1.2. Continuous ChangeFigure F. Isaac Newton (16431727) Did you notice that I sneaked something past you in the example of water filling up a reservoir? The x and \(\dot{x}\) 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. Figure G. On this scale, the graph of (n^{2}+n)/2 appears almost continuous. The physicist Isaac Newton started thinking along these lines in the 1660's, and figured out ways of analyzing x and \(\dot{x}\) functions that were truly continuous. The notation \(\dot{x}\) 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. Figure H. The function x(t)=t^{2}/2, and its tangent line at the point (1,1/2). 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, \(\dot{x}\), 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 7t^{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 \(\dot{x}\) by saying x is a running sum of \(\dot{x}\)'s, or that \(\dot{x}\) 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 \(\dot{x}\), or, if you like, \(\dot{x}\) is the slope of the graph of x. For now we'll concentrate on the slope idea. Figure I. This line isn't a tangent line: it crosses the graph. 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 \(\dot{x}(1)=1\). [3] In the case where the original graph is itself a line, the tangent line simply coincides with the graph, and this also satisfies the definition, because the tangent line doesn't cut across the graph; it lies on top of it. There is one other exceptional case, called a point of inflection, which we won't worry about right now. To prove this, we construct the function representing the line: ℓ(t)=t1/2. We want to prove that this line doesn't cross the graph of x(t)=t^{2}/2. The difference between the two functions, xℓ, is the polynomial t^{2}/2t+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 xℓ 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 DerivativeThat proves that \(\dot{x}(1)=1\), but it was a lot of work, and we don't want to do that much work to evaluate \(\dot{x}\) at every value of t. There's a way to avoid all that, and find a formula for \(\dot{x}\). 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 \(\dot{x}(2)=2\). In general, this scaling argument shows that \(\dot{x}(t)=t\) for any t. Figure J. The function t^{2}/2 again. How is this different from Figure H? This is called differentiating: finding a formula for the function \(\dot{x}\), 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 \(\dot{x}\) 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 \(\dot{x}\), 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, \[x=t^2/2 \dot{x}=t ,\] and our earlier result for discrete ones, \[x=(n^2+n)/2 \dot{x}=n .\] The similarity is no coincidence. A continuous function is just a smoothedout version of a discrete one. For instance, the continuous version of the staircase function shown in Figure B 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 DerivativeIt 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 2t, and the derivative of t^{2}/17 is 2t/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 2The derivative of 5t^{2}+2t is the derivative of 5t^{2} plus the derivative of 2t, since derivatives add. The derivative of 5t^{2} is 5 times the derivative of t^{2}, and the derivative of 2t is 2 times the derivative of t, so putting everything together, we find that the derivative of 5t^{2}+2t is (5)(2t)+(2)(1) = 10t+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 secondorder 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 \begin{align} \dot{a} = (\pi s^2)(2t) \dot{a} = (2\pi s^2) t \end{align}
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 \(\dot{a}\), 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 \(\dot{a}\) by a factor of four. Higherorder PolynomialsSo far, we have the following results for polynomials up to order 2:
Interpreting 1 as t^{0}, we detect what seems to be a general rule, which is that the derivative of t^{k} is kt^{k1}. 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 (in PDF version). It can be proved much more easily using the methods of chapter 2. Example 4◊ If x=2t^{7}4t+1, find \(\dot{x}\). ◊ 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 7t^{6}, so we have \begin{align} \dot{x} &= (2)(7t^6)+(4)(1)+0 \\ &= 14t^64 \end{align}
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 \[\frac{1.1\times 10^{3}}{0.01} = 0.11.\] If the rule for differentiating t^{k} were to hold, then we would have \(\dot{x}=t^{2}\), 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. The Second DerivativeI described how Galileo and Newton found that an object subject to an external force, starting from rest, would have a velocity \(\dot{x}\) 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 \(\dot{x}=at\) 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 \(\ddot x\), with two dots. In our example, \(\ddot x = a\). 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 6Suppose 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 \(\ddot x=3t\), 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 3t. We know that the derivative of t^{2} is 2t, so we need to use a function that's bigger by a factor of 3/2: \(\dot{x}=(3/2)t^2\). In fact, we could add any constant to this, and make it \(\dot{x}=(3/2)t^2+14\), 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 3t^{2}, so we need something half as big as this: x=t^{3}/2. Figure K. The functions 2t, t^{2} and 7t^{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=2t 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 7t^{2} has a bigger second derivative, 14. Figure L. The functions t^{2} and 3t^{2}. 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 3t^{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 3t^{2} it's turned to the right. Figure M. The functions t^{3} has an inflection point at t=0. Figure M shows a third possibility. The function t^{3} has a derivative 3t^{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.
