Chapter 3 · Section 3.3

3.3 Rate Problems

A surprising number of engineering models start from one sentence: the rate of change of a quantity equals the rate at which it comes in minus the rate at which it goes out. Write that sentence in symbols and you have a first-order differential equation. In this section we apply it to radioactive decay and population growth, to a hot part cooling in air, to salt in a stirred tank, to a population that runs out of room, and to a simple chemical reaction.

The mathematics is always a separable or a linear equation from Chapter 2. What changes from problem to problem is the modelling: deciding what the rates are, and using measured data to find the constants.

Method in a nutshell

Name the unknown $x (t)$ and its units, and the time unit.
Write the balance $\frac{d x}{d t} = (rate in) - (rate out)$ , with units checked term by term.
Solve the equation (separable or linear) with the initial condition.
Use extra data (a second measurement) to find any unknown constant.
Answer the question and check that the result makes physical sense, especially as $t \to \infty$ .

Exponential growth and decay

Many quantities change at a rate proportional to the amount present. In a radioactive sample, each nucleus has the same small chance of decaying per second, so the number of decays per second is proportional to the number of nuclei. In a bacterial culture with plenty of food, each cell divides at about the same rate, so the growth rate is proportional to the number of cells. Both lead to

\frac{d x}{d t} = k x, x (0) = x_{0},

with $k > 0$ for growth and $k < 0$ for decay. Separating variables ( $d x / x = k d t$ ) and using the initial condition gives

x (t) = x_{0} e^{k t} .

The constant $k$ is a rate per unit time: $k = - 0.1 h^{- 1}$ means that about $10%$ of the current amount is lost per hour, at that instant. A more intuitive way to describe it is the time needed to halve or double.

Definition 3.3.1 (Half-life and doubling time)

For decay, $x = x_{0} e^{- λ t}$ with $λ > 0$ , the half-life $T_{1/2}$ is the time after which half of any amount remains:

e^{- λ T_{1/2}} = \frac{1}{2} ⟺ T_{1/2} = \frac{ln 2}{λ} .

For growth, $x = x_{0} e^{k t}$ with $k > 0$ , the doubling time is $T_{2} = \frac{ln 2}{k}$ .

The half-life does not depend on $x_{0}$ or on when you start counting. After one half-life $\frac{1}{2}$ remains, after two $\frac{1}{4}$ , after $n$ half-lives $(\frac{1}{2})^{n}$ . So a convenient form of the solution is

x (t) = x_{0} (\frac{1}{2})^{t / T_{1/2}} .

Figure 3.3.1. Decay of

40

mg of an isotope with half-life

6

h (Example 3.3.1). Every

6

hours the amount halves, whatever the starting point.

Example 3.3.1

Technetium-99m, widely used in medical imaging, has a half-life of about $6.0$ h. A hospital receives $40$ mg. Find the decay constant, the amount left after one day, and the time when only $1$ mg remains.

Solution

Decay constant. $λ = \frac{ln 2}{6.0} \approx 0.1155 h^{- 1}$ .
Solution. $x (t) = 40 (\frac{1}{2})^{t /6}$ mg, $t$ in hours.
After 24 h. $24$ h is four half-lives: $x (24) = 40 \cdot \frac{1}{16} = 2.5$ mg.
When is $x = 1$ ? $(\frac{1}{2})^{t /6} = \frac{1}{40}$ . Take logarithms: $\frac{t}{6} ln 2 = ln 40 ⟹ t = \frac{6 ln 40}{ln 2} \approx 31.9 h .$

The short half-life is the point of this isotope: it lasts long enough for imaging and then decays away within a couple of days.

Often $k$ is not given. Instead we have two measurements, and we must find both $x_{0}$ and $k$ from them.

Example 3.3.2

A bacterial culture grows at a rate proportional to its size. Two hours after the start of an experiment it contains $2000$ cells, and six hours after the start $5000$ cells. How many cells were there at the start? What is the doubling time? When will the culture reach $20000$ cells?

Solution

Model. $x (t) = x_{0} e^{k t}$ , $t$ in hours.
Divide the two data. $\frac{x ( 6 )}{x ( 2 )} = \frac{x _{0} e ^{6 k}}{x _{0} e ^{2 k}} = e^{4 k} = \frac{5000}{2000} = 2.5$ . Dividing eliminates $x_{0}$ . So $k = \frac{ln 2.5}{4} \approx 0.2291 h^{- 1} .$
Initial size. $x_{0} = 2000 e^{- 2 k} = \frac{2000}{2.5} \approx 1265$ cells.
Doubling time. $T_{2} = \frac{ln 2}{k} = \frac{4 ln 2}{ln 2.5} \approx 3.03$ h.
20 000 cells. Start from the measurement at $t = 6$ : $5000 e^{k (t - 6)} = 20000$ , so $t - 6 = \frac{ln 4}{k}$ and $t \approx 6 + 6.05 = 12.05$ h.

