7. Applying Rational Functions to Real-World Situations
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Chapter 4
7.
Applying Rational Functions to Real-World Situations
| Student Learning Objectives: |
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Why
Rational functions are more than just mathematical equations; they have practical applications in our daily lives. The lesson delves into how these functions are used in various fields such as 3D printing, drone operations, and even in calculating car fuel efficiency. By understanding the role of rational functions, you can gain insights into how different industries optimize their operations and make informed decisions.
Lesson Settings & Tools
| | 8 Theory slides |
| | 9 Exercises - Grade E - A |
| | Each lesson is meant to take 1-2 classroom sessions |
Applying Rational Functions to Real-World Situations
Slide of 8
Rational functions can be effective tools for modeling real-life situations. Some applications of rational functions involve speed, rate of work, and mixing problems. Rational functions even have applications in medicine and economics. This lesson will investigate the use of rational functions in various situations.
Catch-Up and Review
Here are a few recommended readings before getting started with this lesson.
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Challenge
Finding the Speed of the Wind
Tiffaniqua and Mark designed and 3D printed a drone. A neighbor hears about their feat and needs their help — she lost her dog Quipu! Before taking flight to find Quipo, first, they need to determine the flying conditions. The drone can travel at a maximum speed of 25 kilometers per hour. On a windy day, it travels 5 kilometers against the wind and then returns to the starting location.
External credits: macrovector
Suppose that the drone constantly flies at its maximum rate throughout the search.
a Write a rational function for the time of the flight T(w), where w is the speed of the wind.
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b What is the speed of the wind if the total time of the trip is 25 minutes?
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Example
Combined Miles per Gallon
Tiffaniqua wants to put some of her math skils to use to help her parents. Her mom drives an SUV that travels 15 miles per gallon (mpg) and her father has a hybrid that travels 60 mpg. They travel the same distance every month. Tiffaniqua realizes that she can find a way to improve the combined miles per gallon. She is considering two options.
| Option I | Tune the SUV to increase its mileage by 3 mpg and keep the hybrid as it is. |
|---|---|
| Option II | Buy a new hybrid that can travel 75 mpg and keep the SUV as it is. |
Which option will give a better combined mpg?
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Hint
The combined mpg C is equal to the total miles divided by the total gallons. This can algebraically be written as C = SUV miles + Hybrid milesSUV gallons + Hybrid gallons.
Solution
The combined mpg is equal to the ratio of the total miles to the the total gallons.
C = SUV miles + Hybrid milesSUV gallons + Hybrid gallons
Let x be the number of miles Tiffaniqua's mother drives in a month. Since her parents travel the same distance every month, x also represents the number of miles Tiffaniqua's father drives in a month. Now an expression for each option can be written.
Option I: Tune the SUV, Keep Hybrid as It Is
The SUV is tuned and its mileage is increased by 3 mpg. The hybrid, on the other hand, is kept as it is. SUV:& (15 + 3) =18mpg Hybrid: & 60 mpg The number of gallons consumed by each car can be expressed in terms of x, which is the number of miles that the cars travel in a month. The number of gallons can be found by dividing miles by miles per gallons.
| miles | mpg | gallons=miles/mpg | |
|---|---|---|---|
| SUV | x | 18 | x/18 |
| Hybrid | x | 60 | x/60 |
By substituting these values into the formula, the combined mpg is calculated.
C = SUV miles + Hybrid miles/SUV gallons + Hybrid gallons
SubstituteExpressions
Substitute expressions
C = x + x/x18 + x60
▼
Simplify right-hand side
AddTerms
Add terms
C = 2x/x18 + x60
ExpandFrac
a/b=a * 10/b * 10
C = 2x/10x180 + x60
ExpandFrac
a/b=a * 3/b * 3
C = 2x/10x180 + 3x180
AddFrac
Add fractions
C = 2x/13x180
DivByFracD
a/b/c= a * c/b
C = 360x/13x
ReduceFrac
a/b=.a /x./.b /x.
C = 360/13
CalcQuot
Calculate quotient
C= 27.692307 ...
RoundDec
Round to 1 decimal place(s)
C ≈ 27.7
In this option, the combined mpg is C≈ 27.7.
