# Untitled Page 4

- Page ID
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## 1.2 Continuous change

Did you notice that I sneaked something past you in the example
of water filling up a reservoir? The *x* and x^{·} functions I've
been using as examples have all been functions defined on the integers,
so they represent change that happens in discrete steps, but the flow
of water into a reservoir is smooth and continuous. Or is it? Water is
made out of molecules, after all. It's just that water molecules are so small that
we don't notice them as individuals. Figure g shows
a graph that is discrete, but almost appears continuous because the scale has
been chosen so that the points blend together visually.

The physicist Isaac Newton started
thinking along these lines in the 1660's, and figured out ways of analyzing
*x* and x^{·} functions that were truly continuous. The notation x^{·}
is due to him (and he only used it for continuous functions). Because he was
dealing with the continuous *flow* of change, he called his new set of mathematical
techniques the method of *fluxions*, but nowadays it's known as the calculus.

Newton was a physicist, and he needed to invent the calculus as part of his study
of how objects move. If an object is moving in one dimension, we can specify its
position with a variable *x*, and *x* will then be a function of time, *t*.
The rate of change of its position, x^{·}, is its speed, or velocity.
Earlier experiments by Galileo
had established that when a ball rolled down
a slope, its position was proportional to *t*^{2}, so Newton inferred that
a graph like figure h would be typical for any object
moving under the influence of a constant force. (It could be 7*t*^{2}, or
*t*^{2}/42, or anything else proportional to *t*^{2}, depending on the force
acting on the object and the object's mass.)

Because the functions are continuous, not discrete, we can no longer define
the relationship between *x* and x^{·} by saying *x* is a running sum of
x^{·}'s, or that x^{·} is the difference between two successive *x*'s.
But we already found a geometrical relationship between the two functions
in the discrete case, and that can serve as our definition for the continuous
case: *x* is the area under the graph of x^{·}, or, if you like,
x^{·} is the slope of the graph of *x*. For now
we'll concentrate on the slope idea.

This definition is still a little vague, because we haven't defined what we mean by the “slope” of a curving graph. For a discrete graph like figure d, we could define it as the slope of the line drawn between neighboring points. Visually, it's clear that the continuous version of this is something like the line drawn in figure h. This is referred to as the tangent line.

We still need to convert this intuitive idea of a tangent line into a formal
definition. In a typical example like figure h, the tangent
line can be defined as the line that touches the graph at a certain point,
but, unlike the line in figure i, doesn't cut across
the graph at that point.^{3} By measuring with a ruler on figure
h, we find that the slope is very close to 1, so evidently
x^{·}(1)=1.
To prove this, we construct the function representing the line: ℓ(*t*)=*t*-1/2.
We want to prove that this line doesn't cross the graph of *x*(*t*)=*t*^{2}/2. The difference between the
two functions, *x*-ℓ, is the polynomial *t*^{2}/2-*t*+1/2, and this polynomial will be zero for any
value of *t* where the line touches or crosses the curve. We can use the quadratic formula
to find these points, and the result is that there is only one of them, which is *t*=1.
Since *x*-ℓ is positive for at least some points to the left and right of *t*=1, and
it only equals zero at *t*=1, it must never be negative, which means that the line
always lies below the curve, never crossing it.

### A derivative

That proves that x^{·}(1)=1, but it was a lot of work, and we don't want to do
that much work to evaluate x^{·} at every value of *t*. There's a way to
avoid all that, and find a formula for x^{·}. Compare figures h
and j. They're both graphs of the same function, and they both look the same.
What's different? The only difference is the scales: in figure j, the *t* axis
has been shrunk by a factor of 2, and the *x* axis by a factor of 4. The graph looks the same,
because doubling *t* quadruples *t*^{2}/2. The tangent line here is the tangent line at *t*=2,
not *t*=1, and although it looks like the same line as the one in figure h,
it isn't, because the scales are different. The line in figure h had a slope
of rise/run=1/1=1, but this one's slope is 4/2=2. That means x^{·}(2)=2.
In general, this scaling argument shows that x^{·}(t)=t for any *t*.

