NZ Level 8 (NZC) Level 3 (NCEA) [In development]
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Solve Applications Involving Square Root Functions
Lesson

 

Example 1

On the Earth, when a rock is dropped from a $100$100 metre cliff face the elapsed time $t$t in seconds after it has fallen $d$d metres will be approximately given by $t_e=0.452\sqrt{d}$te=0.452d . On the Moon however, from a $100$100 metre cliff face, the rock would fall so that the elapsed time  after $d$d metres would be given by $t_m=1.11\sqrt{d}$tm=1.11d. Compare the elapsed times for the Earth and Moon for a rock at various points in the fall.

Perhaps the best way to answer this question is to graph both functions. The gravitational force on the Moon is about one-sixth that of the Earth, so we would expect the time to fall $100$100 metres would be longer for the Moon than that for the Earth. 

We can read off the graph that say the Moon rock takes $5$5 seconds to fall $20$20 metres whereas the Earth rock only takes $2$2 seconds. Again the Moon rock takes $10$10 seconds to fall $80$80 metres, where as the Earth rock only takes $4$4 seconds. 

It seems that for both the Earth and the Moon it takes twice as long to fall four times the distance. Isaac Newton discovered a universal law of falling 'bodies'. Ignoring air resistance, for any planet, if a body (such as a rock) takes $t$t seconds to fall $d$d metres, then it will take  $kt$kt seconds to fall $k^2d$k2d metres.

Look carefully at the Earth's curve again. It takes $1$1 second to fall $5$5 metres. Based on Newton's Law, tripling the fall time to $3$3 seconds will allow the rock to fall $3^2=9$32=9 times further. Check that it takes $3$3 seconds to fall $45$45 metres.  

This applet will allow you to compare the falling times from any planet (and the moon) with Earth. Remember that when comparing gravitational attractions, its not really about the size of a planet, but rather more about the density of its matter. The gas giant Uranus for example has almost the same fall curve as the Earth, but there is a vast difference between their sizes.  

Example 2 (extended investigation)

A team of three mathematicians wrote a book called Mathematical Whetstones. They know from their study of economics that the higher they set the sale price, the less books they'll sell. This is a consequence of the Law of Supply and Demand.

After some field research they conclude that the number of books $n$n they'll probably sell is related to the price $$b$b of each book by the equation $n=300-5b$n=3005b . For example, if they put the price up to $$60$60 they are likely to sell $n=300-5\left(60\right)=0$n=3005(60)=0 books. A price of $$30$30 and they might sell $150$150 books etc.

The total revenue $R$R on $n$n books becomes $R=nb=\left(300-5b\right)\times b$R=nb=(3005b)×b (in other words, the number of books sold times the price per book).

There will be costs of production of course! There is the cost of a trademark, advertising and commissions totalling $$1000$1000. There is also the printing costs amounting to $$20$20 per book. The total cost  $C$C therefore, for $n$n books, is $C=1000+20n$C=1000+20n, but since $n=300-5b$n=3005b, we can write:

C $=$= $1000+20n$1000+20n
C $=$= $1000+20(300-5b)$1000+20(3005b)
C $=$= $7000-100b$7000100b
     

The overall profit, say $P$P, will simply be the revenue minus the cost. So:

$P$P $=$= $R-C$RC
$P$P $=$= $\left(300-5b\right)b-\left(7000-100b\right)$(3005b)b(7000100b)
$P$P $=$= $\left(300b-5b^2\right)-\left(7000-100b\right)$(300b5b2)(7000100b)
$P$P $=$= $-5b^2+400b-7000$5b2+400b7000
     

Now the team only want to make a profit of $$900$900, even though they think they could make more money. $$900$900 means an individual profit of $$300$300 each. 

They want a graph of the number of books to sell as a function of the profit $P$P. They proceed by first completing the square on their profit equation:

$P$P $=$= $-5b^2+400b-7000$5b2+400b7000
$P$P $=$= $-5\left(b^2-80b+1400\right)$5(b280b+1400)
$P$P $=$= $-5\left(b^2-80b+1600-200\right)$5(b280b+1600200)
$P$P $=$= $-5\left[\left(b-40\right)^2-200\right]$5[(b40)2200]
$P$P $=$= $1000-5\left(b-40\right)^2$10005(b40)2
     

This new form of the equation will allow the team to get the book price $b$b as a function of the profit $P$P. All they need do is to make $b$b the subject of the formula:

$P$P $=$= $1000-5\left(b-40\right)^2$10005(b40)2
$5\left(b-40\right)^2$5(b40)2 $=$= $1000-P$1000P
$\left(b-40\right)^2$(b40)2 $=$= $\frac{1}{5}\left(1000-P\right)$15(1000P)
$b-40$b40 $=$= $\pm\frac{1}{\sqrt{5}}\sqrt{\left(1000-P\right)}$±15(1000P)
$b$b $=$= $\pm\frac{1}{\sqrt{5}}\sqrt{\left(1000-P\right)}+40$±15(1000P)+40
     

Look carefully and you will see, not one, but two root functions, both of which have a dilation factor ($a=\pm\frac{1}{\sqrt{5}}$a=±15) and both starting off at  $\left(1000,40\right)$(1000,40).

