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A.1 First Calculus Calculations: Simple Motion

For our first Calculus calculations, we need to make a shift: In many math courses before now, you computed the value of a function for a single input value. But Calculus is all about motion and change – how one quantity changes as another varies continuously. For instance, picture how a car’s position changes as time flows from one moment to the next.

Hence a great deal of our focus will be on the rate at which a function changes, rather than solely on the function’s value at a particular moment.

Let’s jump right in and consider an everyday situation and some simple calculations that will – believe it or not – highlight several of the most important ideas we need as we begin our exploration of Calculus.

Scenario 1, Simple Motion: Straight road travel at 9:00 am

simple motion along a long straight roadHannah is driving along a long straight road. At exactly 9:00:00 she passes the 2.0 kilometer (= 2000 meter) marker on the side of the road, and is traveling at a rate of 20 meters/second.

  1. What is Hannah’s change in position in the following 1 second, given that she is traveling at 20 meters/second?
  2. What is Hannah’s position at 9:00:01, after that 1 second has passed? (Remember that at 9:00:00, she was at the 2000 meter marker.)
  3. Can you predict, with certainty, her position 1 hour later, at 10:00:00? If so, where exactly will she be? If not, why can’t you?

(a) Hannah’s change in position over this 1-second interval from 9:00:00 to 9:00:01 is given by that relation you know oh-so-well:
\text{change in position} &= \text{(rate) } * \text{ (time interval)} \\[8px] \text{change in position}_\text{[in 1 sec]} &= \left(20 \, \tfrac{\text{meters}}{\text{second}}\right) * (1 \, \text{second}) \\[8px] &= 20 \, \text{meters} \quad \cmark
(b) Since Hannah started at the 2000-meter marker at 9:00:00, her position at 9:00:01 is
\text{position at 9:00:01} &= \text{position at 9:00:00 } + \text{change in position}_\text{[in 1 sec]} \\[8px] &= 2000 \, \text{meters} + 20 \, \text{meters} \\[8px] &= 2020 \, \text{meters} \quad \cmark
(c) While it’s possible that Hannah continues to travel exactly 20 meters each second for the entire next hour, we can’t say that she does so with any certainty whatsoever. Instead, we have no idea of what happens during that time: maybe she travels more slowly due to traffic, or perhaps she travels more quickly on a highway, or maybe at 9:01 she arrives at her destination and stops altogether. We simply cannot predict her position an hour later based on information we’re given about what’s going on at 9:00.

Believe it or not, this simple scenario illustrates two key points of Calculus:

  1. If we’re given information about (i) a function’s value at a particular instant (Hannah’s position at 9:00:00) and (ii) the function’s rate of change at that instant (her rate of change of position each second), then we can determine the function’s value a short interval later (her position one second later, at 9:00:01).
  2. We cannot extend this reasoning to make a prediction very far out in the future, since we’re not guaranteed that the function’s rate of change (here, her speed of travel) remains the same.

Let’s introduce some symbols to encapsulate the reasoning we used to calculate Hannah’s changing position. First, you’re probably familiar with using t to represent time as it continuously sweeps along.

You’re probably also familiar with using $\Delta t,$ “Delta-t,” with the Greek letter “Delta,” to represent a change in time. For instance, if we think about how Hannah’s position changes from 9:00:00 am to 10:00:00 am, then we have as our time interval $\Delta t$ = 1 hour.

dt refers to a small change in time t
What’s new in our first Calculus calculation here is that we use dt to represent a small change in time: the d should make you think “change” like “Delta,” but the fact that it’s a small “d” should make you think “small change.” How small is “small”? That’s a question we’ll explore as we proceed, and that takes us to the root of Calculus. But for now, it probably feels to you – rightly – that one second is small when you’re talking about traveling at highway speeds. So, in thinking about Hannah’s travel during the interval from 9:00:00 to 9:00:01, we have dt = 1 second.

To indicate Hannah’s position along the straight highway, we’ll follow tradition and use s. (Why “s” to indicate position? We don’t know, but it’s tradition in Math that we’ll follow.) Furthermore, at any instant Hannah has a single identifiable position, and so her position s is a function of time t, which we capture by writing s(t). For instance, the problem statement in Scenario 1 tells us that at t = 9:00:00 her position s is 2000 meters, so
\[s(\text{9:00:00}) = 2000 \, \text{meters}\]

ds refers to a small change in position s
And we’ll use ds to refer to a small change in position s. Here, the small change in position ds that happens over the small 1-second time interval dt is given by
\[ \bbox[10px,border:2px solid blue]{ds = \text{ rate} * dt}\] In words: the small change in position ds equals the rate multiplied by the small change in time dt. This expression is again, simply, the familiar formulation “change equals rate times time-interval,” now applied to the small position change ds over the short time interval dt.

