A.1 First Calculus Calculations: Simple Motion

Let's jump right in and complete a simple, key Calculus calculation drawing on what you already know: use a value and a rate at a moment to compute a small change in an everyday situation.

An everyday scenario: Driving on a straight road

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

simple motion along a long straight road

Hannah 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?

Solution (short version).

(a) Traveling at 20 m/s for 1 second, Hannah's position changes by 20 m.

(b) So her position at 9:00:01 is 2000 + 20 = 2020 m.

(c) We cannot predict her exact position at 10:00:00 from this alone, because her rate may change during the hour. Indeed, it's likely that it does!

This simple scenario highlights two key points:

  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 notation that we'll use throughout Calculus to the same calculation. We use t to represent time as it continuously sweeps along. And we use Δ𝑡 for a general change in time (for example, 1 hour).

dt refers to a small change in time t

In Calculus we use dt for a small change in time. Here, taking dt = 1 second is a reasonable "small step."

We'll use s for position and write s(t) to show position as a function of time. In Scenario 1, 𝑠(9:00:00)=2000meters

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

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 same relation to see that Hannah's rate of travel is

Because a rate can change over time, we indicate the rate at a specific time using a vertical bar: rate at 9:00:00=𝑑𝑠𝑑𝑡∣at 9:00:00 So in this scenario, 𝑑𝑠𝑑𝑡∣at 9:00:00 =20 meterssecond.

And please don't lose sight of the meaning of this abstract-looking expression 𝑑𝑠𝑑𝑡∣at 9:00:00.
For a long time now you've been thinking of speed — the rate at which an object moves — as 𝑣 =Δ𝑥Δ𝑡 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 Δ𝑠 and Δ𝑡.

Encapsulating our calculation for the small change in position ds

Using this notation we can generalize our calculation to apply to the small change in position ds that happens at any time t and occurs over the short-time-interval 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 𝑑𝑠𝑑𝑡 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 𝑑𝑠𝑑𝑡∣at (9:10:00)=28meterssecond 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 speedometer

The 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 𝑑𝑠𝑑𝑡 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 = (your instantaneous rate) ⋅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 rate  =𝑑𝑠𝑑𝑡∣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: position at 9:00:01=position at 9:00:00 +change in position[in 1 sec] That equation is of the general form 𝑠(9:00:00)+ small change 𝑑𝑠final=initial+change The "change" in this case is Hannah's small change in position 𝑑𝑠 =(rate at time 𝑡) ⋅𝑑𝑡.

And since we're using 𝑠(𝑡) to denote position as a function of time, we have: position 𝑠 at time 𝑡+𝑑𝑡⏞𝑠(9:00:01)=position 𝑠 at time 𝑡⏞𝑠(9:00:00)+(rate at time 𝑡)∗𝑑𝑡⏞¯¯¯⏞¯¯¯⏞ small change 𝑑𝑠 Or, generalizing the equation to apply to any time t and short time interval dt:

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 s) =6.00 m above the ground. At that moment it is traveling upward at the rate of 𝑑𝑦𝑑𝑡∣at 0.50 s =5.00 m/s.

  1. Let's set up the calculation to find the ball's height 0.01 s later, 𝑦(0.51 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.
  2. 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, 𝑑𝑦𝑑𝑡∣at 0.80 s, to its rate at the earlier moment t = 0.50 s, 𝑑𝑦𝑑𝑡∣at 0.50 s.

    3.  

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 rate at 9:00:00=𝑑𝑠𝑑𝑡∣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=(𝑑𝑠𝑑𝑡∣at 9:00:00)⋅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, 𝑑𝑠𝑑𝑡∣at time 𝑡, then we can predict its location a short time-interval dt later: 𝑠(𝑡+𝑑𝑡)=𝑠(𝑡)+𝑑𝑠⏞¯¯¯¯⏞¯¯¯¯⏞ (rate at time 𝑡)∗𝑑𝑡
  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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