Option II: Buy a New Hybrid, Keep SUV as It Is
In this option, it is suggested to buy a new hybrid with a mpg of 75 and keep the SUV. As in the procedure followed in the previous option, the information at hand can be organized in a table.
| miles | mpg | gallons=miles/mpg | |
|---|---|---|---|
| SUV | x | 15 | x/15 |
| Hybrid | x | 75 | x/75 |
Substitute these expressions into the formula.
C = SUV miles + Hybrid miles/SUV gallons + Hybrid gallons
SubstituteExpressions
Substitute expressions
C = x + x/x15 + x75
▼
Simplify right-hand side
AddTerms
Add terms
C = 2x/x15 + x75
ExpandFrac
a/b=a * 5/b * 5
C = 2x/5x75 + x75
AddFrac
Add fractions
C = 2x/6x75
DivByFracD
a/b/c= a * c/b
C = 150x/6x
ReduceFrac
a/b=.a /x./.b /x.
C = 150/6
CalcQuot
Calculate quotient
C= 25
In this option, the combined mpg is C=25.
Conclusion
If Option I is chosen, the combined mpg will be about 27.7. If Option II is chosen, it will be 25. Therefore, the first option gives a better combined mpg.
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Example
Adjusting Estimated Time of Arrival
Tiffaniqua's parents want to thank her for solving their car problem and drove to a store 40 miles away in the next city over to buy a new 3D printer that she has dreamed of for years. On the way back, there was a road closure due to a landslide. This caused them to drive 12 mph slower.
This resulted in the return trip taking 3 hours longer. How many hours did it take them to get home from the store?
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Solution
Only the distance to the store is known. The time it takes to go to the store and the speed are unknown. Then, let t and r be the time elapsed and the speed, respectively.
| Speed | Time | Distance | |
|---|---|---|---|
| To the store | r | t | 40 |
On the way home, Tiffaniqua's mother drove 12 mph slower and the trip took 3 hours longer.
| Speed | Time | Distance | |
|---|---|---|---|
| To the store | r | t | 40 |
| From the store | r- 12 | t+ 3 | 40 |
Since the distance traveled is equal to the product of the speed and the time, a system of equations can be written. rt = 40 & (I) (r-12)(t+3)=40 & (II) By manipulating the equations in the system, rational equations can be formed.
rt = 40 (r-12)(t+3)=40
DivEqn
(I): .LHS /t.=.RHS /t.
r = 40t (r-12)(t+3)=40
DivEqn
(II): .LHS /(t+3).=.RHS /(t+3).
r = 40t r-12= 40t+3
Next, since r is isolated in Equation (I), its equivalent expression can be substituted into Equation (II).
Now, the second equation is written only in terms of t. Since only the time is required, this equation will be solved for the time t.
40/t-12=40/t+3
▼
Simplify
NumberToFrac
a = t* a/t
40/t-12t/t=40/t+3
SubFrac
Subtract fractions
40-12t/t=40/t+3
MultEqn
LHS * t=RHS* t
40-12t=40/t+3t
MoveRightFacToNum
a/c* b = a* b/c
40-12t=40t/t+3
MultEqn
LHS * (t+3)=RHS* (t+3)
(40-12t)(t+3)=40t
Distr
Distribute (t+3)
40(t+3)-12t(t+3)=40t
Distr
Distribute 40 & - 12t
40t+120-12t^2-36t=40t
SubEqn
LHS-40t=RHS-40t
120-12t^2-36t=0
CommutativePropAdd
Commutative Property of Addition
- 12t^2-36t+120=0
FactorOut
Factor out - 12
- 12(t^2+3t-10)=0
DivEqn
.LHS /(- 12).=.RHS /(- 12).
t^2+3t-10=0
This equation is now a quadratic equation in terms of t. It can be solved by factoring.
t^2+3t-10 = 0
(t-2)(t+5)=0
▼
Solve using the Zero Product Property
ZeroProdProp
Use the Zero Product Property
lct-2=0 & (I) t+5=0 & (II)
AddEqn
(I): LHS+2=RHS+2
lt=2 t+5=0
SubEqn
(II): LHS-5=RHS-5
lt=2 t=- 5
Since time cannot be negative, the solution t=- 5 is disregarded. Therefore, t=2 is the only solution in this context. Recall that t+3 represents the time to get home from the store. This means that it took Tiffaniqua 2+3=5 hours to get home from the store.
