4. In mammals, resting metabolic rateM is related to body massB as approximately M = 0.8B3/4
a) Find the linear approximation to this function for a body mass of100grams.
b) An animal species that currently averages100 grams in mass evolves to have an average mass of 110grams. Use the linear approximation to estimate how much its metabolic rate would change.
5. The number of species, S, living on an island or habitat fragment of area A can be modeled as S =cAz, where z > 0. We can measure area in units that make c = 1, simplifying the equation into S = Az. A common value for z is approximately 0.45.
Find dSdA, and explain the meaning of this function and its significance for biodiversity conservation.
6. The optimal cruising speedV for a bird in flight is given by the allometric equation V = 12M1/6
whereMis the mass of the bird in kilograms, andV is in meters per second. The average female bald eagle weighs around5.6kg. Find the linear approximation to the equation above at M = 5.6. If a female American golden eagle is about 0.7 kg lighter than a bald eagle, how much faster or slower would you expect its optimal cruising speed to be? (Note: You can check that your answer is in the ballpark by just plugging numbers into the equation above, but the point of this problem is to use a linear approximation to arrive at your answer.)
7. Find the instantaneous rate of change of the functionf(X) = 3 sinX+ lnXatX =π.
8. Find the tangent to the functionf(X) = 5X4+ 3X2−9atX= 2.
9. In this section, we saw that it makes sense to think of differentiation as a function. Is this function linear? Justify your answer.
Exercise 2.6.1 Why is the+c necessary?Find dtd(t2+ 5)and dtd(t2−1).
Exercise 2.6.2 Find the antiderivative off′(X) = 3X2.
But these are very special cases, and in most cases, given the functional form of f′(t), it is impossible to state the antiderivative f(t), and we have to rely on approximations. So, just as in differential equations in Chapter1, symbolic differentiation is usually easy, and symbolic integration is usually hard.
f f’
(easy)
integration (hard !)
Before we go into how to getf back from f′, let’s ask why:why would we want to do this?
One important class of cases involves a given functionf′(t)as a rate of something, and we want to figure out the total amount of the something deposited by that rate.
For example, suppose we are given the speedV(t)of a car from a recording of the speedome- ter. So we haveV(t). ButV(t)is justX′(t), whereXis the car’s position at timet, assuming it starts at position 0 (which makes the constantc in the antiderivative to be 0). Can we recover X(t) fromV(t)?
V(t)
t1 t A
B
0
speedometer
20
40 80
100 0 120
60
t1 0
= X’(t)
For another example, we might knowD′(t), the rate of drug delivery to a patient. This is the readout of the flow meter attached to the intravenous drip. But what we want to know is not D′(t), but ratherD(t), the cumulative amount of drug that was delivered to the patient up to time t.
D’(t)
t I. V. t1
initial level
0 t1
Euler and Riemann: Adding Up Little Rectangles
Let’s look at the case of recovering distance traveled,X(t), given the speedV(t). Suppose we are givenV(t)as a function,V(t) = 3t2. We know from the power rule that ifX(t) =t3+c, then X′(t) = 3t2. So we can say immediately that int seconds, the car has traveled a total of t3+c miles. Since att= 0, it was atc miles, the distance it has covered ist3miles. But what about the case in whichX′is not obviously the derivative of some function?
One way to do this is to use a version of Euler’s method. SupposeV(t) =X′(t)is the velocity of the car. Let’s write down the equation for Euler’s method:
newX=oldX+X′(oldX)·∆t If we make the table to calculate Euler’s method, it looks like t oldX X′(oldX) X′(oldX)·∆t newX=oldX+X′(oldX)·∆t
0 X0 X′(X0) X′(X0)·∆t X∆t=X0+X′(X0)·∆t
∆t X∆t X′(X∆t) X′(X∆t)·∆t X2∆t =X0+X′(X0)·∆t+X′(X∆t)·∆t
2∆t X2∆t X′(X2∆t) X′(X2∆t)·∆t X3∆t =X0+X′(X0)·∆t+X′(X∆t)·∆t+X′(X2∆t)·∆t .
. .
. . .
. . .
. . .
. . .
Exercise 2.6.3 In this table, what are the red entries in the last column? The black entries?
In the differential equation case, the critical step in Euler’s method is the calculation, at each
∆t, of the new value of X′. Since the differential equation gives us X′ as a function of X, X′=f(X), we used that functionf to calculate the newX′.
