5.5: The derivative equation P'(t) = k P(t)

A crucial property of exponential functions established by the \(e ^{kt}\) Rule is

Property 5.5.1 Proportional Growth or Decay

If \(P\) is a function defined by

\[P(t)=C e^{k t}\]

where \(C\) and \(k\) are numbers, then

\[P^{\prime}(t)=k P(t)\]

Proof that \(P(t)=C e^{k t}\) implies that \(P^{\prime}(t)=k P(t)\):

\[P^{\prime}(t)=\left[C e^{k t}\right]^{\prime}=C\left[e^{k t}\right]^{\prime}=C e^{k t} k=k C e^{k t}=k P(t)\]

The reverse implication is also true, and is shown to be true in Chapter 17:

Property 5.5.2 Exponential Growth or Decay

If \(P\) is a function and there is a number \(k\) for which

\[P^{\prime}(t)=k P(t) \quad \text { for all } t \geq 0\]

then there is a number \(C\) for which

\[P(t)=C e^{k t}\]

Furthermore,

\[C=P(0) \quad \text { so that } \quad P(t)=P(0) e^{k t}\]

In the preceding equations, \(k\) can be either positive or negative. When \(k\) is negative, it is more common to emphasize this and write \(-k\) and write \(P(t) = e ^{-k t}\), where in this context it is understood that \(k\) is a positive number.

In Chapter 1, we examined models of population growth, light decay, and penicillin clearance, all of which were of the form

\[P_{t+1}-P_{t}=R P_{t}\]

and found that

\[P_{t}=P_{0} R^{t}\]

These are discrete time models in which the average rate of change of \(P_t\) is proportional to \(P_t\) . The exponential Growth or Decay Property 5.5.2 is simply a continuous time model in which the rate of change of \(P(t)\) is proportional to \(P(t)\), and would be preferred in many instances. Bacterial populations may be visualized as growing continuously (and not in twenty minute bursts), the kidneys clear penicillin continuously (and not in five minute increments), and light decays continuously with depth (and not in one meter increments). Discrete time models are easy to comprehend and with short data intervals give good replications of data, but now that we know the definition of rate of change we can use continuous time or space models.

The equation

\[P^{\prime}=k P \quad \text { or } \quad P^{\prime}(t)=k P(t) \quad \text { or } \quad \frac{d P}{d t}=k P\]

derives from many models of biological and physical processes including population growth, drug clearance, chemical reaction, decay of radio activity – any system which can be described by:

Mathematical Model 5.5.1 Proportional change. The rate of change of a quantity is proportional to the amount of the quantity.

For example in population studies, we commonly assume that

Mathematical Model 5.5.2 Simple population growth. The growth rate of a population is proportional to the size of the population.

Let \(P(t)\) be the size of a population at time, \(t\). The component parts of the sentence in the Mathematical Model of simple population growth are symbolized by

\[\begin{aligned}
\text{a:} & \text{  The growth rate of a population} & : & P^{\prime}(t)\\
\text{b:} & \text{  is proportional to} & : & =k \times\\
\text{c:} & \text{  the size of the population} & : & P(t)\\
\end{aligned}\]

The sentence of the Mathematical Model is then written

\[\underbrace{P^{\prime}(t)}_{a} \underbrace{=k \times}_{b} \underbrace{P(t)}_{c}\]

From the property of Exponential Growth and Decay

\[P(t)=C \times e^{k t} \quad \text { and if } P(0)=P_{0} \text { is known } \quad P(t)=P_{0} e^{k t} \label{5.21}\]

In the event that the rate of decrease of a quantity, \(P(t)\), is proportional to the size of \(P(t)\), then because \(-P ^{\prime} (t)\) is the rate of decrease of \(P(t)\),

\[-P^{\prime}=k P(t), \quad P^{\prime}=-k P(t), \quad \text { and } \quad P(t)=P_{0} e^{-k t},\]

where k is a positive number.

Example 5.5.1 A distinction between discrete and continuous models. Suppose in year 2000 a population is at 5 million people and the population growth rate (excess of births over deaths) is 6 percent per year. One interpretation of this is to let \(P(t)\) be the population size in millions of people at time \(t\) measured in years after 2000 and to write

\[P(0)=5 \quad P^{\prime}(t)=0.06 P(t)\]

Then, from the property of Exponential Growth or Decay 5.5.2, we may write

\[P(t)=P(0) e^{0.06 t}=5 e^{0.06 t}\]

\(P(t)=5 e^{0.06 t}\) does not exactly match the hypothesis that 'population growth rate is 6 percent per year', however. By this equation, after one year,

\[P(1)=5 e^{0.06 \times 1}=5 e^{0.06} \doteq 5 \times 1.0618]

The consequence is that during the first year (and every year) there would be a 6.18 percent increase, a contradiction.

The discrepancy lies with the model equation \(P ^{\prime} (t) = 0.6 P(t)\). Instead, we may write

\[P(0)=5 \quad P^{\prime}(t)=k P(t)\]

where \(k\) is to be determined. Then from Exponential Growth or Decay 5.5.2 we may write

\[P(t)=5 e^{k t}\]

Now impose that \(P(1) = 5 \times 1.06\), a 6 percent increase during the first year, and write

\[P(1)=5 e^{k 1}=5 \times 1.06\]

This leads to

\[e^{k \times 1}=1.06\]

We take the natural logarithm of both numbers and get

\[\begin{aligned}
\ln \left(e^{k}\right) &=\ln 1.06 \\
k &=\ln 1.06 \\
& \doteq 0.05827
\end{aligned}\]

Then

\[P(t)=5 e^{0.05827 t}\]

gives a description of the population \(t\) years after 2000. Each annual population is 6 percent greater than that of the preceding year. The continuous model of growth is actually

\[P(0)=5 \quad P^{\prime}(t)=(\ln 1.06) P(t) \quad P^{\prime}(t)=0.05827 P(t)\]

Explore 5.5.1 Show that for the discrete equations

\[P_{t+1}-P_{t}=r P_{t}, \quad \text { with solution } \quad P_{t}=P_{0}(1+r)^{t}\]

the solution to the continuous equation

\[p^{\prime}(t)=(\ln (1+r)) p(t), \quad p(0)=P_{0}\]

agrees with \(p_t\) at the integers. That is,

\[p(t)=P_{t}, \quad \text { for } \quad t=0,1,2, \cdots .\]

Explore 5.5.2 Bacterial density for v. natriegens grown in a nutrient solution at pH of 6.75 was given in Section 1.1 and are reproduced below. It was found that \(B_{t}=0.022(5 / 3)^{t}\) (Equation 1.1.14) matches the data very well. Find a number, \(k\), such that the solution to \(B(0)=0.022\), \(B^{\prime}(t)=k B(t)\) matches the solution at integer values of \(t\).

The variable \(t\) is used ambiguously here. In \(B_t\) , \(t\) is an index for 16-minute intervals and is \(0, 1, 2, \cdots\), an integer. In \(B(t)\), \(t\) is a number greater than or equal to zero, but is also measuring time in 16-minute intervals. For example, \(B(2)\) is to be the bacterial density at the end of the second 16-minute interval, the same as \(B_2\). \(B(1.5)\) is an estimate of the bacterial density at time 12 minutes, but \(B_{1.5}\) is not defined.

Time (min)	0	16	32	48	64	80
Time index, \(t\)	0	1	2	3	4	5
v. natriegens density, \(B_t\)	0.022	0.036	0.060	0.101	0.169	0.266

The growth of a bank savings account is similar to this simplified model of population growth. If you deposited $5000 in 2000 at a true 6 percent annual interest rate, it may amount to

\[P(t)=5000 e^{0.05827 t}\]

dollars \(t\) years after 2000. On the other hand, some banks advertise and compute interest on the basis of 6% interest with 'instantaneous' compounding, meaning that their model is

\[P^{\prime}(t)=0.06 P(t)\]

leading to

\[P(t)=5000 e^{0.06 t}\]

They will say that their 'APR' (annual percentage rate) is \(100 ~ e ^{0.06} = 6.18\) percent.

The solution to the discrete system

\[P_{0} \quad \text { known } \quad P_{n+1}=K P_{n} \quad \text { is } \quad P_{n}=P_{0} K^{n}\]

The solution of the continuous system

\[p(0)=P_{0}, \quad p^{\prime}(t)=(\ln K) p(t) \quad \text { is } \quad p_{t}=P_{0} e^{(\ln K t)}=P_{0} K^{t}\]

For \(t = n\), an integer, \(p(n) = P_n\).

