# 1.2: Another Limit and Computing Velocity

- Page ID
- 89705

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\(\newcommand{\avec}{\mathbf a}\) \(\newcommand{\bvec}{\mathbf b}\) \(\newcommand{\cvec}{\mathbf c}\) \(\newcommand{\dvec}{\mathbf d}\) \(\newcommand{\dtil}{\widetilde{\mathbf d}}\) \(\newcommand{\evec}{\mathbf e}\) \(\newcommand{\fvec}{\mathbf f}\) \(\newcommand{\nvec}{\mathbf n}\) \(\newcommand{\pvec}{\mathbf p}\) \(\newcommand{\qvec}{\mathbf q}\) \(\newcommand{\svec}{\mathbf s}\) \(\newcommand{\tvec}{\mathbf t}\) \(\newcommand{\uvec}{\mathbf u}\) \(\newcommand{\vvec}{\mathbf v}\) \(\newcommand{\wvec}{\mathbf w}\) \(\newcommand{\xvec}{\mathbf x}\) \(\newcommand{\yvec}{\mathbf y}\) \(\newcommand{\zvec}{\mathbf z}\) \(\newcommand{\rvec}{\mathbf r}\) \(\newcommand{\mvec}{\mathbf m}\) \(\newcommand{\zerovec}{\mathbf 0}\) \(\newcommand{\onevec}{\mathbf 1}\) \(\newcommand{\real}{\mathbb R}\) \(\newcommand{\twovec}[2]{\left[\begin{array}{r}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\ctwovec}[2]{\left[\begin{array}{c}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\threevec}[3]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\cthreevec}[3]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\fourvec}[4]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\cfourvec}[4]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\fivevec}[5]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\cfivevec}[5]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\mattwo}[4]{\left[\begin{array}{rr}#1 \amp #2 \\ #3 \amp #4 \\ \end{array}\right]}\) \(\newcommand{\laspan}[1]{\text{Span}\{#1\}}\) \(\newcommand{\bcal}{\cal B}\) \(\newcommand{\ccal}{\cal C}\) \(\newcommand{\scal}{\cal S}\) \(\newcommand{\wcal}{\cal W}\) \(\newcommand{\ecal}{\cal E}\) \(\newcommand{\coords}[2]{\left\{#1\right\}_{#2}}\) \(\newcommand{\gray}[1]{\color{gray}{#1}}\) \(\newcommand{\lgray}[1]{\color{lightgray}{#1}}\) \(\newcommand{\rank}{\operatorname{rank}}\) \(\newcommand{\row}{\text{Row}}\) \(\newcommand{\col}{\text{Col}}\) \(\renewcommand{\row}{\text{Row}}\) \(\newcommand{\nul}{\text{Nul}}\) \(\newcommand{\var}{\text{Var}}\) \(\newcommand{\corr}{\text{corr}}\) \(\newcommand{\len}[1]{\left|#1\right|}\) \(\newcommand{\bbar}{\overline{\bvec}}\) \(\newcommand{\bhat}{\widehat{\bvec}}\) \(\newcommand{\bperp}{\bvec^\perp}\) \(\newcommand{\xhat}{\widehat{\xvec}}\) \(\newcommand{\vhat}{\widehat{\vvec}}\) \(\newcommand{\uhat}{\widehat{\uvec}}\) \(\newcommand{\what}{\widehat{\wvec}}\) \(\newcommand{\Sighat}{\widehat{\Sigma}}\) \(\newcommand{\lt}{<}\) \(\newcommand{\gt}{>}\) \(\newcommand{\amp}{&}\) \(\definecolor{fillinmathshade}{gray}{0.9}\)Computing tangent lines is all very well, but what does this have to do with applications or the “Real World”? Well - at least initially our use of limits (and indeed of calculus) is going to be a little removed from real world applications. However as we go further and learn more about limits and derivatives we will be able to get closer to real problems and their solutions.