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Example
Concentration of a Mixture
Tiffaniqua and Mark are excited to print parts to make a drone using Tiffaniqua's new 3D printer. They want to make a special mixture for the raw material, called the filament. Tiffaniqua filled a flask with 500 milliliters of water. She then adds 60 grams of water soluble polyester resin.
Mark then adds more water at a rate of 10 milliliters per minute and simultaneously adds more resin at a rate of 6 grams per minute.
a Write a function C(t) for the concentration of resin in the flask after t minutes.
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b After how many minutes will the concentration be 0.2?
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Hint
a How can the amount of water after t minutes be represented? How can the amount of resin after t minutes be represented?
b Substitute the given value into the function found in Part A.
Solution
a Since water increases at a rate of 10 milliliters per minute and resin at 6 grams per minute, these are constant rates of change. Let W(t) be the amount of water and R(t) the amount of resin in the flask after t minutes. Considering that the flask initially has 500 milliliters of water and 60 grams of resin, two functions can be written.
Water: & W(t) = 500 + 10t Resin: & R(t) = 60 + 6t The concentration C(t) is the ratio of grams of resin to milliliters of water. C(t)=R(t)/W(t) ⇕ C(t) = 60+6t/500+10t
b To find after how many minutes the concentration will be 0.2, substitute 0.2 for C(t) into the equation found in Part A. Then, solve it for t.
After 10 minutes, the concentration in the flask will be 0.2.
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Example
Finding the Printing Time of the New 3D Printer
Tiffaniqua and Mark realize they can use their filament with Mark's old 3D printer in addition to the new printer. There is one major difference between the printers, however, it would take the old printer 50 minutes longer than the new printer to create the same parts.
Working together — some parts are printed with the new printer and some with the old printer — they can complete the task in 3 hours and 20 minutes. How long would it take the new printer to create the parts alone?
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Hint
If t is the printing time of the new printer, then t+50 is the printing time of the old printer.
Solution
Let t be the time in minutes it takes the new printer to print the parts Tiffaniqua and Mark need. Then, t+50 will be the printing time for the old printer. In this situation, the work is to print the drone parts, which can be considered as 1. Each machine's rate of work is then 1 over the printing time.
c|c
New Printer's Rate & Old Printer's Rate 1/t & 1/t+50
When both printers work at the same time, the job is completed in 3 hours and 20 minutes. Since t is defined in minutes, this time should also be expressed in minutes.
3 h and20 min = 200 min
The combined rate of the printers is 1 over the combined printing time 200. Therefore, the combined rate is 1200. This rate should be equal to the sum of the individual rates of the printers.
1/t + 1/t+50 = 1/200
The above rational equation can be solved for t. To eliminate the denominators, the equation can be multiplied by the least common denominator, which is the product of all the denominators.
1/t + 1/t+50 = 1/200
MultEqn
LHS * 200t(t+50)=RHS* 200t(t+50)
(1/t + 1/t+50 )(200t(t+50))= 1/200(200t(t+50))
▼
Simplify
Distr
Distribute 200t(t+50)
1/t(200t(t+50))+1/t+50(200t(t+50))= 1/200(200t(t+50))
MoveRightFacToNumOne
1/b* a = a/b
200t(t+50)/t+200t(t+50)/t+50= 200t(t+50)/200
CancelCommonFac
Cancel out common factors
200t(t+50)/t+200t( t+50)/t+50= 200t(t+50)/200
SimpQuot
Simplify quotient
200(t+50)+200t = t(t+50)
▼
Rewrite
Distr
Distribute 200 & t
200t+10 000+200t = t^2+50t
AddTerms
Add terms
400t+10 000= t^2+50t
SubEqn
LHS-400t=RHS-400t
10 000= t^2-350t
SubEqn
LHS-10 000=RHS-10 000
0= t^2-350t-10 000
RearrangeEqn
Rearrange equation
t^2-350t-10 000 = 0
To solve this quadratic equation, the Quadratic Formula can be used.
t^2-350t-10 000 = 0
▼
Solve using the quadratic formula
UseQuadForm
Use the Quadratic Formula: a = 1, b= - 350, c= - 10 000
t=- ( - 350)±sqrt(( - 350)^2 -4( 1) ( - 10 000))/2( 1)
NegNeg
- (- a)=a
t=350±sqrt((- 350)^2 -4(1)(- 10 000))/2(1)
CalcPowProd
Calculate power and product
t=350±sqrt(122 500+40 000)/2
AddTerms
Add terms
t=350±sqrt(162 500)/2
Next, the solutions can be individualized by using the positive and negative signs.
| t=350±sqrt(162 500)/2 | |
|---|---|
| t = 350 + sqrt(162 500)/2 | t = 350 - sqrt(162 500)/2 |
| t ≈ 377 | t ≈ - 27 |
Since time cannot be negative, t≈ - 27 is ignored. The solution t≈ 377 needs to be checked.