Example 5.5.2 Geologists in the early nineteenth century worked out the sequential order of geological layers well before they knew the absolute dates of the layers. Their most extreme

estimates of the age of the earth was in the order of 400 million years⁵, about 1/10 of today’s estimate of \(4.54 \pm 0.05\) billion years⁶ based on decay of radioactive material. Early applications of radiometric dating used the decomposition of uranium-238 first to thorium-234 and subsequently to lead-206. More recently potassium-40 decomposition has been found to be useful (and zircon decay is currently the best available).

Potassium-40 decomposes to both argon-40 and calcium-40 according to

\[9\left({ }^{40} \mathrm{~K}\right) \quad \longrightarrow \quad{ }^{40} \mathrm{Ar}+8\left({ }^{40} \mathrm{Ca}\right)\]

When deposited, volcanic rock contains significant amounts of ⁴⁰K but is essentially free of ⁴⁰Ar because ⁴⁰Ar is a gas that escapes the rock under volcanic conditions. Once cooled, some volcanic rock will become essentially sealed capsules that contain ⁴⁰K and retain the ⁴⁰Ar that derives from decomposition of the ⁴⁰K.

Mathematical Model 5.5.3 Potassium-40 decomposition. The rate of disintegration of ⁴⁰K is proportional to the amount of ⁴⁰K present.

If we let \(K(t)\) be the amount of ⁴⁰K present \(t\) years after deposition of rock of volcanic origin and \(K_0\) the initial amount of ⁴⁰K present, then

\[K(0)=K_{0}, \quad K^{\prime}(t)=-r K(t)\]

where \(r\) is a positive constant. The minus sign reflects the disintegration of ⁴⁰K. From the equation we may write

\[K(t)=K_{0} e^{-r t}\]

The half-life of ⁴⁰K is \(1.28 \times 10^9\) years, meaning that \(1.28 \times 10^9\) years after deposition of the volcanic rock, the amount of ⁴⁰K in the rock will be \(\frac{1}{2} K_0\). We use this information to evaluate \(r\).

\[\begin{aligned}
\frac{1}{2} K_{0} &=K_{0} e^{-r 1280000000} \\
\frac{1}{2} &=e^{-r 1280000000} \\
\ln \frac{1}{2} &=-r 1280000000 \\
r &=\frac{\ln 2}{1280000000} \\
K(t) &=K_{0} e^{-\frac{\ln 2}{128000000} t}
\end{aligned}\]

Problem. Suppose a rock sample is found to have \(5 \times 10^{14}\) ⁴⁰K atoms and \(2 \times 10^{13}\) ⁴⁰Ar atoms. What is the age of the rock?

Solution. It is necessary to assume⁷ that all of the ⁴⁰Ar derives from the ⁴⁰K, and that there has been no leakage of ⁴⁰K or ⁴⁰Ar into or out of the rock. Assuming so, then the number of ⁴⁰K atoms that have decomposed (to either ⁴⁰Ca or ⁴⁰Ar) must be nine times the number of ⁴⁰Ar atoms, or \(9 (2 \times 10^{13}) = 1.8 \times 10^{14}\) atoms. Therefore

\[K_{0}=5 \times 10^{14}+1.8 \times 10^{14}=6.8 \times 10^{14}\]

and

\[K(t)=6.8 \times 10^{14} e^{-\frac{\ln 2}{1280000000} t}\]

We want the value of \(t\) for which \(K(t) = 5 × 10^{14}\). Therefore,

\[\begin{aligned}
5 \times 10^{14} &=6.8 \times 10^{14} e^{-\frac{\ln 2}{1280000000} t} \\
\frac{5}{6.8} &=e^{-\frac{\ln 2}{1280000000} t} \\
\ln \frac{5}{6.8} &=-\frac{\ln 2}{1280000000} t \\
t &=568,000,000
\end{aligned}\]

The rock is about 568 million years old.

 

5.5.1 Two primitive modeling concepts.

Primitive Concept 1. Suppose you have a barrel (which could just as well be a blood cell, stomach, liver, or lake or ocean or auditorium) and \(A(t)\) liters is the amount of water (glucose, plasma, people) in the barrel at time \(t\) minutes. If water is running into the barrel at a rate \(R_1\) liters/minute and leaking out of the barrel at a rate \(R_2\) liters/minute then

\[\begin{array}{cccc}
\begin{array}{c}
\text { Rate of change of water } \\
\text { in the barrel }
\end{array} & = &
\begin{array}{c}\text { Rate water enters } \\
\text { the barrel }
\end{array}
& - & 
\begin{array}{c}
\text { Rate water leaves } \\
\text { the barrel }
\end{array} \\
A ^{\prime} (t)& = & R_{1} & - & R_{2} \\
\frac{\mathrm{L}}{\mathrm{min}} && \frac{\mathrm{L}}{\mathrm{min}} && \frac{\mathrm{L}}{\mathrm{min}}
\end{array}\]

Primitive Concept 2. Similar to Primitive Concept 1 except that there is salt in the water. Suppose \(S(t)\) is the amount in grams of salt in the barrel and \(C_1\) is the concentration in grams/liter of salt in the stream entering the barrel and \(C_2\) is the concentration of salt in grams/liter in the stream leaving the barrel. Then

\[\begin{array}{cccc}
\begin{array}{c}
\text { Rate of change of salt } \\
\text { in the barrel }
\end{array} & = &
\begin{array}{c}\text { Rate salt enters } \\
\text { the barrel }
\end{array}
& - & 
\begin{array}{c}
\text { Rate salt leaves } \\
\text { the barrel }
\end{array} \\
S ^{\prime} (t)& = & C_{1} R_{1} & - & C_{2} R_{2} \\
\frac{\mathrm{g}}{\mathrm{min}} && \frac{\mathrm{g}}{\mathrm{L}} \frac{\mathrm{L}}{\mathrm{min}}&& \frac{\mathrm{g}}{\mathrm{L}}\frac{\mathrm{L}}{\mathrm{min}}
\end{array}\]

Observe that the units are g/m on both sides of the equation. Maintaining a balance in units often helps to find the correct equation.

Example 5.5.3 Suppose a runner is exhaling at the rate of 2 liters per second. Then the amount of air in her lungs is decreasing at the rate of two liters per second. If, furthermore, the CO₂ partial pressure in the exhaled air is 50 mm Hg (approx 0.114 g CO₂/liter of air at body temperature of 310 K)⁸ , then she is exhaling CO₂ at the rate of 0.114 g/liter \(\times\) 2 liters/sec = 0.228 g/sec.

Example 5.5.4 Classical Washout Curve. A barrel contains 100 liters of water and 300 grams of salt. You start a stream of pure water flowing into the barrel at 5 liters per minute, and a compensating stream of salt water flows from the barrel at 5 liters per minute. The solution in the barrel is ‘well stirred’ so that the salt concentration is uniform throughout the barrel at all times. Let \(S(t)\) be the amount of salt (grams) in the barrel \(t\) minutes after you start the flow of pure water into the barrel.

Explore 5.5.3 Draw a graph of what you think will be the graph of \(S(t)\). In doing so consider

• What is \(S(0)\)?
• Does \(S(t)\) increase or decrease?
• Will there be a time, \(t_{*}\), for which \(S(t_{*}) = 0\)? If so, what is \(t_{*}\)?

Solution. First let us analyze \(S\). We use Primitive Concept 2. The concentration of salt in the water flowing into the barrel is 0. The concentration of salt in the water flowing out of the barrel is the same as the concentration \(C(t)\) of salt in the barrel which is

\[C(t)=\frac{S(t)}{100} \mathrm{~g} / \mathrm{L}\]

Therefore

\[\begin{array}{cccc}
\begin{array}{c}
\text { Rate of change of salt } \\
\text { in the barrel }
\end{array} & = &
\begin{array}{c}\text { Rate salt enters } \\
\text { the barrel }
\end{array}
& - & 
\begin{array}{c}
\text { Rate salt leaves } \\
\text { the barrel }
\end{array} \\
S ^{\prime} (t)& = & C_{1} R_{1} & - & C_{2} R_{2} \\
S ^{\prime} (t) & = & 0 \times 5 & = & \frac{S(t)}{100} \frac{\mathrm{gr}}{\mathrm{L}} \times 5 \frac{\mathrm{L}}{\mathrm{min}}
\end{array}\]

Furthermore, \(S(0) = 300\). Thus

\[\begin{aligned}
&S(0)=300 \\
&S^{\prime}(t)=-0.05 S(t)
\end{aligned}\]

From the Exponential Growth and Decay property 5.5.2,

\[S(t)=300 e^{-0.05 t}\]

Explore 5.5.4 . Draw the graph of \(S(t)=300 e^{-0.05 t}\) and compare it with the graph you drew in Explore 5.5.3.

