# Example: Using the derivative

#### Prerequisites

## Understanding the situation

The derivative is a rate of change of something with respect to something, so it will be important in cases where we need to know quantitatively how something is changing. The derivative can help us see when something is changing a lot, when it is changing a little, and when it is changing not at all. In many cases the derivative creates a new function that represents something physical that we will learn to think out as "its own thing" and not just as a rate of change. One example of this is velocity — the rate of change of position. We can think of it in terms of position, but we can also think of it directly, learning to "see" velocity (and speed) of a moving object.

Since we will be spending a lot of time on velocity and acceleration, let's consider something different — an example that shows how we might use the derivative in a symbolic situation where we are interested in how something changes.

The reading What's a derivative? discusses how to use a derivative when we are looking at functions of time, but we will have to consider derivatives with respect to other variables — such as position (in the case of potential energies) -- and, interestingly, we can even consider derivatives of parameters (constants). Consider the following problem:

## Presenting a sample problem

The rate at which energy is used up in pushing a fluid through a pipe depends on a number of factors: the length of the pipe, $L$, the radius of the pipe, $R$, the viscosity of the fluid in the pipe, $\eta$ (eta), and the pressure difference between the ends of the pipe, $\Delta p$. This rate of energy dissipation is called the power, $P$, and is given for this example by the equation

$$P=\frac{\pi (\Delta p)^2R^4}{\eta L}$$

As part of a calculation to understand how the density of blood cells is optimized, we need to know how this rate of energy use depends on the radius of the pipe (artery) and the viscosity of the blood. In order to see which the rate of energy dissipation depends on more strongly, derive the expression for the rate of change of the power dissipation with respect to changes in the radius and viscosity.

## Solving this problem

What we need to do to answer this problem is to find the the derivative of the power dissipation, *$P$*, with respect to $R$ and $\eta$*. *This looks like a mess, but it's really rather straightforward. The trick is to "clear away the clutter" by identifying what's relevant, what's irrelevant, and giving the irrelevant stuff a simpler name.

Here's how it works. We first want to calculate $dP/dR$. If we are only thinking about changing $R$, then all the other stuff on the right side of the equation is being kept fixed so we can treat it as if it were all a constant. We'll call them "*A*". Then we can change the name of the parameter we are considering changing ($R$) to a more familiar one: $x$. Here are the equations that result.

$$P=\frac{\pi (\Delta p)^2R^4}{\eta L}$$

$$A = \frac{\pi (\Delta p)^2}{\eta L}$$

$$P=AR^4$$

$$f(x)=Ax^4$$

The first line is just our original equation. In the second, we take all the mess we are going to keep constant and give it a simple name: *A*. The equation then becomes just the third line. We are considering *P* as a function of *R* so this is just like what we do in math when we consider some function of *x*. We could even change the name of *P* (our function) to *f* and of *R* (our variable) to *x* as in the last line. This is now easy to differentiate and we can then run our substitutions backwards to get the answer.

First we differentiate:

$$\frac{df(x)}{dx} = \frac{d}{dx}(Ax^4) = 4Ax^3$$

Now we change the names of $f$ and $x$ back to their physical names, $P$ and $R$:

$$\frac{dP(R)}{dR} = \frac{d}{dR}(AR^4) = 4AR^3$$

Now we put back the definition of $A$ to get our final result:

$$\frac{dP(R)}{dR} = \frac{4 \pi (\Delta p)^2}{\eta L}R^3$$.

We can do the same thing for differentiating with respect to the viscosity by now taking $\eta = x$ and defining an appropriate constant. Leaving this as an exercise to the reader (all you need to remember is that $\frac{d}{dx} \frac{1}{x} = -\frac{1}{x^2}$)

$$\frac{dP(\eta)}{d\eta} = -\frac{ \pi (\Delta p)^2 R^4}{L}\frac{1}{\eta^2}$$.

This shows us that the power, $P$ changes directly with the third power of $R$ and inversely with the second power of $\eta$. A 5% change in $R$ would, for example, therefore have a bigger effect than a 5% change in $\eta$.

Although these initially look to be a mess, substituting a simple name for the cluster of constants not being changed made things easier. All we had to know was how to differentiate something to the n-th power. This renaming method is a general technique discussed in the section on Building your mathematical toolbelt on the page The repackaging tool: Changing physics equations to math (and back)

Joe Redish 12/27/14

Last Modified: May 22, 2019