Chapter 2
Derivatives and Integrals
Suppose you drive from pointAto pointB. Here, the quantity is “distance,” and the rate of change of distance with respect to time is called “velocity” or “speed.”1
Let’s say the total distance from A to B was10miles, and your trip took a half hour. Then, following common sense, we can define your average speed over the whole trip as
average speedfromAtoB= 10miles
0.5hours or 20miles per hour
But we can look at your average speed over any time interval. If we letX(t)be your progress, the distance covered fromAtoBas a function of timet(Figure2.1), then for any time interval (t1, t2)in that half hour, we can define your average speed over that time interval as
average speed(t1,t2)=change in distance
change in time = X(t2)−X(t1) t2−t1
A B
distance
t1 t2 time
X(t1) X(t2)
t
X
Figure 2.1: An example of distanceX covered fromA toBas a function of time t.
We will use a standard notation: the change int, fromt1 tot2, we will call∆t, So
∆t=t2−t1
and the corresponding change in X,X(t2)−X(t1), we will call∆X:
∆X=X(t2)−X(t1) So
average speed(t1,t2)= X(t2)−X(t1) t2−t1
=∆X
∆t
Exercise 2.2.1 A bowling lane is60feet long. If a bowling ball is released att= 0and reaches the pins2.5seconds later, what is its average speed?
If we now choose a smaller time interval (t1, t3), we get a smaller ∆X over the smaller ∆t (Figure2.2), and a new estimate of average speed, over this shorter interval (shown in red). If we then take t3, t4, etc., closer and closer to t1, then we get a succession of average speeds over shorter and shorter intervals.
1In physics, “velocity” means “speed plus direction,” so velocities can be positive or negative. “Speed” is a more colloquial term, and it is generally thought of as only positive. So, if your car was backing up, we would say that yourspeed going backward was 5 miles/hour, and yourvelocity was−5miles/hour.
A B
distance
t1 t2 time
X(t1) X(t2)
t X
t3
t X X(t3)
Figure 2.2: Choosing a smaller∆t gives an average velocity over a shorter time interval.
Exercise 2.2.2 Let’s say we are given the functional form of the curve in Figure2.2:
f(t) = (B−A) t4 1 +t4 +A
Here we assume A = 0 and B = 1. Calculate estimates of the average speed over several intervals beginning att= 1, say∆t= 0.5,0.2,and0.1.
Clearly, we can compute average speed over any time interval, no matter how short. Now we want to go further and ask what might seem like an odd question: we talked about youraverage speed over any giventime interval. Does it make any sense to talk about your speedat apoint in time? Can we make sense of the concept of yourinstantaneous speed at a timet0?
On the one hand, it makes perfect sense to say “well, atexactly 1:15 p.m., when I was partway there, I was definitely going at some speed or other.” But on the other hand, if you tried to apply the definition of average speed, you would get
instantaneous speedat 1:15 pm= ∆X
∆t = 0 0 which is absurd.
This paradox was known to the ancient Greeks (look up Zeno’s paradoxes), but it wasn’t really answered until the 1600s, with the work of Newton and Leibniz. They realized that the way to approach the idea of instantaneous velocity att0is to look at the average velocity over a small interval, fromt0tot0+ ∆t, and then let that interval get smaller and smaller, approaching zero; that is, let ∆t→0.2
If that process produced an actual number as its limiting value (not 00), then we could well call that limiting value theinstantaneous velocity at t0.
We can now define the instantaneous speed at t0 to be the value that these successive approximations approach as ∆t gets closer and closer to 0, more formally, the limit of these values as∆t approaches0:
instantaneous speedt0= lim
∆t→0
∆X
∆t = lim
∆t→0
X(t0+ ∆t)−X(t0)
∆t
2Actually, Newton and Leibniz tried to reason using the concept of an “infinitesimally small quantity.” It wasn’t until the 1800s that the idea of instantaneous velocity was put on a rigorous foundation using the notion oflimits. In the 1960s, the notion of “infinitesimally small quantity” was made rigorous by UCLA mathematician Abraham Robinson in his nonstandard analysis.