1/t + 1/t+50 = 1/200
t ≈ 377
1/377 + 1/377+50 ? ≈1/200
0.004994 ... ≈ 0.005 ✓
Since a true statement was obtained, t≈ 377 is a solution to the equation. The new printer prints the parts needed in about 377 minutes. The final step is to write this in hours.
377 min
▼
Rewrite
WriteSum
Write as a sum
360 min + 17 min
SplitIntoFactors
Split into factors
6(60 min) + 17 min
HourToMin
1h=60 min
6 (1 h) + 17 min
Multiply
Multiply
6 h + 17 min
The answer is 6 hours and 17 minutes.
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Example
How Far Can the Drone Go?
Drones communicate using radio waves on specific radio frequencies. The drone operates at a frequency of 2.4 gigahertz (GHz). How far a drone can travel depends on a number of factors such as the power of the signal transmitted by the controller.
External credits: macrovector
Let P_t be the power of the radio signal transmitted by the controller, P_r the power of the radio signal received by the drone, λ the wavelength, and d the distance between the drone and its controller. It is known that P_r varies directly with P_t and the square of λ, and inversely with the square of d.
a The controller transmits 300 milliwatts (mW) of power. When the distance between the drone and the controller is 240 meters, the power received by the drone is 4.5 * 10^(- 7) milliwatts. The wavelength of a 2.4 GHz signal is about 0.12 meters. Find the constant of variation.
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b If the power received is less than or equal to 10^(- 9) milliwatts, data exchange is not possible with the drone. How far can the drone go? Round the answer to the nearest integer.
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Hint
a Use the equation of the combined variation z = kxy, where k is the constant of variation.
b Substitute the given value into the equation.
Solution
a It is given that P_r varies directly with P_t and λ^2, and inversely with d^2. Start by labeling the quantities.
P_r: & Power received P_t: & Power transmitted λ: & Wavelength of signal d : & Distance between the drone & and the controller A combined variation equation can then be written as follows. P_r=k P_t λ^2/d^2 Here k is the constant of variation and cannot be 0. To find the value of k, review the given information. P_r: & 4.5 * 10^(- 7) mW P_t: & 300 mW λ: & 0.12 m d : & 240 m Substitute these values into the equation and solve for k.
P_r= k P_t λ^2/d^2
SubstituteValues
Substitute values
4.5 * 10^(- 7)=k( 300)( 0.12^2)/240^2
▼
Solve for k
MovePartNumLeft
a* b/c=a*b/c
4.5 * 10^(- 7)=k(300)(0.12^2)/240^2
CalcPow
Calculate power
4.5 * 10^(- 7)=k(300)(0.0144)/57 600
Multiply
Multiply
4.5 * 10^(- 7)=k4.32/57 600
CalcQuot
Calculate quotient
4.5 * 10^(- 7)=k * 0.000075
Write in scientific notation
4.5 * 10^(- 7)=k * 7.5 * 10^(- 5)
DivEqn
.LHS /(7.5 * 10^(- 5)).=.RHS /(7.5 * 10^(- 5)).
4.5 * 10^(- 7)/7.5* 10^(- 5) = k
WriteProdFrac
Write as a product of fractions
4.5/7.5 * 10^(- 7)/10^(- 5) = k
CalcQuot
Calculate quotient
0.6 * 10^(- 7)/10^(- 5) = k
DivPow
a^m/a^n= a^(m-n)
0.6 * 10^(- 7-(- 5)) = k
SubNeg
a-(- b)=a+b
0.6 * 10^(- 2) = k
CalcPow
Calculate power
0.6 * 0.01 = k
Multiply
Multiply
0.006 = k
RearrangeEqn
Rearrange equation
k=0.006
The constant of variation is 0.006.
b Consider the equation written in Part A.