This is called *differentiating*: finding a formula for the function x^{·}, given a formula
for the function *x*. The term comes from the idea that for a discrete function, the slope is
the difference between two successive values of the function. The function x^{·} is referred to as
the *derivative* of the function *x*, and the art of differentiating is differential
calculus.
The opposite process, computing a formula for *x* when given x^{·}, is called integrating,
and makes up the field of integral calculus;
this terminology is based on the idea that computing
a running sum is like putting together (integrating) many little pieces.

Note the similarity between this result for continuous functions,

and our earlier result for discrete ones,

The similarity is no coincidence. A continuous function is just a smoothed-out version of
a discrete one. For instance, the continuous version of the staircase function shown in figure
b on page 7 would simply be a triangle
without the saw teeth sticking out; the area of those ugly sawteeth is what's represented
by the *n*/2 term in the discrete result *x*=(*n*^{2}+*n*)/2, which is the only thing that makes
it different from the continuous result *x*=*t*^{2}/2.

### Properties of the derivative

It follows immediately from the definition of the derivative that multiplying a function by
a constant multiplies its derivative by the same constant, so for example since we know
that the derivative of *t*^{2}/2 is *t*, we can immediately tell that the derivative of
*t*^{2} is 2*t*, and the derivative of *t*^{2}/17 is 2*t*/17.

Also, if we add two functions, their derivatives add. To give a good example of this, we
need to have another function that we can differentiate, one that isn't just some multiple
of *t*^{2}. An easy one is *t*: the derivative of *t* is 1, since the graph
of *x*=*t* is a line with a slope of 1, and the tangent line lies right on top of the
original line.

##### Example 2

The derivative of 5*t*

^{2}+2

*t*is the derivative of 5

*t*

^{2}plus the derivative of 2

*t*, since derivatives add. The derivative of 5

*t*

^{2}is 5 times the derivative of

*t*

^{2}, and the derivative of 2

*t*is 2 times the derivative of

*t*, so putting everything together, we find that the derivative of 5

*t*

^{2}+2

*t*is (5)(2

*t*)+(2)(1)=10

*t*+2.

The derivative of a constant is zero, since a constant function's graph is a horizontal line, with a slope of zero. We now know enough to differentiate any second-order polynomial.

##### Example 3

◊ An insect pest from the United States is inadvertently released in a village in rural China. The pests spread outward at a rate of*s*kilometers per year, forming a widening circle of contagion. Find the number of square kilometers per year that become newly infested. Check that the units of the result make sense. Interpret the result.

◊ Let *t* be the time, in years, since the pest was introduced.
The radius of the circle is *r*=*st*, and its area is *a*=π *r*^{2}=π(*st*)^{2}.
To make this look like a polynomial, we have to rewrite it as
*a*=(π *s*^{2})*t*^{2}. The derivative is

The units of *s* are km/year, so squaring it gives km^{2}/year^{2}.
The 2 and the π are unitless, and multiplying by *t* gives units
of km^{2}/year, which is what we expect for a^{·}, since
it represents the number of square kilometers per year that become infested.

Interpreting the result, we notice a couple of things. First, the rate
of infestation isn't constant; it's proportional to *t*, so people might not
pay so much attention at first, but later on the effort required to combat the
problem will grow more and more quickly. Second, we notice that the
result is proportional to *s*^{2}. This suggests that anything that could be
done to reduce *s* would be very helpful. For instance, a measure that cut
*s* in half would reduce a^{·} by a factor of four.

### Higher-order polynomials

So far, we have the following results for polynomials up to order 2:

function | derivative |

1 | 0 |

t | 1 |

t^{2} | 2t |

Interpreting 1 as *t*^{0}, we detect what seems to be a general
rule, which is that the derivative of *t*^{k} is *kt*^{k-1}. The proof is straightforward
but not very illuminating if carried out with the methods developed in this chapter,
so I've relegated it to page 140. It can be proved
much more easily using the methods of chapter 2.

##### Example 4

◊ If *x*=2*t*^{7}-4*t*+1, find x^{·}.

