NZ Level 7 (NZC) Level 2 (NCEA)
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Solving for Constant of Proportionality in Power Functions
Lesson

$y=kx$y=kx

In Keeping it in Proportion, we learnt about proportional relationships. The types of relationships we looked at in these previous chapters are actually called directly proportional relationships. If two amounts are directly proportional, it means that as one amount increases, the other amount increases at the same rate.

For example, if you earn $\$18$$18 per hour, your earnings are directly proportional to the number of hours worked because $\text{earnings }=18\times\text{hours worked }$earnings =18×hours worked .

The mathematical symbol for "is directly proportional to..." is the letter alpha, which looks like this:

So, if we use the example above, we could write $e\alpha h$eαh. In other words, "Your earnings ($e$e) are directly proportional to the number of hours you work ($h$h)."

 

The constant of proportionality

The constant of proportionality is the value that relates the two amounts. In the example above, the constant would be $18$18.

We can write a general equation for amounts that are directly proportional.

General Equation for Amounts that are Directly Proportional 

$y=kx$y=kx

where $k$k is the constant of proportionality

Note that this is a linear relationship between $y$y and $x$x.

In this relationship - as $x$x gets larger, $y$y must also get larger. 

Once we solve the constant of proportionality, we can use it to answer other questions in this relationship.

 

Examples

Question 1

Consider the equation $P=90t$P=90t.

a) State the constant of proportionality.

b) Find the value of $P$P when $t=2$t=2.

 

$y=\frac{k}{x}$y=kx

Inverse proportion means that as one amount increases at the same rate that another amount decreases. Mathematically, we write this as $y$y$\alpha$α$\frac{1}{x}$1x. For example, speed and travel time are inversely proportional because the faster you go, the shorter your travel time.

General Equation for Amounts that are Inversely Proportional

We express these kinds of inversely proportional relationships generally in the form

$y=\frac{k}{x}$y=kx

where $k$k is the constant of proportionality and $x$x and $y$y are any variables

Can also be written as $y=kx^{-1}$y=kx1 (a power function)

Note that this is an inverse relationship between $y$y and $x$x

In this relationship - as $x$x gets larger, $y$y must get smaller. 

EXAMPLE 2

Consider the equation $s=\frac{360}{t}$s=360t

a) State the constant of proportionality.

Think: Inversely proportional relationships are written in the form $y=\frac{k}{t}$y=kt, where $k$k is the constant of proportionality.

Do: The constant of proportionality is $360$360.

b) Find the value of $s$s when $t=6$t=6. Give your answer as an exact value.

Think: We need to substitute $t=6$t=6 into the equation.

Do:

$s$s $=$= $\frac{360}{t}$360t
  $=$= $\frac{360}{6}$3606
  $=$= $60$60

 

c) Find the value of $s$s when $t=12$t=12. Give your answer as an exact value.

Think: This is just like (b), except we're going to substitute in $t=12$t=12.

Do:

$s$s $=$= $\frac{360}{t}$360t
  $=$= $\frac{360}{12}$36012
  $=$= $30$30

$y=kx^p$y=kxp

A general power function like $y=kx^p$y=kxp as two constants, the $k$k (constant of proporionality) and the $p$p, the power of the function.

As we saw earlier, the power of the function dictates the overall behaviour of the function and the $k$k dictates the dilation, and is also the proportionality constant.  It depicts the rate at which the relationship between $x^p$xp and $y$y exists.  

Solve for $k$k

To solve for $k$k, we would need to be given a number of other factors and depending on what we know, we will approach the solution differently.

Method 1 - provided 1 point

If we know at least one point, $\left(x_1,y_1\right)$(x1,y1) and the power $p$p, we can find $k$k using algebraic manipulation.

Let's look at this avenue with an example. 

Example 3

When a person sprints, their top speed is in direct proportion with the square of their stride length. If a person can run at a top speed of $12$12 km/h with a stride length of $0.95$0.95 m, then how fast can they run if they increase their stride length to $1.10$1.10 meters?

Think:  Let's start by setting up some variables.  Let $\text{speed}=S$speed=S (in km/h) and $\text{stride length}=l$stride length=l (in m)

Now, the first sentence said...

