2.3 Equations Involving Power Functions

Photo from the US Air Force on Flickr

Introduction

Do skydivers fall at a constant speed or do they go faster and faster the longer they fall? Is there anything they can do to control their speed or is it completely out of their hands?

After we spend a moment outlining the basic solving process for power functions, we will return to these questions.

Techniques for Solving Power Equations

Since most equations involving power functions will be of the form $y = k x^{p}$, the steps for solving all of them are essentially the same. First we make sure that $x^{p}$ is isolated. Once that has been done, we can solve for $x$ by taking the reciprocal power $1/p$ of both sides of the equation. In general it would look like this:

[ \begin{align} y &= k x^{p} \newline \frac{y}{k} &= x^{p} && \text{Divide both sides by }k \newline x &= \left( \frac{y}{k} \right)^{1/p} && \text{Apply the } 1/p \text{ power to both sides} \end{align} ]

Thus, the solution to $y = k x^{p}$ is given by $x = \left( \frac{y}{k} \right)^{1/p}$. This assumes, of course, that $p \neq 0$.

Also keep in mind that if $\frac{y}{k}<0$, then $x = \left( \frac{y}{k} \right)^{1/p}$ may not be a real number and the equation would have "no real number solutions".

Let's consider one concrete example. Suppose we need to solve $80 = 5 x^{2}$. The solving steps will be to first divide by $5$ and then apply the $1/2$ power.

[ \begin{align} 80 &= 5 x^{2} \newline 16 &= x^{2} && \text{Divide both sides by 5.}\newline x &= \pm (16)^{1/2} && \text{Apply the 1/2 power to both sides.} \newline x &= \pm 4 && \text{Simplify.} \end{align} ]

In this case there are two $x$ values that solve the equation. Anytime $p$ is even we might have multiple solutions. And while only positive solutions have significance for many applications, we should not dismiss negative solutions without good cause.

Stopping distance

photo by KOMUnews

When reconstructing accident scenes, police officers rely on a number of tools, including measuring the length of any skid marks made by the tires. With that distance, and an assessment of the road conditions, they can approximate the speed of the vehicle.

So how do they do it? In the prior section we saw how to calculate the stopping distance $d$ using the formula

[ d = \frac{0.03361 v^{2}}{\mu} ]

where $v$ is the speed of the car in feet per second and $\mu$ is the friction coefficient of the road surface.

When police need to find the speed of a car that was involved in an accident they measure the distance of the skid marks, make an observation of the road conditions, and solve this equation for $v$.

Suppose a police investigator finds $286$ foot long skid marks along a snowy road. How fast was the car going? To answer this question we insert the known value for $d$ and the estimated $\mu$ from the table and solve for $v$. The result is

[ \begin{align} 286 &= \frac{0.03361 v^{2}}{0.2} \newline 1701.874 &= v^{2} && \text{Divide both sides by } \frac{0.03361}{0.2}=0.16805 \newline v &= (1701.874)^{1/2} && \text{Apply the } 1/2 \text{ power to both sides} \newline v &\approx\ 41.3 && \text{Evaluate} \end{align} ]

The officer can safely conclude that the car was going a around 41.3 miles per hour.

It is easy to see that the police might need to repeat this process on a regular basis. To simplify the work needed, it makes sense to create a formula by solving the equation for $v$ without inserting values for $d$ and $\mu$. Assuming $v>0$, the result is

[ \begin{align} d &= \frac{0.03361 v^{2}}{\mu} \newline \mu d &= 0.03361 v^2 && \text{Multiply both sides by } \mu \newline \frac{\mu d}{0.03361} &= v^2 && \text{Divide both sides by } 0.03361 \newline v &= \left(\frac{\mu d}{0.03361}\right)^{1/2} && \text{Apply the } 1/2 \text{ power to both sides} \newline \end{align} ]

This formula will give the same results as solving, but is more efficient and less prone to error. In general, when a calculation needs to be repeated frequently, it's best to find a formula rather than repeating the solving process over and over.

Terminal Velocity

Skydiving photo from the US Air Force on Flickr

The force of aerodynamic drag $F_D$ acting on a skydiver is directly proportional to four parameters: the square of the velocity $v$, the density of air $\rho$, the drag coefficient $C$, and the amount of surface area $A$ that is facing the ground, with a constant of proportionality of $1/2$. Translating the statement above, we obtain the equation

[ F_D = \frac{1}{2} C \rho A v^2 ]

From this information, is it possible to find a formula for velocity?

Solving for $v$ gives the following formula:

[ v = \left(\frac{2 D}{ \rho C A}\right)^{1/2} = \sqrt{\frac{2D}{\rho C A}} ]

Of the four values ($D$, $\rho$, $C$ and $A$) that determine velocity, a skydiver has little control over $D$ and $\rho$. The drag force $D$ can never be greater than the skydiver's weight, otherwise they would float instead of fall, and the density of air $\rho$ at lower altitudes is fairly consistent at about $1.1 kg/m^{2}$.