Exercise 2.2.3 If at some instant an object’s speed is30 mih, will it travel30miles in the next hour?
We began this chapter using Euler’s method to get anapproximationforX′(t), and concluded that
X′(t)
instantaneous rate of change
≈ X(t+ ∆t)−X(t) ∆t
average rate of change
Now we can say that X′(t), the left-hand side of this equation, is the instantaneous rate of change of X at time t, and the right-hand side of this equation is exactly the average rate of change of X from timet to timet+ ∆t, as we just defined it. In Euler’s method, we learned that we can make the approximation more accurate by making∆t very close to0.
This connection between average rates of change and instantaneous rates of change is the foundation for the subject that is called “calculus.”
Example: A Falling Object
Legend has it that Galileo dropped balls from the Leaning Tower of Pisa and measured their time of flight. This is not true. The time intervals involved are too short for him to measure using then-existing technology. What he actually did was slow the process down by rolling balls down an inclined plane and measuring the time intervals with a water clock.
He then summarized his findings in a law that is applicable to falling objects.
LetH(t)be the height of the ball above the groundtseconds after we let go of it. According to Galileo, if we ignore air resistance slowing the ball down, its height will be
H(t) =H(0)−16t2 where H(0)is the initial height.3
0.5 1.0 1.5 2.0 2.5
20 40 60 80 100
t (sec) H(t) (ft)
t=0.0 s t=0.5 s t=1.0 s
t=1.5 s
t=2.0 s
t=2.5 s
Figure 2.3: Graph ofH(t) = 100−16t2, representing the heightH of the ball,t seconds after being dropped from an initial height of 100 feet.
3The value “16” results from the assumption thatHis in feet andtis in seconds.
Based on this, can we say how fast the ball is fallingexactly 1.5seconds after we release it from an initial height of100 ft (Figure2.3)?
Since we are asking for the instantaneous velocity at a timet, which is the instantaneous rate of change of the functionH att= 1.5, we are looking forH′(1.5).
How do we compute H′(1.5)? By considering the average rate of change ofH over various time intervals, and then letting the time intervals get smaller and smaller, that is, making ∆t approach 0.
Let’s begin by considering a time interval of0.1 s, fromt= 1.5tot= 1.6, so
∆t= 1.6−1.5 = 0.1 s The average rate of change ofH over this time interval is
average rate of change= H(1.5 + ∆t)−H(1.5)
∆t
= H(1.6)−H(1.5) 0.1
= (100−16·1.62)−(100−16·1.52) 0.1
=−49.6ft s
Exercise 2.2.4 Notice that our calculation results in a negative number. Why does this make sense?
The value∆t= 0.1 s represents a fairly short time interval, so we can consider this to be an approximation ofH′(1.5):
H′(1.5)≈H(1.6)−H(1.5)
0.1 =−49.6ft s
As in Euler’s method, we can make this approximation better by using a smaller ∆t. If we redo the calculation with time interval ∆t= 0.01, we get
H′(1.5)≈H(1.51)−H(1.5)
0.01 = (100−16·1.512)−(100−16·1.52)
0.01 =−48.16ft
s We can get sharper estimates ofH′(1.5)by using even smaller values of the time interval∆t, for example,∆t= 0.001.
H′(1.5)≈ H(1.501)−H(1.5)
0.001 = (100−16·1.5012)−(100−16·1.52)
0.001 =−48.016ft
s
Exercise 2.2.5 ApproximateH′(1.5)using the time interval∆t= 0.0001.
Estimates with smaller and smaller values of ∆t have resulted in a series of estimates of H′(1.5). The actual value of H′(1.5) is the limit of these estimates as∆t approaches0. But what is that limit?
Exercise 2.2.6 Why do we not allow ∆t to reach0?
Exercise 2.2.7 Use successive approximations to find the object’s speed att= 1second.