In the calculator below, enter the amount at $t = 0$ and one later measurement. It finds $k$ , the half-life or doubling time, and answers a query (an amount at a given time, or the time for a given amount). The preset is Example 3.3.2 with the clock started at the first measurement.

Limits of the model

Radioactive decay follows the exponential law extremely well. Population growth follows it only while nothing limits it. A single bacterium that doubled every 3 hours, without limit, would produce more mass than the whole Earth in less than three weeks. Real populations slow down; the logistic model later in this section captures that.

Newton's law of cooling

Put a hot metal part on a bench. It loses heat to the surrounding air, quickly at first and then more and more slowly. Newton's law of cooling states that the rate of change of the body's temperature is proportional to the difference between its temperature $T$ and the ambient (surrounding) temperature $T_{m}$ :

\frac{d T}{d t} = - k (T - T_{m}), k > 0.

The minus sign makes physical sense: if $T > T_{m}$ , the temperature decreases; if $T < T_{m}$ , it increases (the same law describes warming).

The law holds well when heat leaves mainly by convection, the temperature inside the body is nearly uniform, and the temperature difference is moderate. For very hot bodies, radiation (which grows like the fourth power of absolute temperature) becomes important, and the law is only a first approximation.

Solving the equation

New unknown. Let $u = T - T_{m}$ (the "excess temperature"). Since $T_{m}$ is constant, $u^{'} = T^{'}$ , and the equation becomes $u^{'} = - k u$ .
Exponential law. $u = u_{0} e^{- k t}$ , where $u_{0} = T_{0} - T_{m}$ .
Back to $T$ . $T (t) = T_{m} + (T_{0} - T_{m}) e^{- k t} .$

Every body approaches the ambient temperature, and the excess temperature decays exponentially. The constant $k$ depends on the body (its surface area, mass, heat capacity) and on the air flow. It is almost always found from a measurement.

Example 3.3.3

An aluminium casting leaves the mould at $18 0^{\circ}$ C and is left to cool in a workshop at $2 0^{\circ}$ C. After $10$ min its temperature is $12 0^{\circ}$ C. Find its temperature after half an hour, and the time when it is cool enough to handle ( $4 0^{\circ}$ C).

Solution

Model. $T = 20 + 160 e^{- k t}$ , $t$ in minutes.
Find $k$ . $T (10) = 120$ gives $160 e^{- 10 k} = 100$ , so $e^{- 10 k} = \frac{5}{8}$ and $k = \frac{ln 1.6}{10} \approx 0.0470 min^{- 1} .$
After 30 min. No need to compute $k$ numerically: $e^{- 30 k} = (e^{- 10 k})^{3} = (\frac{5}{8})^{3}$ , so $T (30) = 20 + 160 (\frac{5}{8})^{3} = 20 + 39.06 \approx 59. 1^{\circ} C .$
Safe to handle. $20 + 160 e^{- k t} = 40$ gives $e^{- k t} = \frac{1}{8}$ , so $t = \frac{ln 8}{k} = \frac{10 ln 8}{ln 1.6} \approx 44.2 min .$

Figure 3.3.2. Newton's law with

T_{m} = 2 0^{\circ}

C and the

k

of Example 3.3.3. Hot bodies cool and cold bodies warm, always towards

T_{m}

, and the excess temperature halves every

(ln 2) / k \approx 14.7

min.

Example 3.3.4

A temperature sensor stored at $5^{\circ}$ C is brought into a laboratory at $2 5^{\circ}$ C. After $3$ min it reads $1 5^{\circ}$ C. Assuming Newton's law, how long until it reads within $1^{\circ}$ C of the room temperature?