Example 5.5.5 Classical Saturation Curve.

Problem. Suppose a 100 liter barrel is full of pure water and at time \(t = 0\_ minutes a stream of water flowing at 5 liters per minute and carrying 3 g/liter of salt starts flowing into the barrel. Assume the salt is well mixed in the barrel and water overflows at the rate of 5 liters per minute. Let \(S(t)\) be the amount of salt in the barrel at time \(t\) minutes after the salt water starts flowing in.

Explore 5.5.5 Draw a graph of what you think will be the graph of \(S(t)\). In doing so consider

• What is \(S(0)\)?
• Does \(S(t)\) increase or decrease?
• Is there an upper bound on \(S(t)\), the amount of salt in the barrel that will be in the barrel?

Solution: We analyze \(S\); again we use Primitive Concept 2. The concentration of salt in the inflow is 3 g/liter. The concentration \(C(t)\) of salt in the tank at time \(t\) minutes is

\[C(t)=\frac{S(t)}{100}\]

The salt concentration in the outflow will also be \(C(t)\). Therefore

\[\begin{array}{cccc}
\begin{array}{c}
\text { Rate of change of salt } \\
\text { in the barrel }
\end{array} & = &
\begin{array}{c}\text { Rate salt enters } \\
\text { the barrel }
\end{array}
& - & 
\begin{array}{c}
\text { Rate salt leaves } \\
\text { the barrel }
\end{array} \\
S ^{\prime} (t)& = & C_{1} R_{1} & - & C_{2} R_{2} \\
S ^{\prime} (t) & = & 3 \times 5 & = & \frac{S(t)}{100} 5 \\
\frac{g}{\min } & = & \frac{\mathrm{g}}{\mathrm{L}} \frac{\mathrm{L}}{\mathrm{min}} & = & \frac{\mathrm{g}}{\mathrm{L}} \frac{\mathrm{L}}{\mathrm{min}}
\end{array}\]

Initially the barrel is full of pure water, so

\[S(0)=0\]

We now have

\[\begin{array}\
S(0) &=0 \\
S^{\prime}(t) &=15-0.05 S(t)
\end{array} \label{5.22}\]

This equation is not in the form of \(P^{\prime}(t)=k P(t)\) because of the 15. Proceed as follows.

Equilibrium. Ask, 'At what value, \(E\), of \(S(t)\) would \(S ^{\prime} (t) = 0\)?' That would require

\[0=15-0.05 E=0, \text { or } E=300 \mathrm{~g}\]

\(E = 300 \mathrm{~g}\) is the equilibrium level of salt in the barrel. We focus attention on the difference, \(D(t)\), between the equilibrium level and the current level of salt. Thus

\[D(t)=300-S(t) \quad \text { and } \quad S(t)=300-D(t)\]

Now,

\[D(0)=300-S(0)=300-0=300\]

Furthermore,

\[S^{\prime}(t)=[300-D(t)]^{\prime}=-D^{\prime}(t)\]

We substitute into Equations \ref{5.22}

\[\begin{array}{rlrl}
S(0) & =0 & D(0) & =300 \\
S^{\prime}(t) & =15-0.05 S(t) & -D^{\prime}(t) & =15-0.05(300-D(t))
\end{array}\]

The equations for \(D\) become

\[\begin{aligned}
D(0) &=300 \\
D^{\prime}(t) &=-0.05 D(t)
\end{aligned}\]

This is in the form of the Exponential Growth and Decay Property 5.5.2, and we write

\[D(t)=300 e^{-0.05 t}\]

Returning to \(S(t) = 300 - D(t)\) we write

\[S(t)=300-D(t)=300-300 e^{-0.05 t}\]

The graph of \(S(t)=300-300 e^{-0.05 t}\) is shown in Figure \(\PageIndex{1}\). Curiously, the graph of \(S(t)\) is also called an exponential decay curve. \(S(t)\) is not decaying at all; \(S(t)\) is increasing. What is decaying exponentially is \(D(t)\), the remaining salt capacity.

Explore 5.5.6 Show that if \(S(t)=300-300 e^{-0.05 t}\), then

\[S(0)=0, \quad \text { and } \quad S^{\prime}(t)=15-0.05 S(t)\]

Figure \(\PageIndex{1}\): The graph of \(S(t)=300-300 e^{-0.05 t}\) depicting the amount of salt in a barrel initially filled with 100 liters of pure water and receiving a flow of 5 L/m carrying 3 g/L. \(D(t) = 300 - S(t)\).

 

5.5.2 Continuous-space analysis of light depletion.

As observed in Chapter 1, light intensity decreases as one descends from the surface of a lake or ocean. There we divided the water into discrete layers and it was assumed that each layer absorbs a fixed fraction, \(F\), of the light that enters it from above. This hypothesis led to the difference equation

\[I_{d+1}-I_{d}=-F I_{d}\]

Light is actually absorbed continuously as it passes down through a (homogeneous) water medium, not in discrete layers. We examine the light intensity, \(I(x)\), at a distance, \(x\) meters, below the surface of a lake or ocean, assuming that the light intensity penetrating the surface is a known quantity, \(I_0\).

We start by testing an hypothesis about light transmission in water that appears different from the hypothesis we arrived at in Chapter 1:

Mathematical Model 5.5.4 Light Absorbance: The amount of light absorbed by a (horizontal) layer of water is proportional to the thickness of the layer and to the amount of light entering the layer (see Figure \(\PageIndex{2}\)).

Figure \(\PageIndex{2}\): Diagram of light depletion below the surface of a lake or ocean. \(I(x)\) is light intensity at depth \(x\) due to light of intensity \(I_0\) just below the surface of the water.

The mathematical model of light absorbance implies, for example, that

1. The light absorbed by a water layer of thickness 2\(\Delta\) is twice the light absorbed by a water layer of thickness \(\Delta\) and
2. A layer that absorbs 10% of a bright light will absorb 10% of a dim light.

We know from experimental evidence that implication (1) is approximately true for thin layers and for low levels of turbidity. Implication (2) is valid for a wide range of light intensities.

Double Proportionality. From the mathematical model of light absorbance the light absorbed by a layer is proportional to two things, the thickness of the layer and the intensity of the light entering the layer. We handle this double proportionality by assuming that the amount of light absorbed in a layer is proportional to the product of the thickness of the layer and the intensity of the light incident to the layer. That is, there is a number, \(K\), such that if \(I(x)\) is the light intensity at depth \(x\) and \(I(x + \Delta x)\) is the light intensity at depth \(x + \Delta x\), then

\[I(x+\Delta x)-I(x) \doteq-K \Delta x \times I(x) \label{5.23}\]

The product, \(K \Delta x \Delta I(x)\), has the advantage that

1. For fixed incident light intensity, \(I(x)\), the light absorbed, \(I(x + \Delta x) - I(x)\), is proportional to the thickness, \(\Delta x\), (proportionality constant = \(- K ~ I(x)\)) and
2. For fixed thickness \(\Delta x\), the light absorbed is proportional to the incident light, \(I(x)\) (proportionality constant = \(-K \Delta x\)).

Equation \ref{5.23} can be rearranged to

\[\frac{I(x+\Delta x)-I(x)}{\Delta x} \doteq -K I(x)\]

The approximation (\(\doteq\)) improves as the layer thickness, \(\Delta x\), approaches zero.

\[\text{As } \Delta x \rightarrow 0 \quad  \frac{I(x+\Delta x)-I(x)}{\Delta x} \rightarrow I^{\prime}(x)\]

and we conclude that

\[I^{\prime}(x)=-K I(x) \ref{5.24}\]

The Exponential Growth and Decay Property 5.5.2 implies that

\[\text { because } \quad I^{\prime}(x)=-K I(x), \quad I(x)=I_{0} e^{-K x} \ref{5.25}\]

Example 5.5.6 Assume that 1000 \(\mathrm{w/m} ^2\) of light is striking the surface of a lake and that 40% of that light is reflected back into the atmosphere. We first solve the initial value problem

\[\begin{aligned}
I(0) &=600 \\
I^{\prime}(x) &=-K I(x)
\end{aligned}\]

to get

\[I(x)=600 e^{-K x}\]

If we have additional information that, say, the light intensity at a depth of 10 meters is 400 \(\mathrm{W/m}^2\) we can find the value of \(K\). It must be that

\[I(10)=600 e^{-K \times 10}=400\]