Finding H′(t)
In our example, the successive estimates of H′(1.5) are −49.6, −48.16, and−48.016. These estimates look like they are getting closer and closer to48. But how can we be sure that this is the exact value ofH′(1.5)?
We can answer this mathematically by doing a symbolic calculation. Instead of using specific values of∆t, as above, we will do a symbolic calculation using the symbol∆t.
average rate of change
at t = 1.5 = H(1.5 + ∆t)−H(1.5)
∆t (2.1)
We know the functionH, and so we can plug it into equation2.1. SinceH(t) = 100−16t2, we can compute
H(1.5) = 100−16·1.52= 64 The next quantity we need isH(1.5 + ∆t):
H(1.5 + ∆t) = 100−16·(1.5 + ∆t)2
= 100−16·
1.52+ 3∆t+ (∆t)2
= 100−36−48∆t−16·(∆t)2
= 64−48∆t−16·(∆t)2 Substituting these two expressions into equation2.1gives us
average rate of change
at t = 1.5 =
64−48∆t−16·(∆t)2
−64
∆t
Notice that the 64’s in the denominator cancel each other (not a coincidence).
average rate of change
at t = 1.5 =
✚64✚−48∆t−16·(∆t)2
−✚64✚
∆t
=−48∆t−16·(∆t)2
∆t
=✚∆t✚·(−48−16∆t)
✚∆t✚
=−48−16∆t
We have now found a general expression for the average rate of change at t = 1.5 as a function of ∆t, and it is obvious what will happen as∆t approaches0:
−48−16∆t −→ −48 as ∆t−→0 which gives us the exact value ofH′(1.5)as
H′(1.5) = lim
∆t→0
H(1.5 + ∆t)−H(1.5)
∆t = lim
∆t→0(−48−16∆t) =−48
Exercise 2.2.8 Carry out a similar calculation fort= 2.
The procedure that we just applied to findH′(1.5)can be generalized to anyt, and we have now developed a general procedure for findingH′(t):
H′(t) = lim
∆t→0
H(t+ ∆t)−H(t)
∆t Carrying out this calculation, we get
H′(t) = lim
∆t→0
H(t+ ∆t)−H(t)
∆t
= lim
∆t→0
H(0)−16(t+ ∆t)2
−H(0)−16t2
∆t
= lim
∆t→0
✭✭✭✭✭✭
H(0)−16t2−32t·∆t−∆t2
−✭✭✭✭✭✭
H(0)−16t2
∆t
= lim
∆t→0
✚∆t✚·(32t−∆t)
✚∆t✚
= lim
∆t→0(−32t−∆t)
=−32t
So we can now say, for the functionH(t) =H(0)−16t2, that we can calculateH′(t)for any t0.
Exercise 2.2.9 Use this result to find the object’s velocity att= 2.
The derivative of H(t)at the pointt0is the limit as∆t→0of the quantity ∆H∆t. H′(t)
t
0 = lim
∆t→0
∆H
∆t t
0
In fact, this procedure can be carried out for many functionsX(t), including most functions that can be expressed as a formula. (For exceptions, see “Do all functions have derivatives?”)
If X(t) is any function of t, then we can almost always define X′(t)
t0, which is the in- stantaneous rate of change of X at time t0. This is called the derivative of X at the point t0.
Variables Other Than Time
We’ve been talking so far about functions of time, and rates of change with respect to time, like velocity, which is the rate of change of distance with respect to time. But we can also talk about functions ofany variable, not just time. And then we can ask about how the value of the function changes with respect to changes in the variable.
For example, suppose you are climbing a mountain or ascending in an airplane. You observe that as you go higher, the outside air pressure decreases. So air pressure is a function of altitude.
If we letH=height andP =air pressure, then there is some function P =f(H)
In fact, it looks like Figure2.4.
Altitude H (km) Atmospheric
Pressure P (bars)
5 10 15 20
0.2 0.4 0.6 0.8 1.0
0
H P
P =
-H
e
7Figure 2.4: Atmospheric pressureP as a function of altitudeH.
