Chapter 3 Review Problems
4.1 Formal Definition of the Derivative
4.1.2 The Derivative as an Instantaneous Rate of Change: A First Look at Differential Equations
That is, if f(x)= 1x,x =0, then
f(x)= −1
x2, x =0
Looking back at the first three examples, we see that in order to compute f(x) from the formal definition of the derivative, we evaluate the limit
lim
h→0
f(x+h)− f(x) h
Since both limh→0[f(x+h)− f(x)]and limh→0hare equal to 0, we cannot simply evaluate the limits in the numerator and the denominator separately, because this would result in the undefined expression 0/0. It is important to simplify the difference quotient before we take the limit.
4.1.2 The Derivative as an Instantaneous Rate of Change: A First
Theinstantaneous velocityat timet is defined as the limit ofst ast →0, or lim
t→0
s t = lim
h→0
s(t +h)−s(t) h
provided that the limit exists. This quantity is the derivative ofs(t)at timet, which we denote by
ds dt = lim
t→0
s t
Note thatdsdt is an instantaneous rate of change. The instantaneous velocity (or, sim- ply, the velocity) is, then, the slope of the tangent line at a given point of the position functions(t), provided that the derivative at this point exists.
Let’s look at two points on the graph ofs(t), namely,(2,16)and(5,25). We find that the slope of the tangent line is positive at(2,16)and negative at(5,25). The velocity is therefore positive at timet = 2 and negative at timet =5. Att =2 we move away from our starting point, whereas att = 5 we move toward our starting point. At these two times, we move in opposite directions.
There is a difference betweenvelocityandspeed. If you had a speedometer on your bike, it would tell you the speed and not the velocity. Speed is the absolute value of velocity; it ignores direction.
Interpreting the derivative as an instantaneous rate of change will turn out to be extremely important to us. In fact, when you encounter derivatives in your science courses, this will be the interpretation most often used. This interpretation will allow us to describe a quantity in terms of how quickly it changes with respect to another quantity. To illustrate the point, we revisit two previous examples and introduce one new application.
Population Growth At the beginning of this chapter, we described the growth of a population at timetby the continuous functionN(t). If the derivative ofN(t)exists at timet, we can define the instantaneous growth rate of the population by
N(t)= d N
dt = [instantaneous population growth rate at timet]
We are frequently interested in the instantaneous per capita growth rate. This is the growth rate per individual, and it can be obtained by dividing the instantaneous growth rate of the population by the population size at that time. That is,
1 N(t)
d N
dt = [instantaneous per capita growth rate at timet]
In biology textbooks (and in this book), the dependence ont is often not explicitly spelled out, and we write
1 N
d N
dt instead of 1 N(t)
d N dt
The Rate of a Chemical Reaction Another illustration of the use of the derivative as an instantaneous rate of change is in Example 5 of Subsection 1.2.2, where we discussed the reaction rate of the irreversible chemical reaction
A+B→AB
which is proportional to the concentrations of A and B. If the concentration of the product AB is denoted byx, then the reaction rate is equal to
k(a−x)(b−x)
wherea= [A]is the initial concentration of A andb= [B]is the initial concentra- tion of B (Figure 4.14). The reaction rate tells us how quickly the concentration ofx
x a
kab
k(a x)(b x)
Figure 4.14 The reaction rate for a≤b.
changes with time as the reaction proceeds. The concentrationxis thus a function of timet:x =x(t). The reaction rate is an instantaneous rate of change, namely,
lim
t→0
x(t +t)−x(t) t
We can identify the limit ast →0 as the derivative of the functionx(t)with respect tot and therefore write
d x
dt =k(a−x)(b−x) (4.1)
Equation (4.1) is an example of adifferential equation—an equation that contains the derivative of a function. We will discuss such equations extensively in later chapters.
From this point on, when we say “rate of change,” we will always mean “instan- taneous rate of change.” When we are interested in the average rate of change, we will always state this explicitly.
The rate of change in a chemical reaction is described by a differential equation.
It is sometimes possible to solve such differential equations—that is, to state explicitly a function whose derivative satisfies the given equation. We will discuss this situation in detail later. More often, it is not possible (or not necessary) to explicitly find a solu- tion. Without solving the differential equation, we can still obtain useful information about its behavior. We illustrate this property in the next application.
Tilman’s Model for Resource Competition David Tilman (1982) of the University of Minnesota developed a theoretical framework to describe the outcome of competition for limited resources. To test the predictions of his theory, he conducted many experiments on the grassland habitat at Cedar Creek Natural History Area in Minnesota. For this grassland habitat, nitrogen is a limiting resource; that is, adding nitrogen to the soil will result in an increase in biomass. We will discuss the case where one species competes for a single limited resource. We assume that the rate of change of biomass has two components: rate of growth and rate of loss. We write
[rate of biomass change] = [rate of growth] − [rate of loss]
We denote the biomass of the plant population at timetbyB(t)and assume that the rate of growth depends on a single resource whose concentration is denoted by R. We will write an equation for the specific rate of changeof biomass, which is defined as the change of biomass per unit of biomass, or B1d Bdt. We assume that the per-unit rate of loss of biomass is constant and denote this quantity bym. A simple model for how the biomass changes over time is then
1 B
d B
dt = f(R)−m (4.2)
where the function f(R)describes the specific growth rate as a function of resource concentration. A common choice for f(R) is the Monod growth function (or Michaelis–Menten equation) that we considered in Example 6 of Subsection 1.2.3, or
f(R)=a R
k+R (4.3)
wherea and k are positive constants. Let’s graph f(R)and m together in Figure 4.15. Doing so yields the following observations: When 0<m <a, the graphs of the functionsy = f(R)andy =m intersect atR= R∗(read “Rstar”). Consequently, 0
0 R
f(R)
R*
m
ak R R a
Figure 4.15 Growth balances loss whenR=R∗.
at the resource levelR∗, f(R)=m, and thus the specific rate of change 1Bd Bdt is equal to 0. That is, growth balances loss, and the biomass of the species no longer changes.
We say that the biomass is atequilibrium. If the resource levelRwere held at a value less thanR∗, then f(R)−m <0, and the specific rate of growth would be negative;
that is, biomass would decrease. IfR>R∗, then f(R)−m >0, and biomass would increase.
We can compute R∗in the case when f(R)is given by (4.3). SinceR∗satisfies f(R∗)=m, we obtain
a R∗
k+R∗ =m, or R∗= mk a−m