P_r = 0.006 P_t λ^2/d^2 When P_r is less than or equal to 10^(- 9) milliwatts, the connection is lost. To keep the drone under control, the drone must be within a certain distance. This distance can be found by solving the following equation. 10^(- 9) = 0.006 P_t λ^2/d^2 The same values from Part A can be used for P_t and λ, as they are the same for this case as well.
10^(- 9) = 0.006 P_t λ^2/d^2
SubstituteValues
Substitute values
10^(- 9)=0.006( 300)( 0.12^2)/d^2
▼
Solve for d
CalcPow
Calculate power
10^(- 9)=0.006(300)(0.0144)/d^2
Multiply
Multiply
10^(- 9)=0.02592/d^2
MultEqn
LHS * d^2=RHS* d^2
10^(- 9)d^2= 0.02592
DivEqn
.LHS /10^(- 9).=.RHS /10^(- 9).
d^2= 0.02592/10^(- 9)
FracToMultNegExponent
a/b^m=ab^(- m)
d^2= 0.02592 * 10^9
Multiply
Multiply
d^2 = 25 920 000
SqrtEqn
sqrt(LHS)=sqrt(RHS)
sqrt(d^2) = sqrt(25 920 000)
sqrt(a^2)=± a
d = ± sqrt(25 920 000)
UseCalc
Use a calculator
d = ± 5091.168824 ...
RoundInt
Round to nearest integer
d ≈ ± 5091
The negative solution is ignored. The drone can be operated within a distance of about 5091 meters.
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Closure
Calculating the Speed of the Wind
From solving her parents' car issues, to successfully printing a drone, Tiffaniqua and Mark are now trying to locate a missing dog, Quipu! The owner tells them that Quipu was last seen running along the road that leads to the next city where Tiffaniqua bought the printer. They desperately want to find Quipo quickly. The drone can fly at a max speed of 25 kilometers per hour.
External credits: macrovector
Before setting out to find Quipo, they test flew the drone 5 kilometers against the wind before needing to bring it back to the starting point. Suppose that the drone flew at its maximum rate of 25 km/h throughout the trip.
a Write a rational function for the time of the flight T(w), where w is the speed of the wind.
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b What is the speed of the wind if the total time of the trip is 25 minutes?
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Hint
a Write two rational expressions for the times it takes to go upwind and downwind.
b Convert the given time into hours.
Solution
a Recall that the distance is the product of the rate and the time.
d = rt ⇔ t = d/r Since w is the speed of the wind in kilometers per hour, 25-w will be the speed of the drone while going upwind and 25+w will be the speed of the drone while going downwind. The information about the distance, speed, and time can be organized in a table.
| Upwind | Downwind | |
|---|---|---|
| Distance (km) | 5 | 5 |
| Speed (km/h) | 25-w | 25+w |
| Time (h) | 5/25-w | 5/25+w |
Therefore, the flight time of the drone T(w) is the sum of the times. T(w)=5/25-w + 5/25+w
b In the previous part, the unit of T(w) is hours. For this reason, the given time must be converted into hours. To do so, 25min is multiplied by the conversion factor 1 h60 min.
25min * 1 h/60 min = 5/12 h Now substitute this value for T(w) and solve the equation for w.
T(w)=5/25-w + 5/25+w
Substitute
T(w)= 5/12
5/12=5/25-w + 5/25+w
▼
Solve for w
MultEqn
LHS * 12(25-w)(25+w)/5=RHS* 12(25-w)(25+w)/5
5/12(12(25-w)(25+w)/5)=(5/25-w + 5/25+w)(12(25-w)(25+w)/5)
Distr
Distribute (12(25-w)(25+w)/5)
5/12(12(25-w)(25+w)/5)=5/25-w (12(25-w)(25+w)/5)+ 5/25+w(12(25-w)(25+w)/5)
CancelCommonFac
Cancel out common factors
5/12(12(25-w)(25+w)/5)=5/25-w (12( 25-w)(25+w)/5)+ 5/25+w(12(25-w)( 25+w)/5)
SimpQuotProd
Simplify quotient and product
(25-w)(25+w)=12(25+w)+ 12(25-w)
ExpandDiffSquares
(a+b)(a-b)=a^2-b^2
625 - w^2 = 12(25+w) + 12(25-w)
Distr
Distribute 12
625 - w^2 = 300+ 12w + 300 -12w
AddSubTerms
Add and subtract terms
625 - w^2 = 600
SubEqn
LHS-625=RHS-625
- w^2 = - 25
MultEqn
LHS * (- 1)=RHS* (- 1)
w^2 = 25
SqrtEqn
sqrt(LHS)=sqrt(RHS)
sqrt(w^2) = sqrt(25)
sqrt(a^2)=± a
w = ± sqrt(25)
CalcRoot
Calculate root
w = ± 5
The negative solution is ignored because speed cannot be negative. Therefore, the speed of the wind is 5 kilometers per hour. Do not forget to verify the solution in the equation.
