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Online Notes / Calculus I / Derivatives / Interpretation of the Derivative
Calculus I

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The Definition of the Derivative E-Book   Chapter   Section Differentiation Formulas

Before moving on to the section where we learn how to compute derivatives by avoiding the limits we were evaluating in the previous section we need to take a quick look at some of the interpretations of the derivative.  All of these interpretations arise from recalling how our definition of the derivative came about.  The definition came about by noticing that all the problems that we worked in the first section in the chapter on limits required us to evaluate the same limit.

 

Rate of Change

The first interpretation of a derivative is rate of change.  This was not the first problem that we looked at in the limit chapter, but it is the most important interpretation of the derivative.  If  represents a quantity at any x then the derivative  represents the instantaneous rate of change of  at .

 

Example 1  Suppose that the amount of water in a holding tank at t minutes is given by .  Determine each of the following.

(a) Is the volume of water in the tank increasing or decreasing at  minute?   [Solution]

(b) Is the volume of water in the tank increasing or decreasing at  minutes?   [Solution]

(c) Is the volume of water in the tank changing faster at  or  minutes?   [Solution]

(d) Is the volume of water in the tank ever not changing?  If so, when?   [Solution]

 

Solution

In the solution to this example we will use both notations for the derivative just to get you familiar with the different notations.

 

We are going to need the rate of change of the volume to answer these questions.  This means that we will need the derivative of this function since that will give us a formula for the rate of change at any time t.  Now, notice that the function giving the volume of water in the tank is the same function that we saw in Example 1 in the last section except the letters have changed.  The change in letters between the function in this example versus the function in the example from the last section won’t affect the work and so we can just use the answer from that example with an appropriate change in letters.

 

The derivative is.

                                 

 

Recall from our work in the first limits section that we determined that if the rate of change was positive then the quantity was increasing and if the rate of change was negative then the quantity was decreasing.

 

We can now work the problem.

 

(a) Is the volume of water in the tank increasing or decreasing at  minute?   

In this case all that we need is the rate of change of the volume at  or,

                                 

 

So, at  the rate of change is negative and so the volume must be decreasing at this time.

[Return to Problems]

 

(b) Is the volume of water in the tank increasing or decreasing at  minutes?

Again, we will need the rate of change at .

                                   

 

In this case the rate of change is positive and so the volume must be increasing at .

[Return to Problems]

 

(c) Is the volume of water in the tank changing faster at  or  minutes?

To answer this question all that we look at is the size of the rate of change and we don’t worry about the sign of the rate of change.  All that we need to know here is that the larger the number the faster the rate of change.  So, in this case the volume is changing faster at  than at .

[Return to Problems]

 

(d) Is the volume of water in the tank ever not changing?  If so, when?

The volume will not be changing if it has a rate of change of zero.  In order to have a rate of change of zero this means that the derivative must be zero.  So, to answer this question we will then need to solve

                                     

 

This is easy enough to do.

                                       

 

So at  the volume isn’t changing.  Note that all this is saying is that for a brief instant the volume isn’t changing.  It doesn’t say that at this point the volume will quit changing permanently.

 

If we go back to our answers from parts (a) and (b) we can get an idea about what is going on.  At  the volume is decreasing and at  the volume is increasing.  So at some point in time the volume needs to switch from decreasing to increasing.  That time is .

 

This is the time in which the volume goes from decreasing to increasing and so for the briefest instant in time the volume will quit changing as it changes from decreasing to increasing.

[Return to Problems]

 

Note that one of the more common mistakes that students make in these kinds of problems is to try and determine increasing/decreasing from the function values rather than the derivatives.  In this case if we took the function values at ,  and  we would get,

 

 

 

Clearly as we go from  to  the volume has decreased.  This might lead us to decide that AT  the volume is decreasing.  However, we just can’t say that.  All we can say is that between  and  the volume has decreased at some point in time.  The only way to know what is happening right at  is to compute  and look at its sign to determine increasing/decreasing.  In this case  is negative and so the volume really is decreasing at .

 

Now, if we’d plugged into the function rather than the derivative we would have been gotten the correct answer for  even though our reasoning would have been wrong.  It’s important to not let this give you the idea that this will always be the case.  It just happened to work out in the case of  .

 

To see that this won’t always work let’s now look at .  If we plug  and  into the volume we can see that again as we go from  to  the volume has decreases.  Again, however all this says is that the volume HAS decreased somewhere between  and .  It does NOT say that the volume is decreasing at .  The only way to know what is going on right at  is to compute  and in this case  is positive and so the volume is actually increasing at .