So stepping just a little closer to the real world, consider the following problem. You drop a ball from the top of a very very tall building. Let \(t\) be elapsed time measured in seconds, and \(s(t)\) be the distance the ball has fallen in metres. So \(s(0) = 0\text{.}\)

*Quick aside:* there is quite a bit going on in the statement of this problem. We have described the general picture — tall building, ball, falling — but we have also introduced notation, variables and units. These will be common first steps in applications and are necessary in order to translate a real world problem into mathematics in a clear and consistent way.

Galileo ^{1} worked out that \(s(t)\) is a quadratic function:

\begin{align*} s(t) &= 4.9 t^2. \end{align*}

The question that is posed is

How fast is the ball falling after 1 second?

Now before we get to answering this question, we should first be a little more precise. The wording of this question is pretty sloppy for a couple of reasons:

- What we do mean by “after 1 second”? We know the ball will move faster and faster as time passes, so after 1 second it does not fall at one fixed speed.
- As it stands a reasonable answer to the question would be just “really fast”. If the person asking the question wants a numerical answer it would be better to ask “At what speed” or “With what velocity”.

We should also be careful using the words “speed” and “velocity” — they are not interchangeable.

- Speed means the distance travelled per unit time and is always a non-negative number. An unmoving object has speed \(0\text{,}\) while a moving object has positive speed.
- Velocity, on the other hand, also specifies the direction of motion. In this text we will almost exclusively deal with objects moving along straight lines. Because of this velocities will be positive or negative numbers indicating which direction the object is moving along the line. We will be more precise about this later
^{2}.

A better question is

What is the velocity of the ball precisely 1 second after it is dropped?

or even better:

What is the velocity of the ball at the 1 second mark?

This makes it very clear that we want to know what is happening at exactly 1 second after the ball is dropped.

There is something a little subtle going on in this question. In particular, what do we mean by the velocity at \(t=1\text{?.}\) Surely if we freeze time at \(t=1\) second, then the object is not moving at all? This is definitely *not* what we mean.

If an object is moving at a constant velocity ^{3 }in the positive direction, then that velocity is just the distance travelled divided by the time taken. That is

\begin{align*} v &= \frac{\text{distance moved}}{\text{time taken}} \end{align*}

An object moving at constant velocity that moves \(27\) metres in \(3\) seconds has velocity

\begin{align*} v &= \frac{27 m}{3 s} = 9 m/s. \end{align*}

When velocity is constant everything is easy.

However, in our falling object example, the object is being acted on by gravity and its speed is definitely not constant. Instead of asking for *THE* velocity, let us examine the “average velocity” of the object over a certain window of time. In this case the formula is very similar

\begin{align*} \text{average velocity } &= \frac{\text{distance moved}}{\text{time taken}} \end{align*}

But now I want to be more precise, instead write

\begin{align*} \text{average velocity } &= \frac{\text{difference in} \text{ distance}}{\text{difference in time}} \end{align*}

Now in spoken English we haven't really changed much — the distance moved is the difference in position, and the time taken is just the difference in time — but the latter is more mathematically precise, and is easy to translate into the following equation

\begin{align*} \text{average velocity } &= \frac{s(t_2) - s(t_1)}{t_2 - t_1}. \end{align*}

This is the formula for the average velocity of our object between time \(t_1\) and \(t_2\text{.}\) The denominator is just the difference between these times and the numerator is the difference in position — i.e. position at time \(t_1\) is just \(s(t_1)\) and position at time \(t_2\) is just \(s(t_2)\text{.}\)

So what is the average velocity of the falling ball between \(1\) and \(1.1\) seconds? All we need to do now is plug some numbers into our formula

\begin{align*} \mbox{average velocity} &= \frac{\text{difference in}\text{ position}}{\mbox{difference in time}}\\ &= \frac{s(1.1) - s(1)}{1.1-1}\\ &= \frac{4.9 (1.1)^2 - 4.9(1)}{0.1} = \frac{4.9 \times 0.21}{0.1} = 10.29 m/s \end{align*}

And we have our average velocity. However there is something we should notice about this formula and it is easier to see if we sketch a graph of the function \(s(t)\)

So on the left I have drawn the graph and noted the times \(t=1\) and \(t=1.1\text{.}\) The corresponding positions on the axes and the two points on the curve. On the right I have added a few more details. In particular I have noted the differences in position and time, and the line joining the two points. Notice that the slope of this line is

\begin{align*} \text{slope} &= \frac{\text{change in $y$}}{\text{change in $x$}} = \frac{\text{difference in $s$}}{\text{difference in $t$}} \end{align*}

which is precisely our expression for the average velocity.