Viewed differently, we can rearrange the preceding equation to see that Hannah’s rate of travel is the quantity $\dfrac{ds}{dt}:$
\[ \bbox[10px,border:2px solid blue]{\text{rate } = \frac{ds}{dt}}\] The quantity $\dfrac{ds}{dt}$ is pronounced “dee-ess dee-tee,” rather than, for instance “dee-ess over dee-tee” or “dee-ess divided by dee-tee.” Instead, it’s just “dee-ess dee-tee.”

We know that Hannah’s rate of travel changes as her trip proceeds, and we need a way to indicate the specific time for which a given rate applies. To do so, we add a vertical bar with a subscript like so:
\[ \text{rate at 9:00:00} = \left.\frac{ds}{dt}\right|_\text{at 9:00:00} \] That is, since in this scenario at 9:00:00 Hannah is traveling at 20 meters/second, you would write $\left.\dfrac{ds}{dt}\right|_\text{at 9:00:00}=20 \, \tfrac{\text{meters}}{\text{second}},$ and say in Calculus-speak: “Dee-ess dee-tee at 9:00:00 equals 20 meters per second.” The use of the vertical bar with text at its bottom is admittedly awkward initially, but the meaning is clear, and you’ll be using similar notation throughout your studies and use of Calculus.

And please don’t lose sight of the meaning of this abstract-looking expression $\left.\dfrac{ds}{dt}\right|_\text{at 9:00:00}.$ For a long time now you’ve been thinking of speed – the rate at which an object moves – as $v = \dfrac{\Delta x}{\Delta t}$ or something similar. The ideas here are exactly the same, except now (1) we’re using s instead of x, and (2) we’re considering small changes in position and time, and so write ds and dt instead of $\Delta s$ and $\Delta t.$

Encapsulating our calculation for the small change in position ds

Recall that Hannah’s small change in position over the 1-second interval dt starting at 9:00:00 was given by
\[ds_\text{[over 1 second]} = \text{ (rate at 9:00:00)} * dt\] In our new notation, we write the rate as $\dfrac{ds}{dt}:$
\[\text{rate at 9:00:00} = \left.\frac{ds}{dt}\right|_\text{at 9:00:00}=20 \, \tfrac{\text{meters}}{\text{second}} \] Hence we can encapsulate our simple calculation “small change in position = rate multiplied by the small-time-interval” as
\[ds = \left(\left.\frac{ds}{dt}\right|_\text{at 9:00:00} \right) dt \] The notation may be new to you, but this mathematical statement merely represents what you do automatically: it’s a fancy way of writing the calculation you probably quickly did in your head in part (a) of Scenario 1 above.

We can generalize the equation to apply to the small change in position ds that happens at any time t and occurs over the short-time-interval dt:
\[ \bbox[10px,border:2px solid blue]{ds = \left( \left.\frac{ds}{dt}\right|_\text{at time $t$} \right) dt}\] This expression, and variations of it, will be crucially important as we proceed. And because it has ds to refer to a “small” change in position, dt to refer to a “small” change in time, and $\dfrac{ds}{dt}$ to refer to the rate, it is a Calculus expression – you’re doing Calculus calculations already!

Let’s try a quick Check Question to put these ideas to use.

Check Question 1: Hannah's travel at 9:10:00

At 9:10:00, Hannah is traveling faster than she was earlier, and her rate of travel is now
\[\left.\frac{ds}{dt}\right|_\text{at (9:10:00)} = 28 \, \tfrac{\text{meters}}{\text{second}} \] We wish to compute how far she travels in the next 1.5 seconds.

We’re not going to bother doing the multiplication there, and we know this is a (perhaps too) simple question. But we want to keep our focus on using this new notation, and becoming familiar with setting up calculations like this. If it seems straightforward, great: you’re getting it!
Calculus in the Real World: The next time you're on the road. . .

a speedometerThe preceding check question suggests a simple activity you can use to start making Calculus feel more real to you already: the next time you’re in a vehicle, glance at the speedometer – which shows your rate of travel $\dfrac{ds}{dt}$ at the moment you look – and imagine your position changing by the small amount ds over the next second or two. Your entire trip is comprised of the long series of those small changes $ds = \text{ (your instantaneous rate)}\cdot dt$ for each small time interval dt. Starting to think in these terms will help lay a solid foundation to learn Calculus.

