# Analyzing problems involving rates of change in applied contexts

Differential calculus is all about instantaneous rate of change. Let's see how this can be used to solve real-world word problems.

One way to interpret the derivative $f'$ of a function $f$ is that $f'(k)$ is the

*instantaneous rate of change*of $f$ at $x=k$. Let's see how this interpretation can be used to solve word problems.Say a water tank is being filled and the volume, in liters, of the water in the tank after $t$ seconds is given by the linear function $V_1(t)=\dfrac23t$.

The slope of the function, $\dfrac23$, represents its rate of change. In other words, the tank is being filled at a rate of $\dfrac23$ liters per second.

The rate of change of a linear function is always constant, which makes it relatively easy to reason about.

Now say a different tank is being filled, and this time the volume function $V_2(t)=0.1t^2$ isn't linear.

Notice how the graph's growth is gradual at first and becomes more steep towards the end. The rate of change of $V_2$ isn't constant.

If we want to analyze the rate of change of $V_2$, we can talk about its

*instantaneous rate of change*at any given point in time. The instantaneous rate of change of a function is given by the function's*derivative*.For example, $V_2'(5)=1$. Mathematically, this means that the slope of the line tangent to the graph of $V_2$ when $x=5$ is $1$. What does this mean in the context of our water tank?

The tangent line's slope indicates the curve's slope at that particular point in time. Since we already saw how slope gives us the rate of change, we can interpret $V_2'(5)=1$ as follows:

At $t=5$ seconds, the tank is being filled at a rate of $1$ liter per second.

Notice a couple things about this interpretation:

First, the rate is given in

*liters per second*. The units of a derivative are always a ratio of the dependent quantity (e.g. liters) over the independent quantity (e.g. seconds).Second, the rate is given for a specific point in time (e.g. $t=5$ seconds). This is because it's

*instantaneous*. Take another point in time, and the rate might be different. Look at an*interval*of time, and the rate isn't constant.### Common mistake: Forgetting to include units, or using incorrect units

**Remember:**When analyzing problems in applied contexts, remember we should always use units.

For example, in Problem 2, $H$ gets an input that is measured in

*weeks*and gives an output that is measured in*centimeters*. Its derivative $H'$ also gets an input that is measured in weeks, but its output is the rate*centimeters per week*.### Another common mistake: Using phrases that refer to “over an interval of time” rather than “at a point in time”

Derivatives are all about

*instantaneous*rate of change. Therefore, when we interpret the rate of a function given the value of its derivative, we should always refer to the specific point when that rate applies.## Solving problems that involve instantaneous rate of change

Consider the following problem:

Carlos has taken an initial dose of a prescription medication. The amount of medication, in milligrams, in Carlos's bloodstream after $t$ hours is given by the following function:

What is the instantaneous rate of change of the remaining amount of medication after $1$ hour?

The first thing that should pop up to us as we read this problem is that we are asked for the

*instantaneous rate of change*of a quantity. This means we need to use derivatives.The only function whose derivative we can use is $M$, but let's make sure this is what we want: $M$ gives the amount of medication in Carlos's bloodstream over time, and we are asked for the instantaneous rate of change of that amount. So yes, we need $M'$:

We are asked for the instantaneous rate of change

*after $1$ hour*, which means we need to evaluate $M'$ at $t=1$:Finally, we need to remember to use units. Since $M$ gives an amount in

*milligrams*for a given input in*hours*, the units in which we measure $M'$ are*milligrams per hours*.In conclusion, the instantaneous rate of change of the remaining amount of medication after $1$ hour is

**$-7.2$ milligrams per hour**.*Want more practice? Try this exercise.*

### Common mistake: Evaluating the original function rather than the derivative

**Remember:**When we are asked about the rate of change of a function $f$, we want to look for the derivative $f'$. Evaluating $f$ at a point will not provide us any information about the rate of change of $f$ at that point.