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How do we measure velocity?

Calculus can be viewed broadly as the study of change. A natural and important question to ask about any changing quantity is how fast is the quantity changing?

Introduction

Calculus can be viewed broadly as the study of change. A natural and important question to ask about any changing quantity is how fast is the quantity changing?

We begin with a simple problem: a ball is tossed straight up in the air. How is the ball moving? Questions like this one are central to our study of differential calculus.

Exploration

Position and average velocity

Any moving object has a position that can be considered a function of time. When the motion is along a straight line, the position is given by a single variable, which we denote by \(s(t)\). For example, \(s(t)\) might give the mile marker of a car traveling on a straight highway at time \(t\) in hours. Similarly, the function \(s\) described in Preview Activity is a position function, where position is measured vertically relative to the ground.

On any time interval, a moving object also has an average velocity. For example, to compute a car's average velocity we divide the number of miles traveled by the time elapsed, which gives the velocity in miles per hour. Similarly, the value of \(AV_{[0.5,1]}\) in Preview Activity gave the average velocity of the ball on the time interval \([0.5,1]\), measured in feet per second.

In general, we make the following definition:

For an object moving in a straight line with position function \(s(t)\), the average velocity of the object on the interval from \(t = a\) to \(t = b\), denoted \(AV_{[a,b]}\), is given by the formula \[\begin{aligned}\end{aligned}\].

Note well: the units on \(AV_{[a,b]}\) are units of \(s\) per unit of \(t\), such as miles per hour or feet per second.

Instantaneous Velocity

Whether we are driving a car, riding a bike, or throwing a ball, we have an intuitive sense that a moving object has a velocity at any given moment a number that measures how fast the object is moving right now. For instance, a car's speedometer tells the driver the car's velocity at any given instant. In fact, the velocity on a speedometer is really an average velocity that is computed over a very small time interval. If we let the time interval over which average velocity is computed become shorter and shorter, we can progress from average velocity to instantaneous velocity.

Informally, we define the instantaneous velocity of a moving object at time \(t = a\) to be the value that the average velocity approaches as we take smaller and smaller intervals of time containing \(t = a\). We will develop a more formal definition of instantaneous velocity soon, and this definition will be the foundation of much of our work in calculus. For now, it is fine to think of instantaneous velocity as follows: take average velocities on smaller and smaller time intervals around a specific point. If those average velocities approach a single number, then that number will be the instantaneous velocity at that point.

At this point we have started to see a close connection between average velocity and instantaneous velocity. Each is connected not only to the physical behavior of the moving object but also to the geometric behavior of the graph of the position function. We are interested in computing average velocities on the interval \([a,b]\) for smaller and smaller intervals. In order to make the link between average and instantaneous velocity more formal, think of the value \(b\) as \(b = a + h\), where \(h\) is a small (non-zero) number that is allowed to vary. Then the average velocity of the object on the interval \([a,a+h]\) is \[\begin{aligned}\end{aligned}\], with the denominator being simply \(h\) because \((a+h) - a = h\). Note that when \(h \lt 0\), \(AV_{[a,a+h]}\) measures the average velocity on the interval \([a+h,a]\).

To find the instantaneous velocity at \(t = a\), we investigate what happens as the value of \(h\) approaches zero.

Condensed — the full section is in Boelkins, Active Calculus.

Summary

\(s(t)\)average velocity of the object on the interval from \(t = a\) to \(t = b\)\(AV_{[a,b]}\)\[\begin{aligned}\end{aligned}\]
The average velocity on \([a,b]\) can be viewed geometrically as the slope of the line between the points \((a,s(a))\) and \((b,s(b))\) on the graph of \(y = s(t)\), as shown in Figure.
\(t\)\(s\)\([a,b]\)\(AV_{[a,b]} = \frac{s(b) - s(a)}{b-a}\)\([a,b]\)\([a,a+h]\)\(AV_{[a,a+h]} = \frac{s(a+h) - s(a)}{h}\)
The instantaneous velocity of a moving object at a fixed time is estimated by considering average velocities on shorter and shorter time intervals that contain the instant of interest.

Practice (9)

Try each one on paper first. Reveal the answer to check; verified ones can be opened in the solver for every step.