And if you happen to be currently sitting in one place, not moving, then your $\text{rate }=\left.\frac{ds}{dt}\right|_\text{at this moment} = 0.$ Hence your small change in position ds = 0 from moment to moment to moment, just as you are experiencing!

Encapsulating our calculation for position after dt has passed

Let’s extend the use of our new notation to express the calculation we made to find Hannah’s position 1 second after 9:00:00:
\[\text{position at 9:00:01} = \text{position at 9:00:00 } + \text{change in position}_\text{[in 1 sec]}\] That equation is of the general form
& \phantom{s(\text{9:00:00}) + \text{ small change }ds} \\
\text{final} &= \text{initial} + \text{change}
The “change” in this case is Hannah’s small change in position $ds = (\text{rate at time }t) \cdot dt.$
And since we’re using $s(t)$ to denote position as a function of time, we have:
\[\overbrace{s(\text{9:00:01})}^\text{position $s$ at time $t+dt$} = \overbrace{s(\text{9:00:00})}^{\text{position $s$ at time }t} + \overbrace{\text{ small change }ds}^{\text{(rate at time $t$)} * \, dt}\] Or, generalizing the equation to apply to any time t and short time interval dt:
\[ \color{blue}{s(t + dt) = s(t) + \overbrace{\text{ (rate at time t)} * dt}^{ds}} \] Let’s consider a few more check questions regarding simple motion to set up a few more first Calculus calculations and solidify these first ideas.

Check Question 2: Calvin tosses a ball

Gif of person throwing a ball straight up
Calvin tosses a ball straight up. He starts a stopwatch the moment the ball leaves his hand, and time t represents the number of seconds that pass since that initial moment. We’ll measure time in seconds, which we abbreviate “s.”

Let’s use y(t) to denote the ball’s height above ground level as a function of time t. We’ll measure the ball’s height in meters, which we abbreviate “m.”

At t = 0.50 seconds, the ball is at y = 6.00 meters, and is traveling upward at 5.00 meters per second.
At t = 0.50 s, the ball is at a height of $y(0.50 \,\text{s}) = 6.00 \,\text{m}$ above the ground. At that moment it is traveling upward at the rate of $\left.\dfrac{dy}{dt}\right|_\text{at 0.50 s} = 5.00$ m/s.

(a) Let’s set up the calculation to find the ball’s height 0.01 s later, $y(0.51 \,\text{s}).$

We won’t bother to complete the simple math there; instead let’s focus on the Calculus ideas that are probably new to you.
(b) At t = 0.80 s, the ball is still traveling upward but losing speed as it approaches the top of its trajectory. Let’s compare the ball’s rate of travel at t = 0.80 s, $\left.\dfrac{dy}{dt}\right|_\text{at 0.80 s},$ to its rate at the earlier moment t = 0.50 s, $\left.\dfrac{dy}{dt}\right|_\text{at 0.50 s}.$

Have a question or comment about what’s on this screen? Then please head on over to the Forum, and post there!

The upshot

We’re on our way, doing calculus calculations already!

  1. Thinking about Hannah’s trip:
    1. ds represents the small change in an object’s position s(t), and dt represents a short time interval. We write Hannah’s rate of travel at 9:00:00 as
      \[\text{rate at 9:00:00} = \left.\frac{ds}{dt}\right|_\text{at 9:00:00} \]
    2. Her tiny change in position ds over the time interval dt is then given by the familiar formulation “change = rate * time interval”:
      \[ds = \left( \left.\frac{ds}{dt}\right|_\text{at 9:00:00} \right) \cdot dt\]

    (If we’re examining vertical motion, we probably use y in place of s.)

  2. If we know an object’s position $s(t)$ at a given moment, and its rate of travel at time t, $\left.\dfrac{ds}{dt}\right|_\text{at time $t$}$, then we can predict its location a short time-interval dt later:
    \[s(t + dt) = s(t) + \overbrace{\text{ (rate at time $t$)} * dt}^{ds} \]
  3. While we can use this approach to predict position a short time dt later, we can’t extend our prediction out too far without further information since the rate of travel may change.

On the next screen, we’ll extend our thinking about “rates” from time-based rates to more generalized rates.

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