NZ Level 8 (NZC) Level 3 (NCEA) [In development] Solve applications involving Rational Functions
Lesson

## Application 1: Average speeds

Simone averages $60$60 km/h on her trip from point $A$A to point $B$B but then immediately returns to $A$A at a speed of $x$x km/h. Develop a function $A\left(x\right)$A(x) for her average speed for the whole trip. Given that speed limit everywhere between $A$A and $B$B is $120$120 km/h, find the maximum average speed possible for the entire journey.

Suppose we call $d$d the distance between $A$A and $B$B

Since Simone averaged $60$60 km/h over a distance $d$d, the time it took to do so, say $t_1=\frac{d}{60}$t1=d60 hours.

The time $t_2$t2 to come back to $A$A, given an average speed of $x$x, must also be given by  $t_2=\frac{d}{x}$t2=dx hours.

Hence the total time $T=t_1+t_2$T=t1+t2 taken for the entire trip is given by $T=\frac{d}{60}+\frac{d}{x}$T=d60+dx.

Since the total distance travelled was $2d$2d, the average speed $A\left(x\right)$A(x) for the entire journey must be given by the total distance divided by the total time. That is:

 $A\left(x\right)$A(x) $=$= $\frac{2d}{\frac{d}{60}+\frac{d}{x}}$2dd60​+dx​​ $=$= $\frac{2d}{\frac{dx+60d}{60x}}$2ddx+60d60x​​ $=$= $\frac{2d}{d\left(x+60\right)}\times60x$2dd(x+60)​×60x $\therefore$∴   $A\left(x\right)$A(x) $=$= $\frac{120x}{x+60}$120xx+60​     km/h

The function $A\left(x\right)=\frac{120x}{x+60}$A(x)=120xx+60 is a rational function defined for positive values of $x$x only. Note that, on the assumption that Simone returns to $A$A the average speed cannot be zero.  The positive domain implies that we don't have to deal with any discontinuities.

Dividing $120x$120x by $x+60$x+60 shows that $A\left(x\right)=120-\frac{7200}{x+60}$A(x)=1207200x+60, revealing the horizontal asymptote $y=120$y=120. Note that even though $7200$7200 is a large number, as $x\rightarrow\infty$x$\frac{7200}{x+60}\rightarrow0$7200x+600. and this means the the line $y=120$y=120 is the asymptote.

Here is the graph: From the graph, we see that since the fastest possible average speed for the return journey is $120$120 km/h, then the maximum average speed for the entire journey is $80$80 km/h.

### The harmonic mean

In general, if the average speeds from $A$A to $B$B and $B$B to $A$A had been $x$x and $y$y, then the multivariable function $A\left(x,y\right)$A(x,y) giving the average speed for the total journey as $\frac{2xy}{x+y}$2xyx+y, is known as the harmonic mean of $x$x and $y$y

The harmonic mean is always less than or equal to the arithmetic mean (It is equal when $x=y$x=y).

## Application 2: Nicotine levels in the blood

A trial was conducted on a volunteer smoker that involved the measurement of nicotine levels in the blood after smoking one cigarette.

Two series of blood samples were taken.

Series $A$A  blood samples were taken every $4$4 minutes for the first $30$30 minutes only and series $B$B samples were taken every $10$10 minutes for the entire $2\frac{1}{2}$212 hours the experiment was conducted over. The nicotine level for each sample was measured in nanograms per millilitre.

It was found that the maximum nicotine level was $16.4$16.4 ng/ml in both series.

The graph below shows the series $A$A results (as green data points) taken up to this maximum, and series $B$B results (the red data points) after the maximum was reached.  A rational function model was fitted to the data as shown here: Note that the researcher had anticipated an initial rapid increase of nicotine into the blood stream, up to a maximum level. She correctly guessed that this maximum would be reached well inside $30$30 minutes. The researcher knew that biological mechanisms would then begin to reduce the levels at a much more moderate rate. This is why she chose to use two series.

Rational functions of the form $N\left(t\right)=\frac{kt}{t^2+a^2}$N(t)=ktt2+a2, with $k$k,$a$a constants, closely model the way ingested drugs accumulate and then reduce in the blood stream. Through calculus tools we can easily verify that the model shows a maximum of $\frac{k}{2a}$k2a when $t=a$t=a.

This specific model for this experiment becomes $N\left(t\right)=\frac{640t}{t^2+400}$N(t)=640tt2+400 and so, as the curve suggests, the maximum of $16$16 ng/ml predicted by the model to occur at $t=20$t=20, closely matches the data.

We also know that as $t\rightarrow\infty$t,  $N\left(t\right)\rightarrow0$N(t)0. The data points seem to be falling slightly faster than the model is predicting.

#### Worked Examples

##### Question 1

A particular river is about $3900$3900 miles long. It begins as an outlet of a lake at an altitude of $7000$7000 feet above sea level and empties into the sea.

Along the river, the distance, $y$y (in thousands of miles), from the river's origin at lake is a function of the river's height above sea level $x$x (in thousands of feet): $y=\frac{7-x}{0.637x+1.75}$y=7x0.637x+1.75.

1. Find the distance (in thousands of miles) of the river from its origin when it is $7000$7000 feet above sea level.

2. Find the distance (in thousands of miles) of the river from its origin when it is $1200$1200 feet above sea level.

##### Question 2

A cost-benefit model can be used to model the financial cost of removing some pollutant from the environment.

The cost $y$y (in thousands of dollars) of removing $x$x percent of a certain pollutant is modelled by the function $y=\frac{7.2x}{100-x}$y=7.2x100x.

1. Find the cost (in thousands of dollars) of removing $25%$25% of the pollutant.

2. Find the cost (in thousands of dollars) of removing $75%$75% of the pollutant.

### Outcomes

#### M8-7

Form and use trigonometric, polynomial, and other non-linear equations

#### 91575

Apply trigonometric methods in solving problems