The population of a city changes as time goes on.
1. Independent:
2. Dependent:
3. Rate of change:
As you ride in a rising weather balloon, the temperature of the air you encounter decreases.
1. Independent:
2. Dependent:
3. Rate of change:
For a given trip (i.e. a set distance) as the speed of the car increases, the travel time decreases.
1. Independent:
2. Dependent:
3. Rate of change:
As the population of a city decreases, the tax revenue of the city also decreases.
1. Independent:
2. Dependent:
3. Rate of change:
Give your own example of a quantity that changes in response to another quantity, the units of the independent and dependent variables, and units of the rate of change.
A bungee jumper dives from a tower at time \(t=0\). Her height \(s\) (measured in feet) at time \(t\) (in seconds) is given by the graph in Figure. In this problem, you may base your answers on estimates from the graph or use the fact that the jumper's height function is given by \(s(t) = 100\cos(0.75t) \cdot e^{-0.2t}+100\).
1. What is the change in vertical position of the bungee jumper between \(t=0\) and \(t=15\)?
2. Estimate the jumper's average velocity on each of the following time intervals: \([0,15]\), \([0,2]\), \([1,6]\), and \([8,10]\). Include units on your answers.
3. On what time interval(s) do you think the bungee jumper achieves her greatest average velocity? Why?
4. Estimate the jumper's instantaneous velocity at \(t=5\). Show your work and explain your reasoning, and include units on your answer.
5. Among the average and instantaneous velocities you computed in earlier questions, which are positive and which are negative? What does negative velocity indicate?
เปิดเผยคำตอบ
1. Since \(s(0) = 100\cos(0)e^{0}+100 = 200\) and \(s(15) = 100\cos(11.25)e^{-3}+100 \approx 101.25\), the bungee jumper's change in vertical position between \(t=0\) \(t=15\) is \[\begin{aligned}\end{aligned}\].
2. We estimate the jumper's average velocity using the formula \[\begin{aligned}\end{aligned}\]. Using the approximations from part (a) and \(s(1) \approx 159.91, \ \ s(2) \approx 104.74, \ \ s(6) \approx 93.65, \ \ s(8) \approx 119.39, \ \ s(10) \approx 104.69\), we have \[\begin{aligned}AV_{[0,15]} &= \frac{s(15)-s(0)}{15-0} \approx -6.58 \\ AV_{[0,2]} &= \frac{s(2)-s(0)}{2-0} \approx -47.63 \\ AV_{[1,6]} &= \frac{s(6)-s(1)}{6-1} \approx -13.25 \\ AV_{[8,10]} &= \frac{s(10)-s(8)}{10-8} \approx -7.35\end{aligned}\] The units on each of these average velocities are feet per second.
3. On the time interval \([0,4]\) the jumper has the greatest change in position over that period of time, so it is on that interval that her average velocity is greatest in magnitude (but negative in sign). For the same reasons, the largest (positive) average velocity will occur on the interval \([4,8]\).
4. We can approximate the jumper's average velocity on the intervals \([4.9,5]\) and \([5,5.1]\) and find the average of these average velocities. This will give us the slope of the secant line connecting the points \((4.9, s(4.9))\) and \((5.1, s(5.1))\) and be a reasonably good approximation to the instantaneous velocity at \(t = 5\). Now \[\begin{aligned}AV_{[4.9,5]} &= \frac{s(5)-s(4.9)}{5-4.9} \approx 21.31 \\ AV_{[5,5.1]} &= \frac{s(5.1)-s(5)}{5.1-5} \approx 22.25\end{aligned}\] so the jumper's instantaneous velocity at \(t=5\) is about \[\begin{aligned}\end{aligned}\] feet per second.
5. The average velocities we calculated were all negative, while the instantaneous velocity was positive. When the jumper is moving in the downward direction, the average velocity will be negative (indicating a decrease in position), while a positive average velocity indicates an increase in position (moving upwards).
According to the U.S. census, the population of the city of Grand Rapids, MI, was 181,843 in 1980; 189,126 in 1990; and 197,800 in 2000.
1. Between 1980 and 2000, by how many people did the population of Grand Rapids grow?
2. In an average year between 1980 and 2000, by how many people did the population of Grand Rapids grow?
3. \(f\)average rate of change\(f\)\([a,b]\)\[\begin{aligned}\end{aligned}\]\(\frac{f(b)-f(a)}{b-a}\)\(y = f(x)\)\([a,b]\)
4. \(P(t)\)\(t\)\(t\)\(P\)\(t = 0\)\(t = 20\)
5. \[\begin{aligned}\end{aligned}\]\([5,10]\)\([5,9]\)\([5,8]\)\([5,7]\)\([5,6]\)
6. How fast do you think the population of Grand Rapids was changing on January 1, 1985? Said differently, at what rate do you think people were being added to the population of Grand Rapids as of January 1, 1985? How many additional people should the city have expected in the following year? Why?
เปิดเผยคำตอบ
1. The change in population is \(P(2000) - P(1980) = 197 800 - 181 843 = 15 957\) people.
2. Since there were \(15 957\) people added to Grand Rapids over the 20 years from 1980 to 2000, in an average year the population grew by \(\frac{15,957}{20} \approx 798\) people/year.
3. \(\frac{f(b)-f(a)}{b-a}=\frac{\Delta f}{\Delta x}\) measures the slope of a secant line through the points \((a,f(a))\) and \((b,f(b))\).
4. \[\begin{aligned}\end{aligned}\] people per year.