The only unknown in the last equation is \(K\), and we solve

\[\begin{aligned}
600 e^{-K \times 10} &=400 \\
e^{-10 K} &=400 / 600=5 / 6 \\
\ln \left(e^{-10 K}\right) &=\ln (2 / 3) \\
-10 K &=\ln (2 / 3) \\
K & \doteq 0.040557
\end{aligned}\]

Thus we would say that

\[I(x)=600 e^{-0.040557 x}\]

If we know, for example, that 30 \(\mathrm{W/m}^2\) of light are required for a certain species of coral to grow, we can ask for the maximum depth, \(\bar{x}\), at which we might find that species. We would solve

\[\begin{aligned}
I(\bar{x}) &=30 \\
600 e^{-0.040557 \bar{x}} &=30 \\
\ln \left(e^{-0.040557 \bar{x}}\right) &=\ln (30 / 600) \\
-0.040557 \bar{x} &=\ln (1 / 20) \\
\bar{x} &=73.9 \text { meters }
\end{aligned}\]

 

5.5.3 Doubling time and half-life

Suppose \(k\) and \(C\) are positive numbers. The doubling time of \(F(t) = C~ e ^{k t}\) is a number \(t_{dbl}\) such that

\[\text { for any time } t \quad F\left(t+t_{d b l}\right)=2 \times F(t) \text {. }\]

That there is such a number follows from

\[\begin{array}
F\left(t+t_{d b l}\right) &=2 \times F(t) \\
C e^{k\left(t+t_{d b l}\right)} &=2 C e^{k t} \\
C e^{k t} e^{k t_{d b l}} &=2 C e^{k t} \\
e^{k t_{d b l}} &=2 \\
t_{d b l} &=\frac{\ln 2}{k}
\end{array} \label{5.26}\]

The half-life of \(F(t) = C ~ e^{-k t}\) is a number \(t_{\mathrm{half}}\) , usually written as \(t_{1/2}\), such that

\[\text{for any time } t \quad F\left(t+t_{1 / 2}\right)=\frac{1}{2} F(t)\]

Using steps similar to those for the doubling time you can find that

\[t_{1 / 2}=\frac{\ln 2}{k} \label{5.27}\]

Explore 5.5.7 Write the steps similar to those for the doubling time to show that \(t_{1 / 2}=(\ln 2) / k\).

In Figure \(\PageIndex{3}\)A is a graph of the bacterial density from Table 1.1 and of the equation

\[\text { Abs }=0.022 e^{0.0315 t}, \quad t_{d b l}=\frac{\ln 2}{0.0315}=22.0 \quad \text { minutes }\]

The bacterial density doubles every 22 minutes, as illustrated for the intervals [26,48] minutes and [48,70] minutes.

Figure \(\PageIndex{3}\): A. Bacterial population density and \(\mathrm{ABS}=0.022 e^{0.0315 t}\), which has doubling time of 22 minutes. B. Light depletion and \(I_{d}=0.4 e^{-0.196 d}\) which has a 'half life' of 3.5 meters.

In Figure \(\PageIndex{3}\)B is a graph of light intensity decay from Figure 1.3.5 (repeated in Figure \(\PageIndex{4}\)) and of the Equation 1.3.10

\[I_{d}=0.400 \times 0.82^{d}\]

Because \(0.82=e^{\ln 0.82}=e^{-0.198}\),

\[I_{d}=0.400 e^{-0.198 d} \quad \text { and } \quad d_{1 / 2}=\frac{\ln 2}{0.198}=3.5\]

Every 3.5 layers of muddy water the light intensity decays by one-half. \(d_{1/2}\) is a distance and might be called 'half depth' rather than 'half life.' 'Half life' is the term used for all exponential decay, however, and you are well advised to use it.

Example 5.5.7 Problem. Suppose a patient is prescribed to take 80 mg of Sotolol, a drug that regularizes heart beat, once per day. Sotolol has a half-life in the body of 12 hrs. Compute the daily fluctuations of sotolol.

Solution. Let \(s_{t}^{-}\) be the amount of sotolol in the body at time, \(t\), just before the sotolol pill is taken and \(s_{t}^{+}\) be the amount of sotolol in the body at time \(t\) just after the sotolol pill is taken. Then

\[\begin{gathered}
s_{0}^{-}=0, \quad s_{0}^{+}=80, \quad s_{1}^{-}=\frac{1}{4} s_{0}^{+}=\frac{1}{4} 80=20 \\
s_{t}^{+}=s_{t}^{-}+80 \quad s_{t+1}^{-}=\frac{1}{4} s_{t}^{+}= \frac{1}{4}\left(s_{t}^{-}+80\right)=\frac{1}{4} s_{t}^{-}+20
\end{gathered}\]

After some days, \(s_{t}^{-}\) will reach approximate equilibrium, \(E: s_{t}^{-} \doteq E\) and \(s_{t+1}^{-} \doteq E\).

\[\begin{gathered}
s_{0}^{-}=0, \quad s_{t+1}^{-}=\frac{1}{4} s_{t}^{-}+20, \quad E=\frac{1}{4} E+20, \quad E=26.67
\\
\text{Thus, approximately,} \quad s_{t}^{-}=E=26.67, \quad s_{t}^{+}=s_{t}^{-}+80=E+80=106.67.\\
\end{gathered}\]

so the system oscillates between 26.67 mg and 106.67 mg, a four to one ratio.

You are asked to compare this with taking 40 mg of Sotolol twice per day in Exercise 5.5.6

 

5.5.4 Semilogarithm and LogLog graphs.

Functions \(P(t)\) that satisfy an equation \(P ^{\prime} (t) = k  P(t)\) may be written \(P(t) = P_{0} e ^{k t}\) and will satisfy the relation \(\ln P(t)=\left(\ln P_{0}\right)+k t\). The graph of \(\ln P(t) ~ vs~  t\) is a straight line with intercept \(\ln P_{0}\) and slope \(k\). Similarly, if \(P^{\prime}(t)=-k P(t)\), in rectilinear coordinates, the graph of \(\ln P(t) ~ vs~  t\) is a straight line with slope \(-k\). A scientist with data \(t\), \(P(t)\) that she thinks is exponential may plot the graph of \(\ln P(t) ~ vs~  t\). If the graph is linear, then a fit of a line to that data will lead to an exponential relation of the form \(P(t)=A e^{k t}\) or \(P(t)=A e^{-k t}\). She may then search for a biological process that would justify a model \(P^{\prime}(t)=\pm k P(t)\).

Example 5.5.8 In Section 1.3 we showed the results of an experiment measuring the light decay as a function of depth. The data and a semilog graph of the data are shown in Figure \(\PageIndex{4}\).

Depth Layer	\(I_d ~ \mathrm{mW/cm^2}\)
0	0.400
1	0.330
2	0.270
3	0.216
4	0.170
5	0.140
6	0.124
7	0.098
8	0.082
9	0.065

Figure \(\PageIndex{4}\): Data and a semilog graph of the data showing experimental results of measuring light decrease with depth of water.

As shown in the figure,

\[\log _{10} I_{d}=-0.4-0.087 d\]

is a good approximation to the data. Therefore

\[\begin{aligned}
I_{d} & \doteq 10^{-0.4-0.087 d} \\
&=0.4 \times 0.82^{d}
\end{aligned}\]

which is the same result obtained in Section 1.3. As shown in the previous subsection, the relation

\[I^{\prime}(d)=-k I(d)\]

corresponds to a process underlying light depletion in water.

Explore 5.5.8 Do This. In Explore 1.6.1, you were given data for serum penicillin concentration during 30 minutes following a bolus penicillin injection and you may have found that \(P_{t}=200 \times 0.77^{T}\), where \(t\) denotes 5-minute intervals, \(P_{t}=200 \times 0.978^{t}\) where \(t\) denotes minutes, models the data very well. Additional data for the same experiment is shown in Figure \(\PageIndex{5}\).

1. On a copy of Figure \(\PageIndex{5}\) plot the graph of \(\log _{10} P_{t}=\log _{10}\left(200 \times 0.978^{t}\right)\) versus time, \(t\) in minutes, for \(0 \leq t \leq 60\).
2. Why is the previous graph a straight line?
3. The two graphs match well for only the first 30 minutes. It is worth a good bit of your time thinking about an alternate model of penicillin pharmacokinetics that would explain the difference after 30 minutes.

Time (min)	Mezlocillin \(\mu \mathrm{g/ml}\)	Time (min)	Mezlocillin  \(\mu \mathrm{g/ml}\)
0	200	60	21.3
5	151.6	90	12.5
10	118.0	120	8.2
15	93	180	3.7
20	74.4	240	1.7
30	49.8	300	0.8
45	30.7

Figure \(\PageIndex{5}\): Penicillin concentration following a bolus injection of 2 gm of mezlocillin.

