Chapter 3: Instantaneous Rate of Change : The Derivative

Introduction

If we think of a linear function like , then it is easy to see that the ratio is really independent of our choice of the value of as well as the change ; it is simply the slope m of the line. For other functions, life is not so easy!

Example 3.1: The Chord of a Circle

Take, for example, = (the upper semicircle of radius centered at the origin). When , we find that = . Suppose we want to know how much changes when x increases a little, say to or .

Let us look at the ratio for our function = when changes from to . Here = is the change in , and:

=

= .

In general, if we draw the chord from the point ( ) to a nearby point on the semicircle ( ), the slope of this chord is the difference quotient.

slope of chord = =

For example, if x changes only from to , then the difference quotient (slope of the chord) is equal to .

Note that unlike the line, the difference quotient depends on as well as .

Thus, as gets smaller and smaller, the slope of the chord gets closer and closer to the slope of the tangent line.

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In the particular case of a circle, there's a simple way to find the derivative. Namely, the tangent to a circle at a point is perpendicular to the radius drawn to the point of contact, and so its slope is the negative reciprocal of the slope of the radius. The radius joining ( ) to
( ) has slope . Hence, the tangent line has slope .... We write: .... Notice that when is small, such as , the slope of the chord joining ( ) to ( ) is a good approximation to the value of . For example, when , we saw that this difference quotient is -0.2919, which is close to .

Now suppose that we choose a different x, say , and we want to know how fast is changing as increases a little from 15. As an approximation we could take the chord joining ( ) to ( ) when :

=

The limiting value of this difference quotient as becomes smaller and smaller is what we mean by . The value is the slope of the tangent to the circle at the point ( ). As before, we can find this slope exactly, since the tangent to a circle is perpendicular to the radius. The answer is = -0.75. Again we see that, if is small (like ), then the slope of the chord joining ( ) to ( ) is very close to this value (here x =15).

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Exercise 3.2

Suppose we are on an airless planet where the gravitational constant is (the moon has gravitational constant close to this). Then it turns out that an object dropped from rest falls through a distance of exactly meters after seconds have elapsed. (Falling body problems will be taken up more systematically later.) That is, if denotes the distance fallen in meters, we have y = . Suppose that we are interested in the moment sec, and we want to know how much the object falls as increases from 1 to seconds. The distance fallen in that short time interval is , which simplifies to . For example, in the time from sec to sec the object falls a distance m. The difference quotient for this 0.01-sec time interval is = m/sec. It is customary to call this difference quotient the average velocity during the time interval . In other words, it is the distance traveled divided by time in this interval. More generally, the average between and sec. is

= .

Now the derivative of y = at the point is the limiting value of this difference quotient as h approaches zero:

f'(1) = ,

Note that our calculation was independent of the sign of , even though our wording seems to suggest that h is positive. In reality, we need to get the same answer regardless of the sign of . Later, we will discuss what happens when it does depend on the sign of h.

To understand the distinction between average and instantaneous velocity, think of a car: the average velocity during a time interval means the distance traveled divided by time, while the instantaneous velocity at a particular instant means the speed indicated on the speedometer at that moment.

The graph below shows distance fallen as a function of time in this example: . The diagram next to it is an enlargement near the point sec, where meter (i.e., the point (1,1) on the curve). The slope of the line joining ( ) to ( ) on the curve is the average velocity of the falling object during the time interval from 1 to 1.1 sec. The slope of the line which just grazes the parabola at the point (1,1) is the instantaneous velocity of the object at time 1 sec. Thus, we have two interpretations of the derivative: as the slope of the tangent line in the graph, or as the instantaneous velocity.

The curve is a parabola and it also has a geometric construction for the tangent. Let O be the vertex and l the tangent at O. At any point P of the parabola drop a perpendicular from P to L and let Q be the foot of the perpendicular. Then the tangent at P is the line PM where M is the midpoint of OQ. (See picture - provide your own labels {M,O,P,Q} - in the picture P= (1,1) ). Of course, this is far more complicated than the circle. You should try to prove it by using our calculations above.

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As a final example, we look at the reciprocal function y = . Suppose we want to find the derivative f'(x) when .

One way is to approximate. We can join the point (2,0.5) to the point ( ) for small values of h, say 0.1, 0.01, 0.0001, or 0.0000001, and then compute the slope of the chord. We get the following table:

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