Hyperbolas and Other Rational Functions
Solve applications involving hyperbolas
Identify Characteristics of Rational Functions
Transformations of Rational Functions
Graphing Rational Functions
Solve Rational Functions Analytically
Solve applications involving Rational Functions
Level 8 - NCEA Level 3
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)=120−7200x+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+60​→0. 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.
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=7−x0.637x+1.75​.
Find the distance (in thousands of miles) of the river from its origin when it is $7000$7000 feet above sea level.
Give your answer to three decimal places.
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=7.2x100−x​.
Find the cost (in thousands of dollars) of removing $25%$25% of the pollutant.
Form and use trigonometric, polynomial, and other non-linear equations
Apply trigonometric methods in solving problems