5.5.5 Relative Growth Rates and Allometry.

If y is a positive function of time, the relative growth rate of y is

\[\frac{y^{\prime}(t)}{y(t)} \quad \text{Relative Growth Rate.} \label{5.28}\]

The relative growth rate of y is sometimes called the fractional growth rate or the logarithmic growth rate.

Definition 5.5.1 Allometry

Two functions, x and y, of time are said to be allometrically related if there are numbers C and a such that

\[y(t)=C(x(t))^{a} \label{5.29}

If x and y are allometric then

\[\log y=\log \left(C x^{a}\right)=\log C+a \log x , \label{5.30}\]

for any base of log. Therefore if \(\log {y}\) is plotted vs \(\log {x}\) the graph should be a straight line.

Explore 5.5.9 Show that if x and y satisfy Equation \ref{5.30}, \(\log y(t)=\log C+a \log x(t)\), then

\[\frac{y^{\prime}(t)}{y(t)}=a \frac{x^{\prime}(t)}{x(t)}\]

Conclude that if x and y are allometric then the relative growth rate of y is proportional to the relative growth rate of x.

Shown in Figure \(\PageIndex{6}\) is a graph of \(\log _{10}\) of the weight of large mouth bass vs \(\log _{10}\) of their length ⁹. The data appear linear and we conclude that the weight is allometric to the length. An equation of a line close to the data is

\[\frac{\log y-1.05}{\log x-2.0}=\frac{2.6-1.05}{2.5-2.0} , \quad \log y=-5.75+3.1 \log x\]

Then

\[y=10^{-5.75} x^{3.1}\]

The weights of the bass are approximately proportional to the cube of the lengths. This is consistent with the fact that the volume of a cube is equal to the cube of the length of an edge. Many interesting allometric relations are not supported by underlying models, however (Exercises 5.5.29 and 5.5.30).

Figure \(\PageIndex{6}\): Weight vs Length for large mouth bass plotted on a log-log graph.

 

Exercises for Section 5.5, The derivative equation \(P ^{\prime} (t) = k ~P(t)\)

Exercise 5.5.1 Write a solution for each of the following derivative equations. Sketch the graph of the solution. For each, find the doubling time, \(t_{dbl}\), or half life, \(t_{1/2}\), which ever is applicable.

1. \(P(0)=5  \quad P^{\prime}(t)=2 P(t)\)
2. \(P(0)=5 \quad  P^{\prime}(t)=  -2 P(t)\)
3. \(P(0)=2 \quad  P^{\prime}(t)=0.1 P(t)\)
4. \(P(0)=2 \quad  P^{\prime}(t)=  -0.1 P(t)\)
5. \(P(0)=10  \quad P^{\prime}(t)= P(t)\)
6. \(P(0)=10 \quad  P^{\prime}(t)=  -P(t)\)
7. \(P(0)=0 \quad  P^{\prime}(t)=0.01 P(t)\)
8. \(P(0)=0 \quad  P^{\prime}(t)=-0.01 P(t)\)

Exercise 5.5.2 Write a solution for each of the following derivative equations. Sketch the graph of the solution. For each, find the half life, \(t_{1/2}\), which is the time required to 'move half way toward equilibrium.'

Recall the solution in Example 5.5.5 to solve \(S^{\prime}(t)=15-0.05 S(t)\).

1. \(S(0)=0 \quad S^{\prime}(t)=10-2 S(t)\)
2. \(S(0)=2 \quad S^{\prime}(t)=10-2 S(t)\)
3. \(S(0)=5 \quad S^{\prime}(t)=10-2 S(t)\)
4. \(S(0)=10 \quad S^{\prime}(t)=10-2 S(t)\)
5. \(S(0)=0 \quad S^{\prime}(t)=20-S(t)\)
6. \(S(0)=10 \quad S^{\prime}(t)=20-S(t)\)
7. \(S(0)=20 \quad S^{\prime}(t)=20-S(t)\)
8. \(S(0)=30 \quad S^{\prime}(t)=20-S(t)\)

Exercise 5.5.3 Find values of \(C\) and \(k\) so that \(P(t) = C e^{k t}\) matches the data.

1. \(P(0)=5 \quad P(2)=10\)
2. \(P(0)=10 \quad P(2)=5\)
3. \(P(0)=2 \quad P(5)=10\)
4. \(P(0)=10 P(5)=10\)
5. \(P(0)=5 \quad P(2)=2\)
6. \(P(0)=8 \quad P(10)=6\)
7. \(P(1)=5 \quad P(2)=10\)
8. \(P(2)=10 \quad P(10)=20\)

Exercise 5.5.4 Do Explore 5.5.2. In particular, read the paragraph about the ambiguous use of the variable, \(t\).

Exercise 5.5.5 Suppose a barrel has 100 liters of water and 400 grams of salt and at time \(t = 0\) minutes a stream of water flowing at 5 liters per minute and carrying 3 g/liter of salt starts flowing into the barrel, the barrel is well mixed, and a stream of water and salt leaves the barrel at 5 liters per minute. What is the amount of salt in the barrel t minutes after the flow begins? Draw a candidate solution graph for this problem before computing the solution.

Exercise 5.5.6 In Example 5.5.7 it was shown that in a patient who takes 80 mg of sotolol once per day, the daily fluctuation of sotolol is from 26.7 mg to 106.7 mg. Sotolol has a half-life in the body of 12 hours. What is the fluctuation of sotolol in the body if the patient takes two 40 mg of sotolol at 12 hour intervals in the day? Would you recommend two 40 mg per day rather than one 80 mg pill per day?

Exercise 5.5.7 Suppose it is determined that 30 mg of sotolol is sufficient to control irregular heart beats. What size, X, of sotolol pill should a patient take twice per day to insure that the equilibrium value of sotolol immediately before taking each pill is at least 30 mg?

Exercise 5.5.8 A patient takes 10 mg of coumadin once per day to reduce the probability that he will experience blood clots. The half-life of coumadin in the body is 40 hours. What level, \(H\), of coumadin will be accumulated from previous ingestion of pills and what will be the daily fluctuation of coumadin in the body.

Exercise 5.5.9 Plot semilog graphs of the data sets in Table Ex. 5.5.9 and decide which ones appear to be approximately exponential. For those that appear to be exponential, find numbers, \(C\) and \(k\), so that

\[\P(t)=C e^{k t}]

approximates the the data.

Table for Exercise 5.5.9 Data sets for Exercise 5.5.9

a.	b.	c.	d.	e.
t P(t) 0 2.0 1 2.1 2 2.8 3 4.7 4 8.4 5 14.5	t P(t) 0 6.00 2 4.35 4 2.97 6 2.17 8 1.65 10 1.25	t P(t) 0 2.00 5 2.83 10 4.00 15 5.67 20 8.00 25 11.31	t P(t) 0 2.00 3 1.00 6 0.67 9 0.50 12 0.40 15 0.33	t P(t) 0 2.00 3 1.00 6 0.50 9 0.25 12 0.13 15 0.06

Exercise 5.5.10 Shown in Table Ex. 5.5.10 data from V. natriegens growth reported in Chapter 1 on page 4. Find numbers, \(C\) and \(k\), so that

\[P(t)=C e^{k t}\]

approximates the the data. Use your values of \(C\) and \(k\) and compute \(P(0), P(16), P(32), P(48)\), and \(P(64)\) and compare them with the observed values in the Table Ex. 5.5.10.

Table for Exercise 5.5.10 Cell density of V. natriegens measured as light absorbance at 16-minute time increments.

Time (min)	Population Density
0	0.022
16	0.036
32	0.060
48	0.101
64	0.169

Exercise 5.5.11 In Section 1.3 we found from the discrete model of light extinction,

\[I_{d+1}=I_{d}-0.18 I_{d}, \quad \text { that the solution } \quad I_{d+1}=0.82 I_{d}\]

matched the data for light depletion in Figure 1.3.6.

Layer	0	1	2	3	4	5	6	7	8	9
Light Intensity	0.400	0.330	0.270	0.216	0.170	0.140	0.124	0.098	0.082	0.065

Light decrease in water is continuous, however. Find a value of \(k\) for which the solution to the continuous model, \(I^{\prime}(x)=-k I(x)\), matches the data.

Exercise 5.5.12 In Section 1.6 you may have found that the solution, \(y = 200 \times 0.77^{t/5}\), to the difference equation \(P_{t+1}-P_{t}=-0.23 P_{t}\), approximates the data for penicillin concentration from Figure 1.6.1.