5. \[\begin{aligned}AV_{[5,10]} &=\frac{P(10) - P(5)}{10-5} = \frac{189 116.72 - 185 444.20}{10-5} \approx 734.50 \\ AV_{[5,9]} &= \frac{188 376.44 - 185 444.20}{9-5} \approx 733.06 \\ AV_{[5,8]} &= \frac{187 639.06 - 185,444.20}{8-5} \approx 731.62 \\ AV_{[5,7]} &= \frac{186 904.57 - 185,444.20}{7-5} \approx 730.19 \\ AV_{[5,6]} &= \frac{186 172.95 - 185,444.20}{6-5} \approx 728.7535\end{aligned}\] The units on each of these quantities are people per year.
A bungee jumper dives from a tower at time \(t=0\). Her height \(s\) (measured in feet) at time \(t\) (in seconds) is given by the graph in Figure. In this problem, you may base your answers on estimates from the graph or use the fact that the jumper's height function is given by \(s(t) = 100\cos(0.75t) \cdot e^{-0.2t}+100\).
1. What is the change in vertical position of the bungee jumper between \(t=0\) and \(t=15\)?
2. Estimate the jumper's average velocity on each of the following time intervals: \([0,15]\), \([0,2]\), \([1,6]\), and \([8,10]\). Include units on your answers.
3. On what time interval(s) do you think the bungee jumper achieves her greatest average velocity? Why?
4. Estimate the jumper's instantaneous velocity at \(t=5\). Show your work and explain your reasoning, and include units on your answer.
5. Among the average and instantaneous velocities you computed in earlier questions, which are positive and which are negative? What does negative velocity indicate?
เปิดเผยคำตอบ
1. Since \(s(0) = 100\cos(0)e^{0}+100 = 200\) and \(s(15) = 100\cos(11.25)e^{-3}+100 \approx 101.25\), the bungee jumper's change in vertical position between \(t=0\) \(t=15\) is \[\begin{aligned}\end{aligned}\].
2. We estimate the jumper's average velocity using the formula \[\begin{aligned}\end{aligned}\]. Using the approximations from part (a) and \(s(1) \approx 159.91, \ \ s(2) \approx 104.74, \ \ s(6) \approx 93.65, \ \ s(8) \approx 119.39, \ \ s(10) \approx 104.69\), we have \[\begin{aligned}AV_{[0,15]} &= \frac{s(15)-s(0)}{15-0} \approx -6.58 \\ AV_{[0,2]} &= \frac{s(2)-s(0)}{2-0} \approx -47.63 \\ AV_{[1,6]} &= \frac{s(6)-s(1)}{6-1} \approx -13.25 \\ AV_{[8,10]} &= \frac{s(10)-s(8)}{10-8} \approx -7.35\end{aligned}\] The units on each of these average velocities are feet per second.
3. On the time interval \([0,4]\) the jumper has the greatest change in position over that period of time, so it is on that interval that her average velocity is greatest in magnitude (but negative in sign). For the same reasons, the largest (positive) average velocity will occur on the interval \([4,8]\).
4. We can approximate the jumper's average velocity on the intervals \([4.9,5]\) and \([5,5.1]\) and find the average of these average velocities. This will give us the slope of the secant line connecting the points \((4.9, s(4.9))\) and \((5.1, s(5.1))\) and be a reasonably good approximation to the instantaneous velocity at \(t = 5\). Now \[\begin{aligned}AV_{[4.9,5]} &= \frac{s(5)-s(4.9)}{5-4.9} \approx 21.31 \\ AV_{[5,5.1]} &= \frac{s(5.1)-s(5)}{5.1-5} \approx 22.25\end{aligned}\] so the jumper's instantaneous velocity at \(t=5\) is about \[\begin{aligned}\end{aligned}\] feet per second.
5. The average velocities we calculated were all negative, while the instantaneous velocity was positive. When the jumper is moving in the downward direction, the average velocity will be negative (indicating a decrease in position), while a positive average velocity indicates an increase in position (moving upwards).
According to the U.S. census, the population of the city of Grand Rapids, MI, was 181,843 in 1980; 189,126 in 1990; and 197,800 in 2000.
1. Between 1980 and 2000, by how many people did the population of Grand Rapids grow?
2. In an average year between 1980 and 2000, by how many people did the population of Grand Rapids grow?
3. \(f\)average rate of change\(f\)\([a,b]\)\[\begin{aligned}\end{aligned}\]\(\frac{f(b)-f(a)}{b-a}\)\(y = f(x)\)\([a,b]\)
4. \(P(t)\)\(t\)\(t\)\(P\)\(t = 0\)\(t = 20\)
5. \[\begin{aligned}\end{aligned}\]\([5,10]\)\([5,9]\)\([5,8]\)\([5,7]\)\([5,6]\)
6. How fast do you think the population of Grand Rapids was changing on January 1, 1985? Said differently, at what rate do you think people were being added to the population of Grand Rapids as of January 1, 1985? How many additional people should the city have expected in the following year? Why?
เปิดเผยคำตอบ
1. The change in population is \(P(2000) - P(1980) = 197 800 - 181 843 = 15 957\) people.
2. Since there were \(15 957\) people added to Grand Rapids over the 20 years from 1980 to 2000, in an average year the population grew by \(\frac{15,957}{20} \approx 798\) people/year.
3. \(\frac{f(b)-f(a)}{b-a}=\frac{\Delta f}{\Delta x}\) measures the slope of a secant line through the points \((a,f(a))\) and \((b,f(b))\).
4. \[\begin{aligned}\end{aligned}\] people per year.
5. \[\begin{aligned}AV_{[5,10]} &=\frac{P(10) - P(5)}{10-5} = \frac{189 116.72 - 185 444.20}{10-5} \approx 734.50 \\ AV_{[5,9]} &= \frac{188 376.44 - 185 444.20}{9-5} \approx 733.06 \\ AV_{[5,8]} &= \frac{187 639.06 - 185,444.20}{8-5} \approx 731.62 \\ AV_{[5,7]} &= \frac{186 904.57 - 185,444.20}{7-5} \approx 730.19 \\ AV_{[5,6]} &= \frac{186 172.95 - 185,444.20}{6-5} \approx 728.7535\end{aligned}\] The units on each of these quantities are people per year.