◊ This is similar to example 2, the only difference being
that we can now handle higher powers of *t*. The derivative of *t*^{7} is 7*t*^{6}, so
we have

##### Example 5

◊ Calculate 3^{-1}and 3.01

^{-1}. Does this seem consistent with a conjecture that the rule for differentiating

*t*

^{k}holds for

*k*<0?

◊ We have 3^{-1}≈ 0.33333 and 3.01^{-1}≈ 0.332223,
the difference being -1.1× 10^{-3}. This suggests that the graph of *x*=1/*t* has
a tangent line at *t*=3 with a slope of about

If the rule for differentiating *t*^{k} were to hold, then we would have
x^{·}=-t^{-2}, and evaluating this at *x*=3 would give -1/9, which is
indeed about -0.11. Yes, the rule does appear to hold for negative *k*,
although this numerical check does not constitute a proof. A proof is given
in example 10 on p. 27.

### The second derivative

I described how Galileo and Newton found that an object subject to
an external force, starting from rest, would have a velocity x^{·} that was
proportional to *t*, and a position *x* that varied like *t*^{2}. The proportionality
constant for the velocity is called the acceleration, *a*, so that
x^{·}=at and *x*=*at*^{2}/2. For example, a sports car accelerating from
a stop sign would have a large acceleration, and its velocity *at* at a given
time would therefore be a large number. The acceleration can be thought of
as the derivative of the derivative of *x*, written x¨, with two
dots. In our example, x¨=a. In general, the acceleration doesn't need
to be constant. For example, the sports car will eventually have to stop
accelerating, perhaps because the backward force of air friction becomes as
great as the force pushing it forward. The total force acting on the car would
then be zero, and the car would continue in motion at a constant speed.

##### Example 6

Suppose the pilot of a blimp has just turned on the motor that runs its propeller, and the propeller is spinning up. The resulting force on the blimp is therefore increasing steadily, and let's say that this causes the blimp to have an acceleration x¨=3t, which increases steadily with time. We want to find the blimp's velocity and position as functions of time.

For the velocity, we need a polynomial
whose derivative is 3*t*. We know that the derivative of *t*^{2} is 2*t*, so we need
to use a function that's bigger by a factor of 3/2: x^{·}=(3/2)t^{2}. In fact,
we could add any constant to this, and make it x^{·}=(3/2)t^{2}+14, for example,
where the 14 would represent the blimp's initial velocity. But since the blimp has
been sitting dead in the air until the motor started working, we can assume
the initial velocity was zero. Remember, any time you're working backwards
like this to find a function whose derivative is some other function (integrating,
in other words), there is the possibility of adding on a constant like this.

Finally, for the position, we need something whose derivative is (3/2)*t*^{2}.
The derivative of *t*^{3} would be 3*t*^{2}, so we need something half as big
as this: *x*=*t*^{3}/2.

The second derivative can be interpreted as a measure of the curvature of the graph,
as shown in figure k. The graph of the function *x*=2*t* is a line,
with no curvature. Its first derivative is 2, and its second derivative is zero.
The function *t*^{2} has a second derivative of 2, and the more tightly curved
function 7*t*^{2} has a bigger second derivative, 14.

Positive and negative signs of the second derivative indicate concavity.
In figure l,
the function *t*^{2} is like a cup with its mouth pointing up. We say that it's “concave up,” and this
corresponds to its positive second derivative. The function 3-*t*^{2}, with a second derivative less than
zero, is concave down. Another way of saying it is that if you're driving along a road shaped like *t*^{2},
going in the direction of increasing *t*, then your steering wheel is turned to the left, whereas on
a road shaped like 3-*t*^{2} it's turned to the right.

Figure m shows a third possibility. The function *t*^{3} has a derivative 3*t*^{2}, which equals
zero at *t*=0. This called a point of inflection. The concavity of the graph is down on the left, up on the right.
The inflection point is where it switches from one concavity to the other. In the alternative description in
terms of the steering wheel, the inflection point is where your steering wheel is crossing from left to right.