Time min	0	5	10	15	20
Penicillin Concentration	200	152	118	83	74

Penicillin clearance is continuous, however. Find a value of \(k\) for which the solution to \(P^{\prime}(t)=-k P(t)\) matches the data.

Exercise 5.5.13 David Ho and colleagues ¹⁰ published the first study of HIV-1 dynamics within patients following treatment with an inhibitor of HIV-1 protease, ABT-538 which stops infected cells from producing new viral particles. Shown in Exercise Figure 5.5.13A is a graph of plasma viral load before and after ABT-treatment was begun on day 1 for patient number 409 and in Exercise Figure 5.5.13B is a semi-log graph of CD4 cell count following treatment.

1. By what percent is viral load diminished from day-1 to day-12?
2. The line in Figure 5.5.13A has an equation, \(y = 5.9 - 0.19 x.\) Remember that \(y = \log _{10} V\) and x is days. Find \(V_0\) and m so that the graph of \(V(t)=V_{0} e^{-m t}\) in semilog coordinates is the line drawn in Exercise Figure 5.5.13A.
3. What is the half-life of the viral load?
4. From the previous step, \(V^{\prime}(t)=-0.43 V(t)\). Suppose ABT-538 totally eliminates viral production during days 1 to 12. At what rate is the immune system of patient 409 eliminating virus before treatment.
5. Assume that CD4 cell counts increase linearly and the equation in Figure 5.5.13B is \(y = 9.2 + 15.8 x\). At what rate are CD4 cells being produced? Remember that y is CD4 count per \(\mathrm{mm}^{3}\) and there are about \(6 \times 10^{6} \mathrm{mm}^3\) of blood in the human body.

After about 35 days, the HIV virus mutates into a form resistant to ABT-538 and pre-treatment viral loads soon return. Treatment with a protease inhibitor together with drugs that inhibit the translation of HIV RNA to DNA can decrease viral loads to levels below detection for the duration of treatment.

Figure for Exercise 5.5.13 A. Count of HIV viral load during administration of ABT-538. B. Count of helper t-cell during the same period. Note that if \(\log _{10}\) (RNA copies per ml) = 4, for example, then RNA copies per ml = \(10^4\) . See Exercise 5.5.13 Figures adapted by permission at no cost from Macmillan Publishing Group, Ltd. David D. Ho, Avidan U Neumann, Alan S. Perelson, Wen Chen, John M. Leonard & Martin Markowitz, Nature 373 (1995) 123:127 Copyright 1995 http://www.nature.com.

Exercise 5.5.14 You inject two grams of penicillin into the 6 liter vascular pool of a patient. Plasma circulates through the kidney at the rate of 1.2 liters/minute and the kidneys remove 20 per cent of the penicillin that passes through.

1. Draw a schematic diagram showing the vascular pool and kidneys as separate entities, an artery leading from the vascular pool to the kidney and a vein leading from the kidney back to the vascular pool.
2. Let \(P(t)\) be the amount of penicillin in the vascular pool t minutes after injection of penicillin. What is \(P(0)\)?
3. Use Primitive Concept 2 to write an equation for \(P^{\prime}\).
4. Write a solution to your equation.

Exercise 5.5.15 Suppose a rock sample is found to have 8.02 \(\mu \mathrm{g}\) of ⁴⁰K and 7.56 µg of ⁴⁰Ar. What is the age of the rock?

Exercise 5.5.16 Suppose a rock sample is found to have 6.11 mg of ⁴⁰K and 0.05 mg of ⁴⁰Ar. What is the age of the rock?

Exercise 5.5.17 Rubidium-87 decomposes to strontium-87 with a half-life of \(50 \times 10^{9}\) years. Fortunately, rubidium and potassium occur in the same rock types and in the same minerals, usually in the ratio of 1 ⁸⁷Rb atom to approximately 600 ⁴⁰K atoms. Age determined by rubidium-87 to strontium-87 decomposition is an excellent check of ⁴⁰K to ⁴⁰Ar ages. However, ⁸⁷Sr may be lost from the rock or may be present but not derived from ⁸⁷Rb so the ⁸⁷Rb to ⁸⁷Sr age may not be as accurate as the ⁴⁰K to ⁴⁰Ar age.

1. Suppose a rock sample has \(2.5 \times 10^{11}\) atoms of ⁸⁷Rb and \(1.5 \times 10^{10}\) atoms of ⁸⁷Sr. What is the age of the rock?
2. Suppose a rock sample has 6.4 \(\mu \mathrm{g}\) of ⁸⁷Rb and 0.01 \(\mu \mathrm{g}\) of ⁸⁷Sr. What is the age of the rock?

Exercise 5.5.18 A major advancement in archaeology was the development of carbon-14 dating in the 1950’s by an American chemist Willard Libby, for which he received the 1960 Nobel Prize in Chemistry. Carbon-14 develops in the upper atmosphere as neutrons bombard nitrogen, and subsequently combines with oxygen to form carbon dioxide. About 1 in \(10^{12}\) CO₂ atoms is formed with ¹⁴C in today’s atmosphere. Plants metabolize ¹⁴CO₂ (almost) as readily as ¹²CO₂, and resulting sugars are metabolized equally by animals that eat the plants. Consequently carbon from living material is 1 part in \(10^{12}\) carbon-14. Upon death, no additional ¹⁴C is absorbed into the material and ¹⁴C gradually decomposes into nitrogen. Slightly confounding the use of radio carbon dating is the fact that the fraction of atmospheric ¹⁴CO₂ has not been historically constant at 1 molecule per \(10^{12}\) molecules of ¹²CO₂.

Carbon-14 decomposes to nitrogen according to

\[{ }_{6}^{14} \mathrm{C} \longrightarrow{ }_{7}^{14} \mathrm{~N}+\beta^{-}+\bar{\nu}+\text { energy } \label{5.31}\]

where \(\beta^{-}\) denotes an electron and \(\bar{\nu}\) denotes an antineutrino. One of the neutrons of \({ }_{6}^{14} \mathrm{C}\) looses an electron and becomes a proton.

Mathematical Model 5.5.5 Carbon-14 decay. The rate of decomposition of \({ }_{6}^{14} \mathrm{C}\) in a sample is proportional to the size of the sample. One-half of the atoms in a sample will decompose in 5730 years.

1. Write and solve a derivative equation that will show for a sample of ¹⁴C initially of size \(C_0\) what the size will be \(t\) years later.
2. In tissue living today, the amount of ¹⁴C in one gram of carbon is approximately \(10^{-12}\) grams. Assume for this problem that the same ratio in living material has persisted for the last 10,000 years. Also assume that upon death the only change in carbon of any form is the decrease in ¹⁴C due to decomposition to nitrogen. Suppose a 100 gram sample of carbon from bone is found to have \(3 \times 10^{-11}\) grams of ¹⁴C. What is the age of the sample?
3. Suppose that during the time 10,000 years ago until 2,000 years ago the amount of ¹⁴C in one gram of carbon in living tissue was approximately \(1.05 \times 10^{-12}\) grams. Suppose a 100 gram sample of carbon from bone is found to have \(3 \times 10^{-11}\) grams of ¹⁴C. What is the age of the sample?

Exercise 5.5.19 Suppose solar radiation striking the ocean surface is 1250 W/m2 and 20 percent of that energy is reflected by the surface of the ocean. Suppose also that 20 meters below the surface the light intensity is found to be \(\mathrm{W/m}^2\).

1. Write an equation descriptive of the light intensity as a function of depth in the ocean.
2. Suppose a coral species requires 100 \(\mathrm{W/m}^2\) light intensity to grow. What is the maximum depth at which that species might be found?

Exercise 5.5.20 In two bodies of water, \(L_1\) and \(L_2\), the light intensities \(I_{1}(x)\) and \(I_{2}(x)\) as functions of depth \(x\) are measured simultaneously and found to be

\[\I_{1}(x)=800 e^{-0.04 x} \quad \text { and } \quad I_{2}(x)=700 e^{-0.05 x}]

Explain the differences in the two formulas in terms of the properties of water in the two bodies.

Exercise 5.5.21 A spectrophotometer is used to measure bacterial cell density in a growth medium. Light is passed through a sample of the medium and the amount of light that is absorbed by the medium is an indicator of cell density. As cell density increases the amount of light absorbed increases. A standard is established by passing a light beam of intensity \(I_0\) through a 0.5 cm layer of the growth medium without bacteria and measuring the intensity \(I_{St}\) of the beam emerging from the medium. See Figure Ex. 5.5.21.