A skydiver can, however, alter their shape which will affect both surface area $A$ and the drag coefficient $C$. ⊕

For example, suppose a skydiver weighing $734$ Newtons (about 165 lbs) goes into a head-first dive. How fast might they fall?

In this instance we know $D=165$ and $\rho=1.1$, but we'll have to make some assumptions for $A$ and $C$.

Assuming a stable head down position, a surface area of about $A=0.3 \thinspace m^{2}$ is reasonable, and let's choose a value for $C$ that is a little lower than that of a cylinder, say $C=0.7$.

With these values in place, we can evaluate $v$.

[ v = \sqrt{\frac{2D}{\rho C A}} = \sqrt{\frac{(2)(734)}{(1.1)(0.7)(0.3)}} \approx\ 79.7 \thinspace m/s ]

Our model predicts that the maximum speed (sometimes called terminal velocity) of this skydiver in a stable head-first dive is $79.7 \thinspace m/s$, or around $178 \thinspace mph$.

Find the Inverse of a Power Function

The steps we have gone through to solve equations are essentially the same steps needed to find the inverse of a power function. Recall that the inverse of a basic power function $f(x)=x^{p}$ is the power function $f^{-1}(x)=x^{1/p}$ whose power is the reciprocal of the other.

Of course, this assumes $p \neq 0$ and that we have restricted the domain if $f(x)=x^{p}$ is not one-to-one, such as in the case of $f(x)=x^{2}$. Finding the inverse of a standard power function model of the form $f(x) = k x^{p}$ will require the use of the shoes and socks method, since it contains two operations.

To find the inverse of $f(x) = 3 x^{1/5}$, for example, we focus on identifying the operations performed by the function and then invert the list to find the operations of the inverse function.

Now that we know what the inverse function does, we construct its equation as follows:

Notice that the inverse function does the opposite operations in the reverse order, which should always be the case.

Hull Speed

Photo by James Abbot on Flickr

One of the applications in Section 2.2 allowed us to find the hull speed of a boat using the function

[ V = 1.34 L^{1/2} ]

where $V$ is the hull speed in knots and $L$ is the length of the boat in feet at the water line. Since $V$ is a function of $L$, the inverse function, if it exists, will make $L$ a function of $V$. That is to say, if $V=f(L)$, then $L=f^{-1}(V)$. This means that $f^{-1}$, if we can find it, could be used to figure out how long a boat needs to be in order to achieve a particular hull speed.

To find the inverse of $V=f(L) = 1.34 L^{1/2}$ we begin by recognizing that this function first uses the $1/2$ power and then multiplies by $1.34$. The inverse must do the opposite operations in the reverse order: first dividing by $1.34$ and then applying the $2/1=2$ power.

Now that we know what the inverse function does, we construct its equation as follows:

Lastly, to ensure that $f^{-1}$ is one-to-one itself we should add the condition that $V\geq0$.

How might this new function get used? Suppose the Navy needs to design a new aircraft carrier with a hull speed of at least $V= 33$ knots. By evaluating $L=f^{-1}(30)$ the engineers could quickly know that the boat should be at least 606.48 feet long.

Note that the same answer can be found by inserting $V=33$ into the original function and solving for $L$. In other words, evaluating the inverse function is essentially the same as solving the original function.

Dial-up Internet

Let's consider one last example of a power function model and the importance of its inverse. Due to the increasing availability of high-speed internet, the number of subscribers to AOL's dial-up internet service has been decreasing. A 2013 study by the Pew Research Center reported that 3% of Americans still use dial-up connections, primarily in rural areas. That’s more than 9.4 million people.

The number of AOL subscribers (in millions) can be approximated by the power function $S = f(t) = 16.32 t^{-0.84}$ where $t$ is time in years since 2005. What is the inverse of this function, and what significance might it have?

Using the techniques discussed earlier, the inverse function is

[ t = f^{-1}(S) = \left(\frac{S}{16.32}\right)^{1/-0.84} ]

This function is useful because it allows us to approximate the time, in years since 2005, when AOL will have $S$ million subscriptions.

Many companies sell off divisions of their business when that part of the company shrinks. Suppose, for instance, that AOL plans to sell off their dial-up service once the number of subscribers drops to 1,000,000. To find out when that would happen we evaluate the inverse function when $S=1$.

[ f^{-1}(1) = \left(\frac{1}{16.32}\right)^{1/-0.84}\ \approx\ 27.78 ]

This suggest that AOL would continue to offer dial-up internet for about 28 years, until the year 2033. Of course, this assumes that our original model for the number of AOL subscriptions will continue to hold, which is never a sure thing.

Looking Ahead

In the next section we will discuss how to take data, like the AOL data, and create power function models that fit the data closely. Those models can then be used to predict important future behavior.