## Section1.2Displacement and Area

We will now delve into two examples, one concerning displacement and the other area, to explore how the antidifferentiation process should unfold.

###### Example1.2. Object Moving in a Straight Line.

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 $t=0\text{,}$ where is the object at any time $t\text{?}$

Solution

There are two reasonable ways to approach this problem.

Method 1: If $s(t)$ is the position of the object at time $t\text{,}$ we know that $s'(t)=v(t)\text{.}$ Based on our knowledge of derivatives, we therefore know that

\begin{equation*} s(t)=3t^2/2+k\text{,} \end{equation*}

and because $s(0)=10$ we easily discover that $k=10\text{,}$ so

\begin{equation*} s(t)=3t^2/2+10\text{.} \end{equation*}

For example, at $t=1$ the object is at position $3/2+10=11.5$ cm.

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.

Method 2: 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, $t=1\text{?}$ We know that the speed of the object at time $t=0$ is $0\text{;}$ if its speed were constant then in the first second the object would not move and its position would still be 10 when $t=1\text{.}$ In fact, the object will not be too far from 10 at $t=1\text{,}$ but certainly we can do better.

Let's look at the times $0.1\text{,}$ $0.2\text{,}$ $0.3\text{,}$ …, $1.0\text{,}$ 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 $t=0.2\text{,}$ we suppose that the object is travelling at $0.3$ cm/sec, namely, its actual speed at $t=0.1\text{.}$ In this case the object would travel $(0.3)(0.1)=0.03$ centimetres: $0.3$ cm/sec times $0.1$ seconds. Similarly, between $t=0.2$ and $t=0.3$ the object would travel $(0.6)(0.1)=0.06$ centimetres. Continuing, we get as an approximation that the object travels

\begin{equation*} (0.0)(0.1)+(0.3)(0.1)+(0.6)(0.1)+\cdots+(2.7)(0.1)=1.35 \end{equation*}

centimetres, ending up at position 11.35 cm. 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 $11.5\text{,}$ 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:

\begin{equation*} (0.0)(0.01)+(0.03)(0.01)+(0.06)(0.01)+\cdots+(2.97)(0.01)=1.485\text{.} \end{equation*}

We thus approximate the position as $11.485\text{.}$ 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 travelled as $(0.0)(1/n)=0\text{,}$ as before. During the second time interval, from $t=1/n$ to $t=2/n\text{,}$ the object travels approximately $\ds 3(1/n)(1/n)=3/n^2$ centimetres. During time interval number $i\text{,}$ the object travels approximately $\ds (3(i-1)/n)(1/n)=3(i-1)/n^2$ centimetres, that is, its speed at time $(i-1)/n\text{,}$ $3(i-1)/n\text{,}$ times the length of time interval number $i\text{,}$ $1/n\text{.}$ Adding these up as before, we approximate the distance travelled as

\begin{equation*} (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} \end{equation*}

centimetres. 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

\begin{equation*} {3\over n^2}(0+1+2+3+\cdots+(n-1))\text{,} \end{equation*}

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, Gauss's Equation:

\begin{equation*} 1+2+3+\cdots+k={k(k+1)\over2}\text{.} \end{equation*}

In our case we're interested in $k=n-1\text{,}$ so

\begin{equation*} 1+2+3+\cdots+(n-1)={(n-1)(n)\over2}={n^2-n\over2}\text{.} \end{equation*}

This simplifies the approximate distance travelled to

\begin{equation*} {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)\text{.} \end{equation*}

Now this is quite easy to understand: as $n$ gets larger and larger this approximation gets closer and closer to $(3/2)(1-0)=3/2\text{,}$ so that $3/2$ is the exact distance traveled during one second, and the final position is $11.5\text{.}$

So for $t=1\text{,}$ at least, this rather cumbersome approach gives the same answer as the first approach. But really there's nothing special about $t=1\text{;}$ let's just call it $t$ instead. In this case the approximate distance traveled during time interval number $i$ is $\ds 3(i-1)(t/n)(t/n)=3(i-1)t^2/n^2\text{,}$ that is, speed $3(i-1)(t/n)$ times time $t/n\text{,}$ and the total distance traveled is approximately

\begin{equation*} (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}\text{.} \end{equation*}

As before we can simplify this to

\begin{equation*} {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)\text{.} \end{equation*}

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 $\ds (3/2)t^2+10\text{,}$ so the actual position is $\ds (3/2)t^2+10\text{,}$ exactly the answer given by the first method to the problem.