Here we don’t have to do that, because we are given X′(t)explicitly as a function oft, and so we don’t have to recalculate it from the previousX value.
This makes the procedure of approximating X(t) much easier: we can inputX′(t) directly from the formula and do not have to recalculate it using new values ofX for every ∆t. Euler’s method then takes a particularly simple form:
t V(t) V(t)·∆t X(t) 0 V(0) V(0)·∆t X(0) =X0
∆t V(∆t) V(∆t)·∆t X0+V(0)·∆t
2∆t V(2∆t) V(2∆t)·∆t X0+V(0)·∆t+V(∆t)·∆t
3∆t V(3∆t) V(3∆t)·∆t X0+V(0)·∆t+V(∆t)·∆t+V(2∆t)·∆t
... ... ... ...
If we summarize this simplified table, we see that X(t) = X(0)
location at time0
+ V(0·∆t)·∆t
distance covered in first
∆t
+ V(1·∆t)·∆t
distance covered in second
∆t
+ V(2·∆t)·∆t
distance covered in third
∆t
+ · · ·
Notice what this is saying: for each time interval, we are adding a little increment. If we represent time as t= 0,∆t, 2∆t, 3∆t, · · · , k∆t, · · ·, then the little increments are each
V(k·∆t)·∆t
as k ranges from0 ton, where n·∆t is the stopping time. (In other words,n is the number of∆t’s necessary to get to the stopping time.) This has a geometric interpretation that we will discuss a little later.
Each little incrementV(k·∆t)·∆trepresents the distance the car travels in that∆t. We get this by assuming that the velocityV is constant over the short interval∆t, which enables us to use the formula
distance =velocity×time to calculate the distance traveled.
This sum of little increments is called aRiemann sum.
We can summarize the Riemann sum as X(t)≈X(0) +Sumk=n
k=0 V(k·∆t)·∆t where the expression
k=n
Sumk=0 f(k)
means “the sum of all termsf(k), wherek takes on every integer value from0ton.”
Finally, we use the Greek uppercase sigma (Σ) to stand for “sum,” and we have X(t)≈X(0) +
k=n
k=0
V(k·∆t)·∆t
Exercise 2.6.4 Compute:
a)
k=3
k=0
2k b)
k=4
k=0
k3 c)
k=3
k=0
6k+ 2
Procedure for the Riemann Sum (1) Break down the total elapsed time into many small∆t’s.
(2) Assume thatV(t)is constant over each small interval∆t.
(3) Use the equation “distance=velocity×time” to calculate the distance traveled in that∆t.
(4) Add up the little distances.
Exercise 2.6.5 Find the Riemann sum forf(X) =X2+ 5betweenX= 0andX= 2using a step size of0.5.
In this way, we can approximate the function X(t) by a finite sum of little increments that depend on∆t. This approximation gets better and better as∆tgets smaller. So we take the final step of letting∆t approach0. We can then replace the “approximately equals” sign by “exactly equals”:
X(t)=X(0) + lim
∆t→0 k=n
k=0
V(k·∆t)·∆t
We need a new symbol for this infinite limit of the sumΣas∆t approaches0. The standard symbol for this is a big script S shape called the “integral sign”
X(t) =X(0) + t
0
X′·dt (2.2)
The expressiont
0X′·dt is calledthe definite integral of X′(t)from0to t. Equation (2.2) is called the fundamental theorem of calculus.
To get some intuition for the fundamental theorem of calculus, let’s rewrite equation (2.2) as
X(t)−X(0) = t
0
X′·dt (2.3)
Now, let’s think about the IV drip example. To find out how much fluid the patient has received, you could follow the procedure described here: add up the little rectangles. Or you could record how much fluid is in the IV bag at the start of treatment, how much is left at the end, and subtract. The fundamental theorem of calculus tells us that the results will be the same.
An Example: A Speeding Car
Let’s use a Riemann sum approximation to find an integral that is not explicitly known. Suppose the speed of a car, in feet per second, is
V(t) = 2
1000−t3
This corresponds to a car starting at around 45 miles per hour (≈ 65 feet per second) and slowing down to a complete stop over an interval of10seconds.
t
V(t) V(t) = 2√1000-t3
10 0
ft
(
sec(
65
(sec)
If we want to figure out how far the car has traveled at a particular time t, we need to find X(t).
