NZ Level 7 (NZC) Level 2 (NCEA)
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 heαh. In other words, "Your earnings (ee) are directly proportional to the number of hours you work (hh)." ### The constant of proportionality The constant of proportionality is the value that relates the two amounts. In the example above, the constant would be 1818. We can write a general equation for amounts that are directly proportional. General Equation for Amounts that are Directly Proportional y=kxy=kx where kk is the constant of proportionality Note that this is a linear relationship between yy and xx. In this relationship - as xx gets larger, yy 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=90tP=90t. a) State the constant of proportionality. b) Find the value of PP when t=2t=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 yy\alphaα\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 kk is the constant of proportionality and xx and yy are any variables Can also be written as y=kx^{-1}y=kx1 (a power function) Note that this is an inverse relationship between yy and xx In this relationship - as xx gets larger, yy 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 kk is the constant of proportionality. Do: The constant of proportionality is 360360. b) Find the value of ss when t=6t=6. Give your answer as an exact value. Think: We need to substitute t=6t=6 into the equation. Do:  ss == \frac{360}{t}360t​ == \frac{360}{6}3606​ == 6060 c) Find the value of ss when t=12t=12. Give your answer as an exact value. Think: This is just like (b), except we're going to substitute in t=12t=12. Do:  ss == \frac{360}{t}360t​ == \frac{360}{12}36012​ == 3030 ## y=kx^py=kxp A general power function like y=kx^py=kxp as two constants, the kk (constant of proporionality) and the pp, the power of the function. As we saw earlier, the power of the function dictates the overall behaviour of the function and the kk dictates the dilation, and is also the proportionality constant. It depicts the rate at which the relationship between x^pxp and yy exists. ## Solve for kk To solve for kk, 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 pp, we can find kk 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 1212 km/h with a stride length of 0.950.95 m, then how fast can they run if they increase their stride length to 1.101.10 meters? Think: Let's start by setting up some variables. Let \text{speed}=Sspeed=S (in km/h) and \text{stride length}=lstride 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^2S=kl2 (kk 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 $$for kk 1155011550 == \frac{16500}{100^p}\times49^p16500100p​×49p Use$$ 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$49100​p $\log0.7$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 $y$y $1$1 $2$2 $3$3 $4$4 $5$5 $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.

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