###### Example1.3. Area under the Line.

Find the area under the curve $y=3x$ between $x=0$ and any positive value $x=a\text{.}$

Solution

There is here no obvious analogue to the first method in the previous example, but the second method works fine. (Since the function $y=3x$ is so simple, there is another method that works here, but it is even more limited in potential application than is method number one.) How might we approximate the desired area? We know how to compute areas of rectangles, so we approximate the area by rectangles as shown below. Jumping straight to the general case, suppose we divide the interval between 0 and $a$ 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. The height of rectangle number $i$ is then $3(i-1)(a/n)\text{,}$ the width is $a/n\text{,}$ and the area is $\ds 3(i-1)(a^2/n^2)\text{.}$ The total area of the rectangles is

\begin{equation*} (0){a\over n}+3(1){a^2\over n^2}+3(2){a^2\over n^2}+ 3(3){a^2\over n^2}+\cdots+3(n-1){a^2\over n^2}\text{.} \end{equation*}

By factoring out $\ds 3a^2/n^2$ this simplifies to

\begin{equation*} {3a^2\over n^2}(0+1+2+\cdots+(n-1))={3a^2\over n^2}{n^2-n\over2}= {3\over2}a^2\left(1-{1\over n}\right)\text{.} \end{equation*}

As $n$ gets larger this gets closer and closer to $\ds 3a^2/2\text{,}$ which must therefore be the true area under the curve.

What you will have noticed, of course, is that while the problem in the second example appears to be much different than the problem in the first example, and while the easy approach to problem one does not appear to apply to problem two, the “approximation” approach works in both, and moreover the calculations are identical. As we will see, there are many, many problems that appear much different on the surface but turn out to be the same as these problems, in the sense that when we try to approximate solutions we end up with mathematics that looks like the two examples, though of course the function involved will not always be so simple.

Even better, we now see that while the second problem did not appear to be amenable to approach one, it can in fact be solved in the same way. The reasoning is this: we know that problem one can be solved easily by finding a function whose derivative is $3t\text{.}$ We also know that mathematically the two problems are the same, because both can be solved by taking a limit of a sum, and the sums are identical. Therefore, we don't really need to compute the limit of either sum because we know that we will get the same answer by computing a function with the derivative $3t$ or, which is the same thing, $3x\text{.}$

It's true that the first problem had the added complication of the “10”, and we certainly need to be able to deal with such minor variations, but that turns out to be quite simple. The lesson then is this: whenever we can solve a problem by taking the limit of a sum of a certain form, instead of computing the (often nasty) limit we can find a new function with a certain derivative.

##### Exercises for Section 1.2.

Suppose an object moves in a straight line so that its speed at time $t$ is given by $v(t)=2t+2\text{,}$ and that at $t=1$ the object is at position 5. Find the position of the object at $t=2\text{.}$

10
Solution

Let $x(t)$ denote the position at time $t\text{.}$ The position is related to the speed by

\begin{equation*} x'(t) = v(t) = 2t+2\text{,} \end{equation*}

and so we need to find the antiderivative of $v(t)\text{.}$ Take

\begin{equation*} x(t) = t^2 + 2t +C \end{equation*}

for some arbitrary constant $C\text{.}$

Check: $\diff{x}{t} = \diff{}{t}(t^2+2t+C) = 2t+2 = v(t)\text{.}$ To find $C\text{,}$ we use the given data:

\begin{equation*} x(1)=5 \implies 1+2+C = 5 \implies C = 2\text{.} \end{equation*}

Thus, the position of the object at time $t$ is given by

\begin{equation*} x(t) = t^2+2t+2\text{,} \end{equation*}

and at $t=2\text{,}$ we have

\begin{equation*} x(2)=10\text{.} \end{equation*}

Suppose an object moves in a straight line so that its speed at time $t$ is given by $\ds v(t)=t^2+2\text{,}$ and that at $t=0$ the object is at position 5. Find the position of the object at $t=2\text{.}$

$\frac{35}{3}$
Solution

Let $x(t)$ denote the position at time $t\text{.}$ We know that

\begin{equation*} x'(t) = v(t) = t^2+2\text{,} \end{equation*}

and so we need to find the antiderivative of $v(t)\text{.}$ We find

\begin{equation*} x(t) = \frac{t^3}{3}+2t+C \end{equation*}

for some constant $C\text{.}$ The position at $t=0$ is given, so $C$ must satisfy

\begin{equation*} x(0)=C=5\text{.} \end{equation*}

Therefore,

\begin{equation*} x(t)=\frac{t^3}{3}+2t+5 \implies x(2)=\frac{8}{3}+9=\frac{35}{3}\text{.} \end{equation*}