We mentioned earlier that most of the time, it is impossible to find an actual equation for the solution to an integration problem. This is one of those cases. There is no closed-form expression for the antiderivative of2√
1000−t3.8
So how do we find the distance we’ve traveled in the car at some time t? We can use the Riemann sum method to approximate it. So let’s supposeX(0) = 0, and let’s use a step size of
∆t= 0.1to approximateX at time t= 10seconds, the moment at which the speed is 0(i.e., when the car comes to a complete stop). We’ll be able to use this to find the distance required to stop the car, which can be an important safety consideration.
8To state this precisely, the antiderivative of2√
1000−t3is not anelementary function. Elementary functions are those made up of a finite number of power functions, trig functions, exponential functions, and their inverses, combined using addition, subtraction, multiplication, division, and composition—in short, anything for which you can write down a simple formula
The Riemann sum is
t V(t) V(t)·∆t X(t) =V(t)·∆t
0. 63.25 6.325 6.325
0.1 63.25 6.325 12.65
0.2 63.25 6.325 18.97
0.3 63.24 6.324 25.30
0.4 63.24 6.324 31.62
... ... ... ...
9.9 10.90 1.090 535.0
Therefore, the total distance traveled in10seconds is535.0feet.
Exercise 2.6.6 Find the distance traveled by the car in 8 seconds ifV(t) = 4√
500−t3. Use a step size of 0.5. You may want to use SageMath or a spreadsheet to help with the calculation.
The Geometry of the Riemann Sum
There is a geometric visualization of the Riemann sum that gives a lot of significant insight into the concept. When we calculated the Riemann sum, we followed the process below:
(1) We broke down the total elapsed time into many∆t’s.
(2) We assumed thatV(t)was constant over each interval∆t.
(3) Then we used the equation “distance=velocity×time” to calculate the distance traveled in that ∆t.
(4) Then we added up the little distances.
This process can be viewed geometrically (Figure2.20).
t1 t1+ t t
V(t) V(t1)
t
V(t) = 2√1000-t3
V(t1)
true area error
Figure 2.20: The blue rectangle is an approximation to the area underV(t)in a small region of width∆t. The inset illustrates the error in this approximation.
During the brief interval fromt=t1tot=t1+ ∆t, how far did the car move? We approxi- mated its speed over that short interval using the speed at the beginning of the interval, which is V(t1), and assumed that the car was constant in velocity over that interval. Based on this assumption, we then used
distance=velocity×time
to calculate the distance that the car covered during the time intervalt1tot1+ ∆tasV(t1)·∆t.
Now consider the blue rectangle. It has a base that is ∆t wide, and its height isV(t1). So the area of the blue rectangle is given by
area=height×width
or in this case, area =V(t1)·∆t. But this is the same calculation that we just did for the distance traveled:
distance = velocity × time
V(t1)·∆t = V(t1) × ∆t
area = height × width
In other words, the area of the little blue rectangle is the distance traveled during that∆t.
Then what is the total distance traveled by the car? It must be approximately the sum of the little distances. But then the total distance that the car taveled from 0 tot is approximately equal to the sum of these little rectangles, or
V(t)·∆t (Figure2.21).
t V(t)
V(t) = 2√1000-t3
t
Figure 2.21: We can estimate the area under the curve V(t) by adding up the areas of all the blue rectangles of heightV(t)and width∆t.
The sum of the little rectangles is approximately the area under the curve, and the sum of the little rectangles is approximately equal to the total distance the car has driven. In the limit as∆t approaches 0, the two are the same:distance is the area under the curve of velocity as a function of time.
Thus, the distance the car traveled from point A at t = 0 tot =t1 is the area under the velocity curve betweent= 0andt=t1(Figure2.22).
V(t)
t1 t A
B
0
speedometer
20
40 80
100 0 120
60
t1 0
= X’(t)
Figure 2.22: Left: The car’s speedometer tells us the car’s velocity V at any timet. Right: We can graph this Velocity data asV(t). The shaded blue area is equal to the distance the car has traveled fromt= 0tot=t1.
And similarly, in the drug drip problem, the cumulative amount of drug delivered fromt = 0 tot=t1 is equal to the area under the flow rate curveD′(t)(Figure2.23).
D’(t)
t I. V. t1
initial level
0 t1
Figure 2.23: Left: The intravenous drug delivery system has a flow meter that tells us, at any time, how fast the drug is flowing into the patient, that is, D′(t), where D(t) is the total amount of drug delivered. Right: The shaded area shows the total amount of drug that has been delivered by time t1, while the pink region shows the amount of drug left in the bag.
