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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.**

Challenge

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

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 $60mpg.$ 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. |

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The combined mpg $C$ is equal to the total miles divided by the total gallons. This can algebraically be written as $C=SUV gallons+Hybrid gallonsSUV miles+Hybrid miles .$

The combined mpg is equal to the ratio of the total miles to the the total gallons.
#### 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.

By substituting these values into the formula, the combined mpg is calculated.
In this option, the combined mpg is $C≈27.7.$ #### Option II: Buy a New Hybrid, Keep SUV as It Is

Substitute these expressions into the formula.
In this option, the combined mpg is $C=25.$ #### Conclusion

$C=SUV gallons+Hybrid gallonsSUV miles+Hybrid miles $

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. $SUV:Hybrid: (15+3)=18mpg60mpg $

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=mpgmiles $ | |
---|---|---|---|

SUV | $x$ | $18$ | $18x $ |

Hybrid | $x$ | $60$ | $60x $ |

$C=SUV gallons+Hybrid gallonsSUV miles+Hybrid miles $

SubstituteExpressions

Substitute expressions

$C=18x +60x x+x $

▼

Simplify right-hand side

AddTerms

Add terms

$C=18x +60x 2x $

ExpandFrac

$ba =b⋅10a⋅10 $

$C=18010x +60x 2x $

ExpandFrac

$ba =b⋅3a⋅3 $

$C=18010x +1803x 2x $

AddFrac

Add fractions

$C=18013x 2x $

DivByFracD

$b/ca =ba⋅c $

$C=13x360x $

ReduceFrac

$ba =b/xa/x $

$C=13360 $

CalcQuot

Calculate quotient

$C=27.692307…$

RoundDec

Round to $1$ decimal place(s)

$C≈27.7$

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=mpgmiles $ | |
---|---|---|---|

SUV | $x$ | $15$ | $15x $ |

Hybrid | $x$ | $75$ | $75x $ |

$C=SUV gallons+Hybrid gallonsSUV miles+Hybrid miles $

SubstituteExpressions

Substitute expressions

$C=15x +75x x+x $

▼

Simplify right-hand side

AddTerms

Add terms

$C=15x +75x 2x $

ExpandFrac

$ba =b⋅5a⋅5 $

$C=755x +75x 2x $

AddFrac

Add fractions

$C=756x 2x $

DivByFracD

$b/ca =ba⋅c $

$C=6x150x $

ReduceFrac

$ba =b/xa/x $

$C=6150 $

CalcQuot

Calculate quotient

$C=25$

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.

Example

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?{"type":"text","form":{"type":"math","options":{"comparison":"1","nofractofloat":false,"keypad":{"simple":true,"useShortLog":false,"variables":["x"],"constants":["PI"]}},"text":"<span class=\"katex\"><span class=\"katex-html\" aria-hidden=\"true\"><\/span><\/span>"},"formTextBefore":null,"formTextAfter":"hours","answer":{"text":["5"]}}

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 $12mph$ slower and the trip took $3$ hours longer.

Speed | Time | Distance | |
---|---|---|---|

To the store | $r$ | $t$ | $40$ |

From the store | $r−12$ | $t+3$ | $40$ |

${rt=40(r−12)(t+3)=40 (I)(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=t40 (r−12)(t+3)=40 $

DivEqn

$(II):$ $LHS/(t+3)=RHS/(t+3)$

${r=t40 r−12=t+340 $

$t40 −12=t+340 $

▼

Simplify

NumberToFrac

$a=tt⋅a $

$t40 −t12t =t+340 $

SubFrac

Subtract fractions

$t40−12t =t+340 $

MultEqn

$LHS⋅t=RHS⋅t$

$40−12t=t+340 t$

MoveRightFacToNum

$ca ⋅b=ca⋅b $

$40−12t=t+340t $

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

$LHS/(-12)=RHS/(-12)$

$t_{2}+3t−10=0$

$t_{2}+3t−10=0$

$(t−2)(t+5)=0$

▼

Solve using the Zero Product Property

ZeroProdProp

Use the Zero Product Property

$t−2=0t+5=0 (I)(II) $

AddEqn

$(I):$ $LHS+2=RHS+2$

$t=2t+5=0 $

SubEqn

$(II):$ $LHS−5=RHS−5$

$t=2t=-5 $

Example

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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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.

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:Resin: W(t)=500+10tR(t)=60+6t $

The concentration $C(t)$ is the ratio of grams of resin to milliliters of water.
$C(t)=W(t)R(t) ⇕C(t)=500+10t60+6t $

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 $t.$

Example

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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If $t$ is the printing time of the new printer, then $t+50$ is the printing time of the old printer.

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.
To solve this quadratic equation, the Quadratic Formula can be used.
Next, the solutions can be individualized by using the positive and negative signs.

Since time cannot be negative, $t≈-27$ is ignored. The solution $t≈377$ needs to be checked.
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.
The answer is $6$ hours and $17$ minutes.

