Instantaneous Rate of Change From a Math Geek

instantaneous rate of changeIf you’re a math geek like me, you love calculus.  If you’re like most, the word makes you cringe.  But it really doesn’t have to be that way.  Really.  All that calculus is involves how things change.  Geometry looks at shapes and angles.  Calculus examines how certain things change by using graphs which are beautifully translated into equations that we can work with.  Finding a car’s instantaneous rate of change of speed is an often looked at problem.  As with any subject, learning extra math skills to give you an edge might just be what you need to succeed in math. 

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Rates of Change

Let’s take a look at at the instantaneous rate of change problem.

Here we are going to consider a function, f(x), that represents some quantity that varies as x varies.  For instance, maybe f(x) represents the amount of water in a holding tank after x minutes.  Or maybe f(x) is the distance traveled by a car after x hours.  In both of these examples we use x to represent time.  Of course x doesn’t have to represent time, but it makes for examples that are easy to visualize.

What we want to do here is determine just how fast f(x) is changing at some point, say x=a . This is called the instantaneous rate of change or sometimes just rate of change of f(x) at x=a .

As with the tangent line problem all that we’re going to be able to do at this point is to estimate the rate of change.  So let’s continue with the examples above and think of f(x) as something that is changing in time and x being the time measurement.  Again x doesn’t have to represent time but it will make the explanation a little easier.  While we can’t compute the instantaneous rate of change at this point we can find the average rate of change.

To compute the average rate of change of f(x) at x=a, all we need to do is to choose another point, say x, and then the average rate of change will be:

Average rate of change = [change in  f(x)]/change in x.

In other words = [f(x)-f(a)]/[x-a]

Then to estimate the instantaneous rate of change at x=a all we need to do is to choose values of x getting closer and closer to x=a (don’t forget to chose them on both sides of x=a) and compute values of A.R.C.  We can then estimate the instantaneous rate of change from that.

Lets take a look at an example.

Suppose that the amount of air in a balloon after t hours is given by

V(t) = t^3 – 6t^2 + 35

Solution

Okay.  Here is a table of values of t and the average rate of change for those values.

t

A.R.C.

t

A.R.C.

625.047.0
5.519.754.510.75
5.115.914.914.11
5.0115.09014.9914.9101
5.00115.0090014.99914.991001
5.000115.000900014.999914.99910001

So, from this table it looks like the average rate of change is approaching 15 and so we can estimate that the instantaneous rate of change is 15 at this point.

So, just what does this tell us about the volume at this point?  Let’s put some units on the answer from above.  This might help us to see what is happening to the volume at this point.  Let’s suppose that the units on the volume were in cm3.  The units on the rate of change (both average and instantaneous) are then cm3/hr.

We have estimated that at t=5 the volume is changing at a rate of 15 cm3/hr.  This means that at  the volume is changing in such a way that, if the rate were constant, then an hour later there would be 15 cm3 more air in the balloon than there was at t=5.

We do need to be careful here however.  In reality there probably won’t be 15 cm3 more air in the balloon after an hour.  The rate at which the volume is changing is generally not constant and so we can’t make any real determination as to what the volume will be in another hour.  What we can say is that the volume is increasing, since the instantaneous rate of change is positive, and if we had rates of change for other values of t we could compare the numbers and see if the rate of change is faster or slower at the other points.

For instance, at  the instantaneous rate of change at t=4 is 0 cm3/hr and at  the instantaneous rate of change at t=3 is -9 cm3/hr.  I’ll leave it to you to check these rates of change.  In fact, that would be a good exercise to see if you can build a table of values based on these rates of change.

Anyway, back to the example.  At t=4  the rate of change is zero and so at this point in time the volume is not changing at all.  That doesn’t mean that it will not change in the future.  It just means that exactly at t=4  the volume isn’t changing.  Likewise at t=3 the volume is decreasing since the rate of change at that point is negative.  We can also say that, regardless of the increasing/decreasing aspects of the rate of change, the volume of the balloon is changing faster at t=5  than it is at t=3 since 15 is larger than 9.

For more Calculus help on rates and other derivative problems, check out Calculus I – Applications of Derivatives!