In fact, we can generalize the car velocity V(t)to be any functionf(t). In general, letf be any reasonably well behaved function oft.
Let’s define a new function oft, calledF(t),
F(t) =the area underf(t)from0tot
So, for example, the hatched area is the value ofF(t+ ∆t), and the pink area is the value of F(t) (Figure2.24). Now consider the green rectangle. It has height f(t) and width ∆t, so its area is f(t)·∆t. But now let’s look at the green rectangle from the point of view of the area-under-f functionF. It is clear that the green rectangle is approximately equal to the area underf from0tot+ ∆t (the hatched area) minus the area underf from0tot(the pink area).
This approximation gets better and better as∆t→0.
F(t+ ∆t)
hatched area
− F(t)
pink area
≈ f(t)·∆t
green rectangle
t t t t
f(t)
f(t)
t
+
Figure 2.24: The green rectangle is approximately equal to the area underf(t)fromttot+ ∆t, which is the shaded area, minus the area underf(t)from 0 to t, which is the pink area.
If we divide both sides by∆t,
F(t+ ∆t)−F(t)
∆t ≈f(t) then as ∆t→0, the left-hand side is just the definition ofF′, so
F′=f
We can now say that the area underf froma tob is just the area underf from0tobminus the area underf from0toa.
F(b)−F(a) = b
a
f(t)·dt This is another version of the fundamental theorem of calculus.
Exercise 2.6.7 According to CDC data, the average American six-year-old girl weighs 42.5 pounds, and the average ten-year-old girl weighs75pounds. What is the area under the growth (rate of change of weight) function betweent= 6andt= 10?
Example: A Drug Drip
We can illustrate the general principle of integration by Riemann sums using the example of an IV drip delivering drug to a patient.
D’(t)
t I. V.
t1 initial level
t2
0 t1 t2
Figure 2.25: The cross-hatched area is the amount of drug that was delivered from t1 tot2. It is equal to the total amount delivered from t = 0tot =t2 minus the amount delivered from t= 0tot=t1(white shaded area).
The flow meter on the drip gives usD′(t), the flow rate as a function of time. But what we need to know is how much drug was delivered to the patient between t1 and t2 (Figure2.25).
Consulting the figure, we see that this is the cross-hatched part of the area under the curve D′(t).
Then the cross-hatched area can be found by realizing that
cross-hatched area = area underD′(t)from
0tot2 − area underD′(t)from 0tot1
t2
t1 D′(t)·dt = t2
0 D′(t)·dt − t1
0 D′(t)·dt
t2
t1 D′(t)·dt = D(t2) − D(t1)
This is an application of the fundamental theorem of calculus.
But it is important to remember that in the typical case, we will be given D′(t) as data, the readout of the flow meter, that has been stored as a digital record. We have no idea what mathematical function of time that is, and we can be pretty sure that it isn’t some simple mathematical function that happens to have a handy antiderivative function.
Therefore, the expression t2
t1 D′(t)·dt is not going to be very helpful in these real-world cases, and the only technique open to us is to add up the little rectangles.
Further Exercises 2.6
1. What is the area under the graph of the functionf(x) = cosX−X betweenX= 0and X=π?
2. a) Approximate the area under the graph of f(X) =X2 betweenX = 2and X = 4 using a∆X of0.5.
b) Find the exact area underf(X) =X2betweenX= 2andX= 4.
3. a) Approximate the area under the graph of f(X) = x4 + 1 between X = 1 and X= 3using∆t= 0.5.
b) Find the exact area under the graph off(X) =X4+ 1betweenX= 1andX= 3.
c) How could you make the answer to part (a) closer to the exact value you found in part (b)?
4. You are studying a plant population whose age distribution is given by X(a) = 109 a12, where a is age in years. The smallest individuals you can reliably identify are one year old, so the age distribution starts at 1, and the plant can live no longer than ten years.
What fraction of the population is between 3 and 6 years old?
5. A building has solar panels on the roof. The graph below shows the amount of power generated by the solar panels. Assume you have the data used to generate the graph, sampled so frequently that it may be regarded as continuous. Describe how you would compute the total amount of electricity generated between9a.m. and11a.m.
08:00 10:00 12:00 14:00 16:00 18:00 0
5 10 15 20
time power (kw)