We can now ask: how much does pressure change with respect to height? We could even ask:
“what is the rate at whichP is changing with respect toH?”
There are many examples in science where we are looking at one variable as a function of another variable. In fact, science is all about looking for relationships that show one variable as a function of another.
In chemistry, we study the properties of gases, such as their pressures and volumes. We know that if we put a gas under pressure, say from a piston, the volume of the gas decreases. So there is a function that gives the volumeV for a given value of pressureP,V =f(P). (This is called Boyle’s law.) And again, we can talk about the rate at whichV is changing with respect toP.
Astrophysics studies the properties of stars, such as their distance from us and their velocities (which can be figured out from the color of the light they emit). There is a famous law, called Hubble’s law, that says that the velocity with which a star is receding from us is a function of its distance from us.
In biology, the subject ofallometrystudies the basic physical measurements that can be made on animals, like body length, skull size, heart rate, and metabolic rate. It studies how these characteristics are related to other physical characteristics, such as body mass. It is interesting to look at how these scale with each other over a very wide range of sizes, from ants to whales.
For example, let’s say we are looking at how heart rateH scales with body massM. We look at the data from thousands of species and find that they lie on a curved line, giving usH=f(M).
Depending on the shape of the curve, we can talk about the rate at which H is changing with respect toM.
So any time one quantity Y can be expressed as a function of some other quantityX, we can ask: if X changes, how much willY change in response? Let’s define this concept of rate precisely.
For example, let’s go back to our example of the relation between air pressureP and altitude H (Figure2.4). We can define the concept of the rate of change ofP with respect toHexactly as we did in defining a rate of change with respect to time. First we define an average rate of change over some interval as ∆P∆H.
For example, if we went from an altitude of H = 2 km to H = 5 km, then the change in altitude is
∆H= 5−2 = 3 km
We will see the atmospheric pressure drop, from 0.75 bars at H = 2 km down to0.49 bars at H= 5 km. Therefore, thechange in pressure is
∆P = 0.49−0.75 =−0.26bar s
We then define the average rate of change, over the interval fromH = 2 km toH = 5 km, of atmospheric pressure (P) with respect to altitude (H), by
average rate of change ofPwith respect toH= ∆P
∆H =−0.26 bar
3 km =−0.09bar km This is the average rate of change over an interval.
It then makes perfect sense to do exactly what we did with respect to time: pick an arbitrary point H0, and define theinstantaneous rate of change of P with respect toH at the point H0
as the limit of the average rate of change as the interval∆H approaches to zero:
instantaneous rate of change ofPwith respect toHatH0= lim
∆H→0
∆P
∆H
H0
Exercise 2.2.10 The function describing how air pressure varies with elevation is P(H) = 101352e−0.28H2396
where P is measured in pascals and H in meters. Approximate the rate of change of P with respect toH at a height of 2000 meters.
Notation
We have now defined a concept: for anyY =f(X), at any pointX0, the instantaneous rate of change of Y with respect toX is
∆Xlim→0
∆Y
∆X X
0
= lim
X→X0
f(X)−f(X0) X−X0
The only question is: what to call this? We have been using Y′ to mean the derivative with respect to time, and we can extend this to allowY′to mean the derivative with respect to some other variable, but the only problem is that we have no way of saying what that other variable is. We don’t have a terminology yet for the instantaneous rate of change with respect to some arbitrary variable.
We are rescued by Gottfried Leibniz, the co-inventor of calculus in the late 1600s, along with Isaac Newton. Newton had favored a notation something like ourX′ (actually an X with a dot over it, X), so˙ X′ is called the Newtonian form. Leibniz, on the other hand, wanted to stress that this is a ratio of two quantities, ∆X and∆t, so he adopted somewhat odd notation. He looked at the ratio ∆X∆t and decided to refer tolim∆t→0∆X
∆t as a ratio he called dXdt.