5/12=5/25-w + 5/25+w
Substitute
w= 5
5/12 ? =5/25- 5 + 5/25+ 5
▼
Evaluate right-hand side
AddSubTerms
Add and subtract terms
5/12 ? = 5/20 + 5/30
ExpandFrac
a/b=a * 3/b * 3
5/12 ? = 15/60 + 5/30
ExpandFrac
a/b=a * 2/b * 2
5/12 ? = 15/60 + 10/60
AddFrac
Add fractions
5/12 ? = 25/60
ReduceFrac
a/b=.a /5./.b /5.
5/12 = 5/12 ✓
They begin flying the drone going downwind, making the drone and speed of the search faster. Over a few hours, they have no luck or clue about Quipu. Tiffaniqua's got it — she remembers the landslide along the road Quipu was last seen. Since the drone can only cover about 5000 meters from the controller, they start closer to the landslide.
External credits: macrovector
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Applying Rational Functions to Real-World Situations
Exercises
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We want to solve the given equation for the variable E. m=2E/V^2 We will apply inverse operations to both sides of the equation to isolate E.
We have solved the equation for E!
We want to solve the given equation for the variable T.
m/T^2=k/4 π^2
Let's use the Properties of Equality to isolate T on one side of the equation.
We have solved the equation for T!
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We want to solve the given equation for the variable m. To do so, we will apply inverse operations to both sides of the equation.
We have successfully isolated m!
We can use the equation we found in Part A to find the number of 3D printers m that need to be produced in order for the average cost of the printer C to be $80. To do so, let's substitute 80 for C and simplify.
Therefore, 250 printers must be produced in order for the printers to cost $80.
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Bronze is an alloy composed of copper and tin. There are different bronze alloys, but bronze is typically 88 % copper and 12 % tin. Ignacio has 220 grams of copper. How many grams of tin does Ignacio need to make bronze?
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Let x be the weight of tin in grams. We are asked to find x if we know that bronze is an alloy composed of 88 % copper and 12 % tin. Therefore, the percent of copper in the mixture must be equal to the ratio of the weight of the copper in the mixture to the total weight of the mixture. Percent of copper=Weight of copper/Total weight of mixture In our case, the percent of copper in the mixture is 88 % that can be written as 88100, the weight of copper in the mixture is 220 grams, and the total weight of the mixture is 220+x. Percent of copper=Weight of copper/Total weight of mixture ⇓ 88/100=220/220+x Next, we will solve the equation for x.
Therefore, Ignacio needs 30 grams of tin to turn all of his copper to bronze.
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The taper per foot is the rate at which the carving tapers, and the formula below can be used to find it. T = 24(R-r)/L All lengths are measured in inches.
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Let's rearrange the given formula to solve for L. We will apply inverse operations to both sides of the formula.
We have isolated L on one side of the equation!
Using the equation from Part A, let's find L for T= 0.75. We are also given the values of R= 6 and r= 5.
Therefore, the length of the carving is 32 inches.
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The average hourly wage of workers in an industry in Dallas is modeled by the function W(t), where t is the number of years that have passed since 1960. W(t) = 15.25t/0.06t+36.5 In what year does the model predict that wages will be $ 28 per hour?
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We will first find the value of t. To do so, let's substitute 28 for W(t) and then solve the equation for t.
After about 75 years, the wages will be $28 per hour. Recall that t represents the number of years that have passed since 1960. This means that we need to add 75 to the year 1960. 1960+75=2035
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Applying Rational Functions to Real-World Situations
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