The two graphs are drawn together here:

The connection between the root function and the parabola is made very clear. We can see the two root functions starting from $\left(1000,40\right)$(1000,40) and moving toward the '$b$b' axis.

The maximum profit possible is $$1000$1000 made by setting the price per book at $$40$40

However based on a profit  of $$900$900 they wanted to make, they have two choices. They can set the price a little bit below $$40$40, or a little bit above $$40$40 as indicated by the dashed lines in the diagram.

To work out the actual book price in each case, simply substitute $P=900$P=900 into each function so that:

$b_1$b1 $=$= $\frac{1}{\sqrt{5}}\sqrt{1000-\left(900\right)}+40=44.47$151000(900)+40=44.47
$b_2$b2 $=$= $-\frac{1}{\sqrt{5}}\sqrt{1000-\left(900\right)}+40=35.53$151000(900)+40=35.53
     

After much discussion, the team decide to sell the Mathematics books for $45 even. Their decision to round up will slightly effect their net profit.

If we return to the first equations, the team are thus hoping to sell $n=300-5(45)=75$n=3005(45)=75 books and bring in as revenue $$3375$3375. The cost of production is given as $C=7000-100(45)=2500$C=7000100(45)=2500, and so the net profit on $75$75 books is the difference $$875$875.

More Worked Examples

QUESTION 1

The length of a blue whale calf in its first few months is modelled approximately by the equation $l=1.5\sqrt{t+4}$l=1.5t+4, where $l$l represents its length in meters at $t$t months of age.

  1. Complete the table of values.

    Months ($t$t) $0$0 $60$60 $96$96
    Length ($l$l) $\editable{}$ $\editable{}$ $\editable{}$
  2. Graph the length function.

    Loading Graph...

  3. In this practical context, what are the possible values of $t$t?

  4. Is the calf’s length increasing by the same amount each month?

    Yes

    A

    No

    B

    Yes

    A

    No

    B
  5. According to the model, what will be the length of the whale at $7$7 years of age? Give your answer correct to one decimal place.

  6. An orca whale calf is born shorter than a blue whale calf. If its growth can be modelled by the equation $L=k\sqrt{t+4}$L=kt+4, which of the following could be a possible value of $k$k?

    $0.7$0.7

    A

    $1.5$1.5

    B

    $2.3$2.3

    C

    $0.7$0.7

    A

    $1.5$1.5

    B

    $2.3$2.3

    C

QUESTION 2

On Earth, the time (in seconds) taken for an object to fall through a distance of $d$d feet is given approximately by the equation $T=\frac{\sqrt{d}}{5}$T=d5.

  1. At a bungee jumping facility, you can choose to jump from a height of $441$441 feet or $225$225 feet above a river. Noah and Deborah jump simultaneously from heights of $441$441 feet and $225$225 feet respectively. How many seconds before Deborah must Noah jump so that they both reach the bottom of the jump at the same time?

  2. Under water, objects will fall much more slowly and the time taken to fall through a certain distance is modelled by function $Q$Q. Which of the following could be the graphs of $T$T and $Q$Q?

    Loading Graph...

    A

    Loading Graph...

    B

    Loading Graph...

    C

    Loading Graph...

    D

    Loading Graph...

    A

    Loading Graph...

    B

    Loading Graph...

    C

    Loading Graph...

    D

QUESTION 3

An electron is constantly moving, and its position with respect to the fixed point ($x=0$x=0) can be modelled by the equation $x=-\left(t-4\right)^2+4$x=(t4)2+4, where $x$x is its position after $t$t seconds. After $4.5$4.5 seconds, it is at $x=3.75$x=3.75.

  1. Rearrange the equation to make $t$t the subject.

  2. Using the result of the previous part, determine the time $t$t at which the electron is at the fixed position.

  3. Determine how long it takes the electron to get from $x=-5.1$x=5.1 to $x=-6.3$x=6.3.

    Give your answer to the nearest tenth of a second.

Outcomes

M8-7

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

91575

Apply trigonometric methods in solving problems

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