Figure for Exercise 5.5.21 Diagram of spectrophotometer. A light beam of intensity \(I_0\) enters the standard solution and the intensity \(I_{St}\) of the emerging beam is measured. A light beam of the same intensity \(I_0\) enters the sample solution and the intensity \(I_{Sm}\) of the emerging beam is measured. See Exercise 5.5.21.

A light beam of the same intensity \(I_0\) enters the sample solution and the intensity \(I_{Sm}\) of the emerging beam is measured.

In the mathematical model of light absorbance (the amount of light absorbed by a layer of water is proportional to the thickness of the layer and to the amount of light entering the layer), the proportionality constant \(K\) is a measure of the opacity of the water. Recall that the solution Equation \ref{5.25} is \(I(x)=I_{0}, e^{-K x}\).

The bacteria in the sample placed in the spectrophotometer increase the turbidity and therefore the opacity of the solution. Explain why cell density is proportional to

\[\ln \left(\frac{I_{S m}}{I_{S t}}\right)\]

The number \(\ln \left(I_{S m} / I_{S t}\right)\) is called absorbance.

Exercise 5.5.22 A patient comes into your emergency room and you start a penicillin infusion of 0.2 gms/min into the 6 liter vascular pool. Plasma circulates through the kidney at the rate of 1.2 liters/minute and the kidneys remove 20 per cent of the penicillin that passes through.

1. Draw a schematic diagram showing the vascular pool and kidneys as separate entities, an artery leading from the vascular pool to the kidney and a vein leading from the kidney back to the vascular pool.
2. Let \(P(t)\)be the amount of penicillin in the vascular pool \(t\) minutes after injection of penicillin. What is \(P(0)\)?
3. Use Primitive Concept 2 to write an equation for \(P^{\prime}\).
4. Compute the solution to your equation and draw the graph of \(P\).
5. The saturation level of penicillin in this problem is critically important to the correct treatment of your patient. Will it be high enough to control the infection you wish to control? If not, what should you do?
6. Suppose your patient has impaired kidney function and that plasma circulates through the kidney at the rate of 0.8 liters per minute and the kidneys remove 15 percent of the penicillin that passes through. What is the saturation level of penicillin in this patient, assuming you administer penicillin the same as before?

Exercise 5.5.23 An egg is covered by a hen and is at \(37^{\circ}C\). The hen leaves the nest and the egg is exposed to \(17^{\circ}C\) air.

1. Draw a graph representative of the temperature of the egg \(t\) minutes after the hen leaves the nest.

Mathematical Model 5.5.6 Egg cooling. During any short time interval while the egg is uncovered, the change in egg temperature is proportional to the length of the time interval and proportional to the difference between the egg temperature and the air temperature.

2. Let \(T(t)\) denote the egg temperature t minutes after the hen leaves the nest. Consider a short time interval, \([t, t + \Delta t]\), and write an equation for the change in temperature of the egg during the time interval \([t, t + \Delta t]\).
3. Argue that as ∆t approaches zero, the terms of your previous equation get close to the terms of \[T^{\prime}(t)=-k(T(t)-17) \label{5.32}\]
4. Assume \(T(0) = 37\) and find an equation for \(T(t)\).
5. Suppose it is known that eight minutes after the hen leaves the nest the egg temperature is \(35 ^{\circ}C\). What is \(k\)?
6. Based on that value of \(k\), if the coldest temperature the embryo can tolerate is \(32 ^{\circ}C\), when must the hen return to the nest?

Note: Equation \ref{5.32} is referred to as Newton’s Law of Cooling.

Exercise 5.5.24 Consider the following osmosis experiment in biology laboratory.

Material: A thistle tube, a 1 liter flask, some salt water, and some pure water, a membrane that is impermeable to the salt and is permeable to the water.

The bulb of the thistle tube is filled with salt water, the membrane is place across the open part of the bulb, and the bulb is inverted in a beaker of pure water so that the top of the pure water is at the juncture of the bulb with the stem. See the diagram.

Because of osmotic pressure the pure water will cross the membrane pushing water up the stem of the thistle tube until the increase in pressure inside the bulb due to the water in the stem matches the osmotic pressure across the membrane.

Our problem is to describe the height of the water in the stem as a function of time.

Mathematical Model 5.5.7 Osmotic diffusion across a membrane. The rate at which pure water crosses the membrane is proportional to the osmotic pressure across the membrane minus the pressure due to the water in the stem.

Assume that the volume of the bulb is much larger than the volume of the stem so that the concentration of salt in the thistle tube may be assumed to be constant.

Introduce notation and write a derivative equation with initial condition that will describe the height of the water in the stem as a function of time. Solve your derivative equation.

Exercise 5.5.25 2 kilos of a fish poison, rotenone, are mixed into a lake which has a volume of \(100 \times 20 \times 2 = 4000\) cubic meters. A stream of clean water flows into the lake at a rate of 1000 cubic meters per day. Assume that it mixes immediately throughout the whole lake. Another stream flows out of the lake at a rate of 1000 cubic meters per day.

What is the amount (\(p_t\) for discrete time or \(P(t)\) for continuous time) of poison in the lake at time \(t\) days after the poison is applied?

1. Treat the problem as a discrete time problem with one-day time intervals. Solve the difference equation \[p_{0}=2 \quad p_{t+1}-p_{t}=-\frac{1000}{4000} p_{t}\]
2. Let \(t\) denote continuous time and \(P(t)\) the amount of poison in the lake at time \(t\). Let \([t, t + \Delta t]\) denote a short time interval (measured in units of days). An equation for the mathematical model is \[P(t+\Delta t)-P(t)=-\frac{P(t)}{4000} \Delta t 1000\] 
3. Argue that \[P(0)=2, \quad P^{\prime}(t)=-0.25 P(t)\]
4. Compute the solution to this equation.
5. Compare the solution to the discrete time problem, \(p_t\) , with the solution to the continuous time problem, \(P(t)\).
6. For what value of \(k\) will the solution, \(Q(t)\), to \[Q(0)=2, \quad Q^{\prime}(t)=k Q(t) \quad \text { satisfy } \quad Q(t)=p_{t}, \quad \text { for } \quad t=0,1,2, \cdots ?\]
7. Which of \(P(t)\) and \(Q(t)\) most accurately estimates rotenone levels?
8. On what day, \(\bar{t}\) will \(P(\bar{t}) = 4g\)?

Exercise 5.5.26 Continuous version of Chapter Exercise 1.11.7. Atmospheric pressure decreases with increasing altitude. Derive a dynamic equation from the following mathematical model, solve the dynamic equation, and use the data to evaluate the parameters of the solution equation.

Mathematical Model 5.5.8 Mathematical Model of Atmospheric Pressure. Consider a vertical column of air based at sea level and assume that the temperature within the column is constant, equal to \(20 ^{\circ} C\). The pressure at any height is the weight of air in the column above that height divided by the cross sectional area of the column. In a ’short’ section of the column, by the ideal gas law the the mass of air within the section is proportional to the product of the volume of the section and the pressure within the section (which may be considered constant and equal to the pressure at the bottom of the section). The weight of the air above the lower height is the weight of air in the section plus the weight of air above the upper height.

Sea-level atmospheric pressure is 760 mm Hg and the pressure at 18,000 feet is one-half that at sea level.

Exercise 5.5.27 When you open a bottle containing a carbonated soft drink, CO2 dissolved in the liquid turns to gas and escapes from the liquid. If left open and undisturbed, the drink goes flat (looses its CO2). Write a mathematical model descriptive of release of carbon dioxide in a carbonated soft drink. From your model, write a derivative equation descriptive of the carbon dioxide content in the liquid minutes after opening the drink.

Exercise 5.5.28 Decompression illness in deep water divers. In the 1800’s technology was developed to supply compressed air to under water divers engaged in construction of bridge supports and underwater tunnels. While at depth those divers worked without unusual physical discomfort. Shortly after ascent to the surface, however, they might experience aching joints, numbness in the limbs, fainting, and possible death. Affected divers tended to walk bent over and were said to have the “bends”.

It was believed that nitrogen absorbed by the tissue at the high pressure below water was expanding during ascent to the surface and causing the difficulty, and that a diver who ascended slowly would be at less risk. The British Navy commissioned physician and mathematician J. S. Haldane¹¹ to devise a dive protocol to be followed by Navy divers to reduce the risk of decompression illness. Nitrogen flows quickly between the lungs and the plasma but nitrogen exchange between the plasma and other parts of the body (nerve, brain tissue, muscle, fat, joints, liver, bone marrow, for example) is slower and not uniform. Haldane used a simple model for nitrogen exchange between the plasma and the other parts of the body.