Symbols used here

\sin,\ \cos,\ \tan

sine, cosine, tangent
Ratios of sides in a right triangle; coordinates on the unit circle.

P(A),\ P(A \mid B)

probability, conditional probability
Chance of A; chance of A given that B happened.

Euler's number
2.71828…, the base whose exponential is its own derivative.

imaginary unit
i² = −1.

\pi

pi
Ratio of a circle's circumference to its diameter, 3.14159…

\infty

infinity
Not a number: "grows without bound" in limits and intervals.

\log_b x,\ \ln x

logarithm, natural log
The exponent b must be raised to for x; ln uses base e.

\sum_{k=1}^{n} a_k

summation
Add a_k for k = 1 up to n.

\lim_{x \to a} f(x)

limit
The value f(x) approaches as x approaches a.

f'(x),\ \frac{dy}{dx}

derivative
Instantaneous rate of change; slope of the graph.

\int f(x)\,dx,\ \int_a^b

integral
Antiderivative (indefinite) or signed area from a to b (definite).

y',\ y''

first and second derivative of y
Prime notation for derivatives with respect to x (or t).

C,\ C_1,\ C_2

arbitrary constants
Constants of integration fixed by initial conditions.

How to: How do we measure velocity?

How is the average velocity of a moving object connected to the values of its position function?
How do we interpret the average velocity of an object geometrically on the graph of its position function?
How is the notion of instantaneous velocity connected to average velocity?

Questions people ask

What is a derivative in one sentence?

The slope of the graph at a point — the rate at which the output is changing there. Speed is the derivative of position.

What is an integral in one sentence?

The accumulated total of a rate — the area under the curve. Distance travelled is the integral of speed.

Why are derivatives and integrals opposites?

That is the fundamental theorem of calculus: accumulating a rate and then measuring how fast the accumulation grows gets you back the rate. Integration undoes differentiation up to a constant.

When do I use substitution and when integration by parts?

Substitution when part of the integrand is the derivative of another part (u and du both present). Parts when the integrand is a product of two unrelated kinds of function — a polynomial times an exponential, log or trig function.

ลองดูสิ

Parts of this page are adapted from Boelkins, Active Calculus (CC BY-SA 4.0). Condensed and re-explained here; errors are ours.

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