \text{ hrs} \notag \]
("But De Wolf," you say, "even with a brick on the gas pedal, you won't have a constant speed due to other road factors!" Shush, yes I will, this is I-70 driving across Kansas, it's as straight and flat as they come.)
The cost of printing 10 pages is $5.50, and we know the cost-per-page is a constant rate, so cost is directly proportional to the number of pages needed. How much will it cost to print 100 pages?
- Answer
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Name cost \(C\) and number of pages \(n\). "Cost is directly proportional to number of pages" is code for \(C = kn\). We know that \(n=10\) goes with cost \(C = 5.5\), so we can determine \(k\).
\[ 5.5 = k(10) \quad \longrightarrow \quad k = \frac{5.5}{10} = 0.55 \notag \]
Now I want to know what \(C\) is if \(n = 100\). Simply plug in:
\[ C = (0.55)(100) = $55.00 \notag \]
Inverse Variation
Sometimes two quantities are related in such a way that if I raise one, the other gets smaller.
If the quantities \(x\) and \(y\) are related by an equation of the form \( y = \dfrac{k}{x} \), for some constant \(k \neq 0\), then we say \(y\) varies inversely as \(x\), or \(y\) is inversely proportional to \(x\). The constant (plain ol' number\) \(k\) is still the constant of proportionality.
This equation is NOT the equation of a straight line. Think about what happens when \(x\) gets bigger. Since \(k\) is constant, staying the same, when the denominator of the fraction gets bigger, the fraction as a whole gets smaller. Then \(y\), who is defined by this fraction, will be getting smaller. On a graph, this relationship looks like this when \(k\) is positive:
We see as we go to the right, the curve falls downhill, showing that as \(x\) increases, the \(y\)-value decreases.
1. A perpetuity is a bond that pays the same amount every year forever. For example, you could pay a price, denoted \(P\), today and in return, receive $1000 every year. The current price of that perpetuity depends on the market interest rate, written as a decimal and denoted \(i\), according to the equation \( P = \dfrac{1000}{i} \). If interest rates go up, what will happen to the prices of such bonds? If the price is currently sitting at $10,000, what must the market interest rate be?
2. Boyle's Law states that, if the temperature and amount of an ideal gas are held constant, the pressure \(P\) exerted by the gas is inversely proportional to the volume it occupies. If I compress a gas into a smaller volume without letting any escape, the pressure will increase. Boyle's Law is often stated in math symbols as \(PV = nRT\), where \(n\) and \(R\) are constants specific to the situation. Find a way of writing that equation so that it matches the form in the definition of inverse variation.
Solution
1. Price and interest rate are inversely proportional, so if interest rates go up, the prices will go down. We can plug in the price $10,000 and solve for \(i\) to determine the associated rate.
\[ 10000 = \frac{1000}{i} \quad \longrightarrow \quad i = \frac{1000}{10000} = 0.1 \notag \]
The market interest rate is currently 10%.
2. Since temperature \(T\) is being held constant, the right hand side of the equation all boils down to some plain ol' number, and we could nickname it \(k\). Then with a little rearrangement, we can make the equation look like the formula for inverse variation between two variables \( y = \frac{k}{x}\), seeing the model as
\[ PV = k \quad \longrightarrow \quad P = \frac{k}{V} \notag \]
If I move house all by myself, it takes me 12 hours to pack everything. If I ask my mother to help, it takes only 6 hours. The more people I ask to help, the less time it takes, so the time \(t\) is inversely proportional to the number of people working, \( n\). If I ask my whole family of 6 people to help, how long will it take to pack?
- Answer
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We write the model with inverse variation, \(t = \dfrac{k}{n} \). I said if I work alone, meaning \(n = 1\), it takes me 12 hours. Plugging these values in, we get \(12 = \dfrac{k}{1}\), which means \(k = 12\). We can confirm because if my mom helps, so that \(n = 2\), we should get 6 hours, and yep, \(t = \frac{12}{2} = 6\). Great. Now if I have \(n = 6\), the time will be \(t = \frac{12}{6} = 2\) hours!
More Than Two Variables
In the Boyle's Law example, we held temperature constant, but in real life, of course it could vary, in which case \(T\) would be seen as a variable in the equation \(PV = (nR)T \). There may be more than two quantities involved in a model that can all vary, giving us multiple variables in our equation. In the info box below, \(k\) always represents a constant number.