When a person sprints, their top speed is in direct proportion with the square of their stride length

 This tells us that $S=kl^2$S=kl2 ($k$k is the constant of proportionality we will need to find)

We are also given one point $$

Do: If we substitute in the values for $S$S and $l$l into the equation we will be able to find $k$k

$S$S $=$= $kl^2$kl2
$12$12 $=$= $k\times0.95^2$k×0.952
$\frac{12}{0.95^2}$120.952 $=$= $k$k
$k$k $=$= $13.30$13.30 (to 2dp)

So now we have the equation that dictates this relationship for this person.  $S=13.30l^2$S=13.30l2

The second part of the question just requires us to use this formula.  

$S$S $=$= $13.30l^2$13.30l2
$S$S $=$= $13.30\times1.10^2$13.30×1.102
$S$S $=$= $16.09$16.09 km/h

So the key tasks involved here were to develop the raw equation detailing the relationship, then use the point provided to solve for $k$k.  

 

Method 2 - provided 2 points

If we know two points, $\left(x_1,y_1\right)$(x1,y1) and $\left(x_2,y_2\right)$(x2,y2), we can solve simultaneously for $p$p and $k$k.

NOTE that this method uses Logarithms.  If you have not yet studied logarithms, then stick with method 1 for now.  

Example 4

Image source - San diego Zoo Image source - San diego Zoo  

It is known that a $49$49 kg mountain lion has an oxygen consumption of $11550$11550ml/hr and an African lion ($100$100 kg) has an oxygen consumption of $16500$16500 ml/hr.  Using this information:

a) develop a power function that fits the data

b) determine the oxygen consumption of a $3.2$3.2 kg house cat.  

Think: before starting we need to define some variables.  Let $W$W = weight of the cat in kilograms and $C$C = oxygen consumption measured in ml/hr. 

We don't know $k$k or $p$p at this stage, so our raw equation looks like

$C=kW^p$C=kWp

We have two sets of data

$\left(W_1,C_1\right)=\left(49,11550\right)$(W1,C1)=(49,11550)

$\left(W_2,C_2\right)=\left(100,16500\right)$(W2,C2)=(100,16500)

Do: knowing the points and raw equation, we create $2$2 equations and solve simultaneously.

$C$C $=$= $kW^p$kWp
$11550$11550 $=$= $k49^p$k49p $\left(1\right)$(1)
$16500$16500 $=$= $k100^p$k100p $\left(2\right)$(2)
$k$k $=$= $\frac{16500}{100^p}$16500100p Rearrange $$ for $k$k
$11550$11550 $=$= $\frac{16500}{100^p}\times49^p$16500100p×49p Use $$ with $k$k value replaced
$\frac{11550}{16500}$1155016500 $=$= $\frac{49^p}{100^p}$49p100p Solve for $p$p
$0.7$0.7 $=$= $\frac{49}{100}^p$49100p
$\log0.7$log0.7 $=$= $\log0.49^p$log0.49p
$\log0.7$log0.7 $=$= $p\times\log0.49$p×log0.49
$p$p $=$= $\frac{\log0.7}{\log0.49}$log0.7log0.49
$p$p $=$= $0.5$0.5  

Once we have $p$p we can then solve for $k$k

$k=\frac{16500}{100^p}=\frac{16500}{100^{0.5}}=\frac{16500}{10}=1650$k=16500100p=165001000.5=1650010=1650

And thus our full equation is $C=1650W^{0.5}$C=1650W0.5

Part b is a substitution exercise, substituting in the value of $3.2$3.2 for $W$W and evaluating $C$C.

$C$C $=$= $1650W^{0.5}$1650W0.5
  $=$= $1650\times3.2^{0.5}$1650×3.20.5
  $=$= $2951.61$2951.61

So the $3.2$3.2 kg house cat would have oxygen consumption of $2951.61$2951.61 ml/hr.

Further Examples

Example 5

If $\left(4,512\right)$(4,512) is on the curve, $P=kQ^4$P=kQ4, solve for $k$k.

Example 6

Consider the relationship where $y$y is directly proportional to the cube of $x$x.

  1. Using $k$k as a constant of proportionality, state the equation relating $x$x and $y$y.

  2. The following table of values shows the relationship between $x$x and $y$y.

    $x$x $1$1 $2$2 $3$3 $4$4 $5$5
    $y$y $3$3 $24$24 $81$81 $192$192 $375$375

    Solve for $k$k, the constant of proportionality.

Example 7

For the quadratic pictured here, determine the constant of proportionality $k$k.

Loading Graph...

  1. Give your answer in exact form.

Outcomes

M7-2

Display the graphs of linear and non-linear functions and connect the structure of the functions with their graphs

91257

Apply graphical methods in solving problems

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