Leibniz dX dt = lim
∆t→0
∆X
∆t t
0
Newton X′= lim
∆t→0
∆X
∆t
t0
There is a clear drawback to the Leibniz notation. What is “dX”? What is “dt”? How can we take their ratio if we don’t know what the individual terms mean? And why can’t we divide top and bottom byd?
The answer to all these questions is that the Leibniz notation can’t really be read as the ratio of two anythings, and the termsdX anddt don’t really mean anything by themselves.4
Rather, the whole expression
“dX dt
t0
” means “ lim
∆t→0
∆X
∆t
t0
”
The big advantage of the Leibniz notation is that we can now state explicitly both of the variables in the limit. This makes it possible to return to our original question what to call lim∆X→0∆Y
∆X
X0, whereX is arbitrary. We will call it d Y
dX X
0
or df dX
X
0
So everywhere in this text, when we sayX′, we usually mean dXdt, but we will sometimes use the notationY′for convenience when the relevant variable is obvious. And when we want to refer to a function of some other variable, such as P =f(H), we will usually call the instantaneous rate of change dHdP.
“Sensitivity”
The quantity we just defined as the “instantaneous rate of change of Y with respect toX” can also be seen as the definition of the concept of “sensitivity.” When we are talking about the
“sensitivity ofY toX,” we are really talking about the quantity d YdX.
Suppose, for example, we are looking at a drug for cancer chemotherapy. We run experiments and determine what percent of cancer cells are still alive when we give D amount of drug. If we call the percentage of cancer cells still alive P, then our experiments give usP =f(D). A typical graph might look like this:
D (mg) P
0 % 100 %
10 5
4In fact, Leibniz was criticized by many of his fellow mathematicians for this, and for centuries his notion of dX as an “infinitesimal” quantity was frowned upon. The philosopher Bishop Berkeley ridiculeddXdt as “the ratio of the ghosts of two departed quantities.” It was not until the 1960s, nearly 300 years later, that Leibniz was fully vindicated when UCLA math professor Abraham Robinson came up with an idea called nonstandard analysis, which provided a mathematically rigorous foundation for infinitesimals.
We can then talk about the sensitivity of the cancer cells to increasing drug dosages. What we mean is
dP dD
D0
So we can say, for example, that for dosages below2milligrams, the cancer cells are highly sensitive to the drug, because dDdP
D0 is more negative whenD0<2.
Further Exercises 2.2
1. The rate of change of the position of a car at some timet0 is given by dXdt = 55. What does this mean in plain English?
2. You are studying the athletic performance of runners. You have two motion-triggered cameras that produce time-stamped photographs.
a) The runner reaches the first camera, at the 500 m mark, at 9:03:05 a.m. and the second camera, at the 600 m mark, at 9:03:25 a.m. What is her average speed over that time interval?
b) When is she running at that speed?
c) How could you change your measurement setup (without getting new equipment) to better approximate the runner’s instantaneous speed at 500 m?
3. You are an ecologist studying bottom-dwelling stream invertebrates. You need to mea- sure the speed at which the water is flowing at a particular point you have chosen to study. You have a stopwatch, a long measuring tape, a supply of Ping-Pong balls (which float and are easy to see), and brightly colored flags that can be used to mark points along the shore or in the water. How would you use this equipment to estimate the instantaneous speed of the water? (You may want to include a diagram with your response.)
4. Use successive approximations to approximate the derivative of the functions below at the points specified.
a)f(X) = 6X5 atX= 2 b)f(x) = 7X3+ 2atX= 3.5
c)f(X) = sinX atX=−3(use radians) d)f(X) = sin
ln(X3+ 1)
atX= 4
5. You are studying a new blood-pressure-lowering drug. You find that blood pressure is not very sensitive to the drug at low doses, very sensitive at intermediate doses, and not very sensitive at high doses. Rephrase this statement in terms of dMdP, where P is blood pressure andM is the drug dosage.