- If \(x,y,\) and \(z\) are related by the equation \(z = kxy\), we say \( z\) is proportional to the product of \(x\) and \(y\), or \(z\) is jointly proportional to \(x\) and \(y\), or \(z\) varies jointly as \(x\) and \(y\).
- Looking at an equation solved for one variable, like the above or like \(z = k \dfrac{x}{y}\), we know that \(z\) is directly proportional to any variables in the numerator, and inversely proportional to any variables in the denominator.
- You can also be proportional to expressions other than a variable alone! For example, if \(z = k(x+y) \), we say \(z\) is directly proportional to the sum of \(x\) and \(y\). Or if \(z = \dfrac{k}{x^2} \), we say that \(z\) is inversely proportional to the square of \(x\).
1. Newton's Law of Gravity tells us the gravitational force \(F\) between two objects with masses \(m_1\) and \(m_2\) that are a distance \(r\) apart, computed using the model \( F = G\dfrac{m_1m_2}{r^2} \), where \(G\) is the universal gravitational constant (\(6.67430\times 10^{-11}\) \(\frac{N\cdot m^2}{kg^2}\)). Describe the relationships between \(F\) and the other variables \(m_1, m_2, r\). If the distance between the masses goes up, what happens to the gravitational force they exert on each other?
2. If I have a spherical balloon with radius \(r\), the volume inside is given by \(V = \frac{4}{3}\pi r^3 \). Describe the variation described by this equation, and identify the constant of proportionality. If the radius increases by a factor of 2, what happens to the volume of the balloon?
Solution
1. \(F\) is directly proportional to the product of \(m_1\) and \( m_2\). \(F\) is inversely proportional to the square of the distance \(r\). If the distance \(r\) increases, then certainly the the square of the distance, \(r^2\), increases, which makes the fraction as a whole get smaller. Thus the force will decrease.
2. \( V\) is directly proportional to the cube of \(r\). Remember that \(\pi\) is just a plain ol' number even though he looks like a letter, so he gets lumped in with the \(\frac{4}{3}\) as part of the constant of proportionality. If the radius increases by a factor of 2, the new radius is \(2r\). Let's plug that in to the model to see what happens.
\[ V_{\text{new}} = \frac{4}{3} \pi (2r)^3 = \frac{4}{3}\pi ( 8r^3 )= 8 \left( \frac{4}{3} \pi r^3 \right) = 8 V_{\text{original}} \notag \]
The volume also increases (it'd better) but by a factor of 8.
1. Think back to that perpetuity price vs. market interest rates example... I picked a specific payment of $1000 per year for that example, but of course you have a lot of different payment options to choose from. Let's nickname the payment amount with a variable, \($X\). Rewrite the relationship between price, payment, and interest rate, and describe the relationships. If the payment you want increases, what happens to the price of the bond?
2. The arithmetic mean (average) of a set of \(k\) sample values \( \{x_1, x_2, ..., x_k\} \) is calculated
\[ \overline{x} = \frac{ x_1 + x_2 + ... + x_k}{k} \notag \]
If the number of sample values is kept the same, but the values themselves change, describe the variation. Also, give the constant of proportionality.
- Answer
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1. The price appeared in the numerator of the fraction, so the equation is now \(P = \dfrac{X}{i}\), and we're seeing all the letters as variables now. \(P\) is directly proportional to \(X\) and inversely proportional to \(i\). If the interest rates stay the same and you want to buy a perpetuity that pays a higher amount each year, then the price will also increase.
2. The average \( \overline{x}\) is directly proportional to the sum of the values, \(x_1 + x_2 + ... + x_k\). Even though this doesn't currently look quite right, remember that we can write the division like this
\[ \overline{x} = \frac{1}{k} ( x_1 + x_2 + ... + x_k ) \notag\]
to see that the constant of proportionality is \(\frac{1}{k}\), still just a plain ol' number!
Solving for a variable in terms of other variables
If you have an equation relating multiple variable quantities, and it's solved for the guy you're curious about, fantastic. It was very easy to calculate the price \(P\) given an interest rate \(i\) in one go, because we had \(P\) isolated on one side of the equation. All we had to do was plug in a number for \(i\) and calculate! On the other hand, solving for an interest rate when given a price was a little more involved, because the variable was NOT isolated, but rather stuck in a denominator. If we wanted to find interest rates associated with a whole bunch of different price options, this constant repeated solving would become tedious.