Mathematical Model 5.5.9 Nitrogen absorption and release in tissue. The rate at which nitrogen is absorbed by a tissue is proportional to the difference in the partial pressure of nitrogen in the plasma and the partial pressure of nitrogen in the tissue.

Air is 79 percent nitrogen. Assume that the partial pressure of nitrogen in the lungs and the plasma are equal at any depth. At depth \(d\),

\[\text { Plasma pp } \mathrm{N}_{2}=\text { Lung pp } \mathrm{N}_{2}=0.79\left(1+\frac{d}{10}\right) \quad \text { atmospheres. }\]

1. What is the partial pressure of nitrogen in a diver’s lungs at the surface?
2. Suppose a diver has not dived for a week. What would you expect to be the partial pressure of nitrogen in her tissue?
3. A diver who has not dived for a week quickly descends to 30 meters. What is the nitrogen partial pressure in her lungs after descending to 30 meters?
4. Let \(N(t)\) be the partial pressure of nitrogen in a tissue of volume, \(V \), \(t\) minutes into the dive. Use the Mathematical Model 5.5.9 Nitrogen absorption and release in tissue and Primitive Concept 2, to write an equation for \(N ^{\prime}\).
5. Check to see whether (\(k\) is a proportionality constant) \[N(t)=0.79\left(1+\frac{d}{10}\right)-0.79 \frac{d}{10} e^{-\frac{k}{V} t} \label{5.33}\) solves your equation from the previous step.
6. Assume \frac{k}{V} in Equation \ref{5.33} is 0.0693 and \(d = 30\). What is \(N(30)\)?
7. Haldane experimented on goats and concluded that on the ascent to the surface, \(N(t)\) should never exceed two times Lung pp N₂. A diver who had been at depth 30 meters for 30 minutes could ascend to what level and not violate this condition if \(\frac{k}{V} = 0.0693\)?

Haldane supposed that there were five tissues in the body for which \(\frac{k}{V} = 0.139, 0.0693, 0.0347, 0.0173, 0.00924\), respectively, and advised that on ascent to the surface, \(N(t)\) should never exceed two times pp N₂ in any one of these tissues.

Exercise 5.5.29 E. O. Wilson, a pioneer in study of area-species relations on islands, states in Diversity of Life, p 221, :

"In more exact language, the number of species increases by the area-species equation, \(S = C ~ A^{z}\), where \(A\) is the area and \(S\) is the number of species. \(C\) is a constant and \(z\) is a second, biologically interesting constant that depends on the group of organisms (birds, reptiles, grasses). The value of \(z\) also depends on whether the archipelago is close to source ares, as in the case of the Indonesian islands, or very remote, as with Hawaii \(\cdots\) It ranges among faunas and floras around the world from about 0.15 to 0.35."

Discuss this statement as a potential Mathematical Model.

Exercise 5.5.30 The graph of Figure 5.5.30 showing the number of amphibian and reptile species on Caribbean Islands vs the areas of the islands is a classic example from P. J. Darlington, Zoogeography: The Geographical Distribution of Animals, Wiley, 1957, page 483, Tables 15 and 16.

1. Treat Trinidad as an unexplained outlier (meaning: ignore Trinidad) and find a power law, \(S = C ~ A^z \), relating number of species to area for this data.
2. Darlington observes that "\(\cdots\) within the size range of these islands \(\cdots\), division of the area by ten divides the amphibian and reptile fauna by two \(\cdots\), but this ratio is a very rough approximation, and it might not hold in other situations." Is your power law consistent with Darlington’s observation?
3. Why might Trinidad (4800 km²) have nearly twice as many reptilian species (80) as Puerto Rico (8700 km²) which has 41 species?

Figure for Exercise 5.5.30 The number of amphibian and reptile species on islands in the Caribbean vs the areas of the islands. The data is from Tables 15 and 16 in P. J. Darlington, Zoogeography: The Geographical Distribution of Animals, Wiley, 1957, page 483.

Exercise 5.5.31 The graph in Figure Ex. 5.5.31 relates surface area to mass of a number of mammals. Assume mammal population densities are constant (each two mammal populations are equally dense), so that the graph also relates surface area to volume.

1. Find an equation relating the surface area, S, of a cube to the volume, V , of the cube.
2. Find an equation relating the surface area, \(S = 4 \pi r^{2} \), of a sphere of radius \(r\) to the volume, \(V = \frac{4}{3} \pi r^{3} \), of the sphere.
3. Find an equation relating the surface area of a mammal to the mass of the mammal, using the graph in Figure Ex. 5.5.31. Ignore the dark dots; they are for beech trees.
4. In what way are the results for the first three parts of this exercise similar?

Figure for Exercise 5.5.31 Graph for Exercise 5.5.31 relating surface area to mass of mammals. From A. M. Hemmingson, Energy metabolism as related to body size, and its evolution, Rep. Steno Mem. Hosp. (Copenhagen) 9:1-110. With permission from Dr. Peter R. Rossing, Director of Research, Steno Diabetes Center S/A. All rights reserved.

Exercise 5.5.32 Body Mass Indices. The Body Mass Index,

\[\mathrm{BMI}=\frac{\text { Mass }}{\text { Height }^{2}}\]

was introduced by Adolphe Quetelet, a French mathematician and statistician in 1869. The Center for Disease Control and Prevention (CDC) notes that BMI is a helpful indicator of overweight and obesity in adults.

From simple allometric considerations, \(\text { BMI3 }=\text { Mass } / \text { Height }^{3}\) should be approximately a constant, \(C\). If

\[B M I 3=\text { Mass } / \text { Height }^{3}=C \text { then } B M I=\text { Mass }^{\text {Height }}{ }^{2}=C \text { Height. }\]

so that BMI should increase with height. CDC also states that "\(\cdots\) women are more likely to have a higher percentage of body fat than men for the same BMI." If a male and a female both have BMI = 23 and are of average height for their sex (1.77 meters for males and 1.63 meters for females), then

\[\text { BMI3 for the male }=\frac{23}{1.77}=13.0 \quad \text { and } \quad \text { BMI3 for the female }=\frac{23}{1.63}=14.1\]

Thus BMI3 is larger for the female than for the male and may indicate a larger percentage of body fat for the female.

Shown are four Age, and 50th percentile Weight, Height data points for boys and for girls. Compute BMI and BMI3 for the four points and plot the sixteen points on a graph. Which of the two indices, BMI or BMI3, remains relatively constant with age? Data are from the Centers for Disease Control and Prevention, http://www.cdc.gov/growthcharts/data...l/cj41c021.pdf and \(\cdots\) cj41c022.pdf.

Age (Boys)	8	12	16	20
Weight (kg) (50 percentile)	26	41	62	71
Height (cm) (50 percentile)	128	149	174	177
BMI kg/m^2
BMI3 kg/m^3
Age (Girls)	8	12	16	20
Weight (kg) (50 percentile)	26	41	54	58
Height (cm) (50 percentile)	128	151	162	163
BMI kg/m^2
BMI3 kg/m^3

We suggest that BMI3 might be more useful than BMI as an index of body fat. Other indices of body fat that have been suggested include \(M / H, M^{1 / 3} / H, H / M^{1 / 3},\) and \(c M^{1.2} / H^{3.3}\). The interested reader should visit the web site cdc.gov/nccdphp/dnpa/bmi and read the references there.

 

 

 

⁵ Charles Darwin wrote in the Origin of Species that Earth was several hundred million years old, but he was opposed in 1863 by a dominant physical scientist, William Thompson (later to become Lord Kelvin) who estimated that Earth was between 24 and 400 million years old. His estimate was based on his calculation of the time it would take for Earth to cool from molten rock to today’s temperatures in the upper layers of the Earth. See article by Philip England, Peter Molnar, and Frank Richter, GSA Today, 17, 1 (January 1, 2007).

⁶ Wikipedia

⁷ Because ⁴⁰Ar is a gas at the temperatures that the rock was formed, no ⁴⁰Ar was originally in the rock.

⁸ \[\frac{\frac{50}{760} \mathrm{~A} \times 1 \text { liter } \times 44 \mathrm{~mol} \mathrm{~wt}}{0.08206 \text { gas const } \times 310 \mathrm{~K}}=0.114 \mathrm{~g}\]

⁹ Data from Robert Summerfelt, Iowa State University.

¹⁰ David D. Ho, et al, Nature 373 (1995) 123:127.

¹¹ J. S. Haldane was the father of J. B. S. Haldane who, along with R. A. Fisher and Sewall Wright developed the field of population genetics.