$New Printer’s Ratet1 Old Printer’s Ratet+501 $

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.
$3h and20min=200min $

The combined rate of the printers is $1$ over the combined printing time $200.$ Therefore, the combined rate is $2001 .$ This rate should be equal to the sum of the individual rates of the printers.
$t1 +t+501 =2001 $

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.
$t1 +t+501 =2001 $

MultEqn

$LHS⋅200t(t+50)=RHS⋅200t(t+50)$

$(t1 +t+501 )(200t(t+50))=2001 (200t(t+50))$

▼

Simplify

Distr

Distribute $200t(t+50)$

$t1 (200t(t+50))+t+501 (200t(t+50))=2001 (200t(t+50))$

MoveRightFacToNumOne

$b1 ⋅a=ba $

$t200t(t+50) +t+50200t(t+50) =200200t(t+50) $

CancelCommonFac

Cancel out common factors

$t 200t (t+50) +t+50 200t(t+50) =200200t(t+50) $

SimpQuot

Simplify quotient

$200(t+50)+200t=t(t+50)$

▼

Rewrite

Distr

Distribute $200&t$

$200t+10000+200t=t_{2}+50t$

AddTerms

Add terms

$400t+10000=t_{2}+50t$

SubEqn

$LHS−400t=RHS−400t$

$10000=t_{2}−350t$

SubEqn

$LHS−10000=RHS−10000$

$0=t_{2}−350t−10000$

RearrangeEqn

Rearrange equation

$t_{2}−350t−10000=0$

$t_{2}−350t−10000=0$

▼

Solve using the quadratic formula

UseQuadForm

Use the Quadratic Formula: $a=1,b=-350,c=-10000$

$t=2(1)-(-350)±(-350)_{2}−4(1)(-10000) $

NegNeg

$-(-a)=a$

$t=2(1)350±(-350)_{2}−4(1)(-10000) $

CalcPowProd

Calculate power and product

$t=2350±122500+40000 $

AddTerms

Add terms

$t=2350±162500 $

$t=2350±162500 $ | |
---|---|

$t=2350+162500 $ | $t=2350−162500 $ |

$t≈377$ | $t≈-27$ |

$t1 +t+501 =2001 $

$t≈377$

$3771 +377+501 ≈?2001 $

$0.004994⋯≈0.005✓$

$377min$

▼

Rewrite

WriteSum

Write as a sum

$360min+17min$

SplitIntoFactors

Split into factors

$6(60min)+17min$

HourToMin

$1h=60min$

$6(1h)+17min$

Multiply

Multiply

$6h+17min$

Example

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 $2.4GHz$ 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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a Use the equation of the combined variation $z=ykx ,$ where $k$ is the constant of variation.

b Substitute the given value into the equation.

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}:P_{t}:λ:d: Power receivedPower transmittedWavelength of signalDistance between the droneand the controller $

A combined variation equation can then be written as follows.
$P_{r}=d_{2}kP_{t}λ_{2} $

Here $k$ is the constant of variation and cannot be $0.$ To find the value of $k,$ review the given information.
$P_{r}:P_{t}:λ:d: 4.5×10_{-7}mW300mW0.12m240m $

Substitute these values into the equation and solve for $k.$
$P_{r}=d_{2}kP_{t}λ_{2} $

SubstituteValues

Substitute values

$4.5×10_{-7}=240_{2}k(300)(0.12_{2}) $

▼

Solve for $k$

MovePartNumLeft

$ca⋅b =a⋅cb $

$4.5×10_{-7}=k240_{2}(300)(0.12_{2}) $

CalcPow

Calculate power

$4.5×10_{-7}=k57600(300)(0.0144) $

Multiply

Multiply

$4.5×10_{-7}=k576004.32 $

CalcQuot

Calculate quotient

$4.5×10_{-7}=k⋅0.000075$

Write in scientific notation

$4.5×10_{-7}=k⋅7.5×10_{-5}$

DivEqn

$LHS/(7.5×10_{-5})=RHS/(7.5×10_{-5})$

$7.5×10_{-5}4.5×10_{-7} =k$

WriteProdFrac

Write as a product of fractions

$7.54.5 ⋅10_{-5}10_{-7} =k$

CalcQuot

Calculate quotient

$0.6⋅10_{-5}10_{-7} =k$

DivPow

$a_{n}a_{m} =a_{m−n}$

$0.6⋅10_{-7−(-5)}=k$

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

$k=0.006$

b Consider the equation written in Part A.

$P_{r}=d_{2}0.006P_{t}λ_{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}=d_{2}0.006P_{t}λ_{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}=d_{2}0.006P_{t}λ_{2} $

SubstituteValues

Substitute values

$10_{-9}=d_{2}0.006(300)(0.12_{2}) $

▼

Solve for $d$

CalcPow

Calculate power

$10_{-9}=d_{2}0.006(300)(0.0144) $

Multiply

Multiply

$10_{-9}=d_{2}0.02592 $

MultEqn

$LHS⋅d_{2}=RHS⋅d_{2}$

$10_{-9}d_{2}=0.02592$

DivEqn

$LHS/10_{-9}=RHS/10_{-9}$

$d_{2}=10_{-9}0.02592 $

FracToMultNegExponent

$b_{m}a =ab_{-m}$

$d_{2}=0.02592⋅10_{9}$

Multiply

Multiply

$d_{2}=25920000$

SqrtEqn

$LHS =RHS $

$d_{2} =25920000 $

$a_{2} =±a$

$d=±25920000 $

UseCalc

Use a calculator

$d=±5091.168824…$

RoundInt

Round to nearest integer

$d≈±5091$

Closure

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.

b What is the speed of the wind if the total time of the trip is $25$ minutes?

a Write two rational expressions for the times it takes to go upwind and downwind.

b Convert the given time into hours.

a Recall that the distance is the product of the rate and the time.

$d=rt⇔t=rd $

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) | $25−w5 $ | $25+w5 $ |

$T(w)=25−w5 +25+w5 $

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 $60min1h .$

$25min⋅60min1h =125 h $

Now substitute this value for $T(w)$ and solve the equation for $w.$
$T(w)=25−w5 +25+w5 $

Substitute

$T(w)=$