Good news, life doesn't have to be that way! If we had solved the equation for the variable \(i\) in terms of the other variables, we could easily calculate results with one step of a calculator.
To solve an equation for a desired variable in terms of the other variables, simply use algebra to isolate the desired variable on one side, doing inverse operations as necessary to move other variables over. This can be done on any equation, not just models using variation.
1. Suppose I know the volume that I want inside my spherical balloon, and I want to compute the necessary radius. Solve \(V = \frac{4}{3}\pi r^3\) for \(r\). If I want the volume to be \(36\) cubic inches, what should the radius be?
2. Solve the equation \( F = G\dfrac{m_1m_2}{r^2} \) for the variable \(m_1\). Also, solve for \(r\) in terms of the other variables.
3. The monthly cost of running De Wolf's De Washing Machines depends on several factors: my facility's rent, \(C_r\); the cost of my utilities which depends on temperature outside, \(T\), and the number of machines I make, \(x\); and the materials and labor costs, which also depend on \(x\). The total cost is given by \( C = C_r + (k_1T + k_2 x) + 250x \), where \(k_1\) and \(k_2\) are just constants. Solve this equation for \(x\).
Solution
1. To start, I use whatever inverse operations are needed to get all the clutter away from the \(r^3\). I need to multiply both sides of the equation by \(\frac{3}{4\pi} \):
\[ \frac{3}{4\pi} \cdot V = \frac{4}{3}\pi r^3 \cdot \frac{3}{4\pi} \quad \longrightarrow \quad r^3 = \frac{3V}{4\pi} \notag \]
Then to finish, we need to undo the cubing by taking a cube root on both sides, resulting in \( r = \sqrt[3]{\dfrac{3V}{4\pi}}\). If we want \(V = 36\), we just plug that in and simplify to find \(r\):
\[ r = \sqrt[3]{\dfrac{3(36)}{4\pi}} = \sqrt[3]{\dfrac{3(9)}{\pi}}= \sqrt[3]{\dfrac{27}{\pi}}=\frac{3}{\sqrt[3]{\pi}} \notag \]
2. To isolate \(m_1\), I need to divide away the things he's multiplied by, and multiply over the things he's divided by.
\[ \frac{ r^2}{G m_2} \cdot F = G\dfrac{m_1m_2}{r^2} \cdot \frac{ r^2}{G m_2} \quad \longrightarrow \quad m_1 = \frac{ F r^2}{G m_2} \notag \]
To solve instead for \(r^2\), use your favorite trick for getting a variable out of a denominator, maybe cross-multiplication? Or "multiplying over" the \(r^2\). However you want to think about it.
\[ F = G\dfrac{m_1m_2}{r^2} \quad \longrightarrow \quad F r^2 = Gm_1m_2 \notag \]
Then to isolate \(r\), we need to divide both sides by \(F\) and then take a square root.
\[ r = \sqrt{ \frac{Gm_1m_2}{F}} \notag \]
3. To start isolating \(x\), I need to bring together my like terms to make sure I only have one \(x\) term. I also need to subtract all the non-\(x\) terms over to the same side.
\[ C = C_r + k_1T + k_2 x + 250x \quad \longrightarrow \quad k_2 x + 250 x = C - C_r - k_1 T \quad \longrightarrow \quad (k_2+250)x = C - C_r - k_1T \notag \]
Having factored \(x\) out, I am ready to isolate him by dividing over the expression in parentheses. I get \(x= \dfrac{C-C_r-k_1T}{k_2+250} \).
- The surface area of a cylinder-shaped tin can, with radius \(r\) and height \(h\), is given by \(A = 2\pi r^2 + 2\pi r h \). Solve this equation for \(h\).
- The volume of a cone with base radius \(r\) and height \(h\) is given by \(V = \pi r^2 \dfrac{h}{3} \). Solve this equation for \(r\). Also solve it for \(h\).
- Answer
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- \( h = \dfrac{A - 2\pi r^2}{2\pi r} \)
- \( r = \sqrt{ \dfrac{3V}{\pi h}} \) and \(h = \dfrac{ 3V }{\pi r^2} \)