Example 7.1.1 An object moves in a straight line so that its speed at time $t$ is given by $v(t)=3t$ in, say, cm/sec. If the object is at position $10$ on the straight line when $t=0$, where is the object at any time $t$?

There are two reasonable ways to approach this problem. If $s(t)$ is the position of the object at time $t$, we know that $s'(t)=v(t)$. Because of our knowledge of derivatives, we know therefore that $\ds s(t)=3t^2/2+k$, and because $s(0)=10$ we easily discover that $k=10$, so $\ds s(t)=3t^2/2+10$. For example, at $t=1$ the object is at position $3/2+10=11.5$. This is certainly the easiest way to deal with this problem. Not all similar problems are so easy, as we will see; the second approach to the problem is more difficult but also more general.

We start by considering how we might approximate a solution. We know that at $t=0$ the object is at position 10. How might we approximate its position at, say, $t=1$? We know that the speed of the object at time $t=0$ is $0$; if its speed were constant then in the first second the object would not move and its position would still be 10 when $t=1$. In fact, the object will not be too far from 10 at $t=1$, but certainly we can do better. Let's look at the times $0.1$, $0.2$, $0.3$, …, $1.0$, and try approximating the location of the object at each, by supposing that during each tenth of a second the object is going at a constant speed. Since the object initially has speed 0, we again suppose it maintains this speed, but only for a tenth of second; during that time the object would not move. During the tenth of a second from $t=0.1$ to $t=0.2$, we suppose that the object is traveling at $0.3$ cm/sec, namely, its actual speed at $t=0.1$. In this case the object would travel $(0.3)(0.1)=0.03$ centimeters: $0.3$ cm/sec times $0.1$ seconds. Similarly, between $t=0.2$ and $t=0.3$ the object would travel $(0.6)(0.1)=0.06$ centimeters. Continuing, we get as an approximation that the object travels $$ (0.0)(0.1)+(0.3)(0.1)+(0.6)(0.1)+\cdots+(2.7)(0.1)=1.35 $$ centimeters, ending up at position 11.35. This is a better approximation than 10, certainly, but is still just an approximation. (We know in fact that the object ends up at position $11.5$, because we've already done the problem using the first approach.) Presumably, we will get a better approximation if we divide the time into one hundred intervals of a hundredth of a second each, and repeat the process: $$ (0.0)(0.01)+(0.03)(0.01)+(0.06)(0.01)+\cdots+(2.97)(0.01)=1.485. $$ We thus approximate the position as $11.485$. Since we know the exact answer, we can see that this is much closer, but if we did not already know the answer, we wouldn't really know how close.

We can keep this up, but we'll never really know the exact answer if we simply compute more and more examples. Let's instead look at a "typical'' approximation. Suppose we divide the time into $n$ equal intervals, and imagine that on each of these the object travels at a constant speed. Over the first time interval we approximate the distance traveled as $(0.0)(1/n)=0$, as before. During the second time interval, from $t=1/n$ to $t=2/n$, the object travels approximately $\ds 3(1/n)(1/n)=3/n^2$ centimeters. During time interval number $i$, the object travels approximately $\ds (3(i-1)/n)(1/n)=3(i-1)/n^2$ centimeters, that is, its speed at time $(i-1)/n$, $3(i-1)/n$, times the length of time interval number $i$, $1/n$. Adding these up as before, we approximate the distance traveled as $$ (0){1\over n}+3{1\over n^2}+3(2){1\over n^2}+ 3(3){1\over n^2}+\cdots+3(n-1){1\over n^2} $$ centimeters. What can we say about this? At first it looks rather less useful than the concrete calculations we've already done. But in fact a bit of algebra reveals it to be much more useful. We can factor out a 3 and $\ds 1/n^2$ to get $$ {3\over n^2}(0+1+2+3+\cdots+(n-1)), $$ that is, $\ds 3/n^2$ times the sum of the first $n-1$ positive integers. Now we make use of a fact you may have run across before: $$ 1+2+3+\cdots+k={k(k+1)\over2}. $$ In our case we're interested in $k=n-1$, so $$ 1+2+3+\cdots+(n-1)={(n-1)(n)\over2}={n^2-n\over2}. $$ This simplifies the approximate distance traveled to $$ {3\over n^2}{n^2-n\over2}={3\over2}{n^2-n\over n^2}= {3\over2}\left({n^2\over n^2}-{n\over n^2}\right)= {3\over2}\left(1-{1\over n}\right). $$ Now this is quite easy to understand: as $n$ gets larger and larger this approximation gets closer and closer to $(3/2)(1-0)=3/2$, so that $3/2$ is the exact distance traveled during one second, and the final position is $11.5$.

So for $t=1$, at least, this rather cumbersome approach gives the same answer as the first approach. But really there's nothing special about $t=1$; let's just call it $t$ instead. In this case the approximate distance traveled during time interval number $i$ is $\ds 3(i-1)(t/n)(t/n)=3(i-1)t^2/n^2$, that is, speed $3(i-1)(t/n)$ times time $t/n$, and the total distance traveled is approximately $$ (0){t\over n}+3(1){t^2\over n^2}+3(2){t^2\over n^2}+ 3(3){t^2\over n^2}+\cdots+3(n-1){t^2\over n^2}. $$ As before we can simplify this to $$ {3t^2\over n^2}(0+1+2+\cdots+(n-1))={3t^2\over n^2}{n^2-n\over2}= {3\over2}t^2\left(1-{1\over n}\right). $$ In the limit, as $n$ gets larger, this gets closer and closer to $\ds (3/2)t^2$ and the approximated position of the object gets closer and closer to $\ds (3/2)t^2+10$, so the actual position is $\ds (3/2)t^2+10$, exactly the answer given by the first approach to the problem. $\square$

Example 7.1.2 Find the area under the curve $y=3x$ between $x=0$ and any positive value $x$. There is here no obvious analogue to the first approach in the previous example, but the second approach works fine. (Because the function $y=3x$ is so simple, there is another approach that works here, but it is even more limited in potential application than is approach number one.) How might we approximate the desired area? We know how to compute areas of rectangles, so we approximate the area by rectangles. Jumping straight to the general case, suppose we divide the interval between 0 and $x$ into $n$ equal subintervals, and use a rectangle above each subinterval to approximate the area under the curve. There are many ways we might do this, but let's use the height of the curve at the left endpoint of the subinterval as the height of the rectangle, as in figure 7.1.1. The height of rectangle number $i$ is then $3(i-1)(x/n)$, the width is $x/n$, and the area is $\ds 3(i-1)(x^2/n^2)$. The total area of the rectangles is $$ (0){x\over n}+3(1){x^2\over n^2}+3(2){x^2\over n^2}+ 3(3){x^2\over n^2}+\cdots+3(n-1){x^2\over n^2}. $$ By factoring out $\ds 3x^2/n^2$ this simplifies to $$ {3x^2\over n^2}(0+1+2+\cdots+(n-1))={3x^2\over n^2}{n^2-n\over2}= {3\over2}x^2\left(1-{1\over n}\right). $$ As $n$ gets larger this gets closer and closer to $\ds 3x^2/2$, which must therefore be the true area under the curve. $\square$