Let us examine what happens to the average velocity as we look over smaller and smaller time-windows.

\begin{align*} \text{time window} && \text{average velocity}\\ 1 \leq t \leq 1.1 && 10.29\\ 1 \leq t \leq 1.01 && 9.849\\ 1 \leq t \leq 1.001 && 9.8049\\ 1 \leq t \leq 1.0001 && 9.80049 \end{align*}

As we make the time interval smaller and smaller we find that the average velocity is getting closer and closer to \(9.8\text{.}\) We can be a little more precise by finding the average velocity between \(t=1\) and \(t=1+h\) — this is very similar to what we did for tangent lines.

\begin{align*} \text{average velocity} &= \frac{s(1+h) - s(1)}{(1+h)-1}\\ &= \frac{4.9(1+h)^2 - 4.9}{h}\\ &= \frac{9.8h + 4.9h^2}{h}\\ &= 9.8 + 4.9h \end{align*}

Now as we squeeze this window between \(t=1\) and \(t=1+h\) down towards zero, the average velocity becomes the “instantaneous velocity” — just as the slope of the secant line becomes the slope of the tangent line. This is our second limit

\begin{align*} v(1) &= \lim_{h \to 0} \frac{s(1+h)-s(1)}{h} = 9.8 \end{align*}

More generally we define the instantaneous velocity at time \(t=a\) to be the limit

\begin{align*} v(a) &= \lim_{h \to 0} \frac{ s(a+h) - s(a) }{h} \end{align*}

We read this as

The velocity at time \(a\) is equal to the limit as \(h\) goes to zero of \(\frac{s(a+h)-s(a)}{h}\text{.}\)

While we have solved the problem stated at the start of this section, it is clear that if we wish to solve similar problems that we will need to understand limits in a more general and systematic way.

## Exercise

###### Stage 1

As they are used in this section, what is the difference between speed and velocity?

Speed can never be negative; can it be zero?

Suppose you wake up in the morning in your room, then you walk two kilometres to school, walk another two kilometres to lunch, walk four kilometres to a coffee shop to study, then return to your room until the next morning. In the 24 hours from morning to morning, what was your average velocity? (In CLP-1, we are considering functions of one variable. So, at this stage, think of our whole world as being contained in the \(x\)-axis.)

Suppose you drop an object, and it falls for a few seconds. Which is larger: its speed at the one second mark, or its average speed from the zero second mark to the one second mark?

The position of an object at time \(t\) is given by \(s(t)\text{.}\) Then its average velocity over the time interval \(t=a\) to \(t=b\) is given by \(\dfrac{s(b)-s(a)}{b-a}\text{.}\) Explain why this fraction also gives the slope of the secant line of the curve \(y=s(t)\) from the point \(t=a\) to the point \(t=b\text{.}\)

Below is the graph of the position of an object at time \(t\text{.}\) For what periods of time is the object's velocity positive?

###### Stage 2

Suppose the position of a body at time \(t\) (measured in seconds) is given by \(s(t)=3t^2+5\text{.}\)

- What is the average velocity of the object from 3 seconds to 5 seconds?
- What is the velocity of the object at time \(t=1\text{?}\)

Suppose the position of a body at time \(t\) (measured in seconds) is given by \(s(t)=\sqrt{t}\text{.}\)

- What is the average velocity of the object from \(t=1\) second to \(t=9\) seconds?
- What is the velocity of the object at time \(t=1\text{?}\)
- What is the velocity of the object at time \(t=9\text{?}\)