Solution

Model. $T = 25 + (5 - 25) e^{- k t} = 25 - 20 e^{- k t}$ .
Find $k$ . $25 - 20 e^{- 3 k} = 15$ gives $e^{- 3 k} = \frac{1}{2}$ : the gap halves every $3$ minutes.
Within 1 °C. We need $20 e^{- k t} = 1$ , i.e. the gap must shrink by a factor $20$ : $(\frac{1}{2})^{t /3} = \frac{1}{20} ⟹ t = \frac{3 ln 20}{ln 2} \approx 13.0 min .$

Engineers call $1/ k$ the time constant of the sensor (here $1/ k = 3/ ln 2 \approx 4.3$ min). A sensor needs several time constants to "catch up" with a sudden change. This lag must be taken into account in any temperature control loop.

Try your own data in the cooling calculator. Give the ambient temperature, the initial temperature and one later reading; you can ask for the temperature at a given time or the time at which a given temperature is reached.

Mixing problems

A tank holds a solution of salt (or a pollutant, a dye, a chemical) in water. A solution of known concentration flows in, the tank is kept well stirred, and the mixture flows out (Figure 3.3.3). How much salt is in the tank at time $t$ ?

Figure 3.3.3. A well-stirred tank. Liquid enters at

r_{in}

L/min with concentration

c_{in}

kg/L and leaves at

r_{out}

L/min with the tank's own concentration

x / V

Let $x (t)$ be the amount of salt (kg) and $V (t)$ the volume of liquid (L). Salt enters with the inflow and leaves with the outflow:

rate in $= r_{in} (L/min) \times c_{in} (kg/L) = r_{in} c_{in}$ kg/min;
rate out $= r_{out} (L/min) \times \frac{x}{V} (kg/L)$ , because "well stirred" means the outflow has the same concentration as the whole tank.

The volume changes at the net flow rate, so $V (t) = V_{0} + (r_{in} - r_{out}) t$ . The balance is

\frac{d x}{d t} = r_{in} c_{in} - \frac{r _{out}}{V _{0} + ( r _{in} - r _{out} ) t} x .

This is a linear first-order equation in $x$ . When $r_{in} = r_{out}$ , the volume is constant and the equation is also separable.

Common mistake

The outflow carries the concentration of the tank, $x / V$ , not the inflow concentration $c_{in}$ . And when the flow rates differ, $V$ is a function of $t$ ; writing the initial volume $V_{0}$ in the denominator is wrong.

Example 3.3.5

A $500$ L tank contains brine with $20$ kg of dissolved salt. Brine containing $0.1$ kg of salt per litre flows in at $10$ L/min, and the well-stirred mixture flows out at the same rate. Find the amount of salt after one hour, the long-term amount, and the time when there are $40$ kg of salt in the tank.

Solution

Rates. In: $10 \times 0.1 = 1$ kg/min. The volume stays $500$ L, so out: $10 \cdot \frac{x}{500} = \frac{x}{50}$ kg/min.
Equation. $x^{'} = 1 - \frac{x}{50}$ , $x (0) = 20$ .
Solve. Write it as $x^{'} = - \frac{1}{50} (x - 50)$ . This has the same form as Newton's law of cooling with "ambient value" $50$ , so $x = 50 + (20 - 50) e^{- t /50} = 50 - 30 e^{- t /50} .$
After 60 min. $x (60) = 50 - 30 e^{- 1.2} \approx 50 - 9.04 = 40.96$ kg.
Long term. $x \to 50$ kg. This is $c_{in} V = 0.1 \times 500$ : in the end the tank has the inflow concentration.
40 kg. $30 e^{- t /50} = 10$ , so $t = 50 ln 3 \approx 54.9$ min.

Figure 3.3.4. Salt in the tank of Example 3.3.5 (solid) and for two other initial amounts (dashed). With equal flow rates every solution approaches

c_{in} V = 50

kg with time constant

V / r = 50

min.

Example 3.3.6

A tank with a capacity of $300$ L initially holds $100$ L of water in which $5$ kg of salt is dissolved. Brine with $0.2$ kg/L flows in at $6$ L/min, and the stirred mixture flows out at $4$ L/min. How much salt is in the tank at the moment it starts to overflow, and what is the concentration then?

Solution

Volume. The tank gains $6 - 4 = 2$ L/min, so $V (t) = 100 + 2 t$ . It is full when $100 + 2 t = 300$ , i.e. at $t = 100$ min.
Rates. In: $6 \times 0.2 = 1.2$ kg/min. Out: $4 \cdot \frac{x}{100 + 2 t} = \frac{2 x}{50 + t}$ kg/min.
Equation. $x^{'} + \frac{2}{50 + t} x = 1.2$ , $x (0) = 5$ . This is linear.
Integrating factor. $μ = e^{\int 2/ (50 + t) d t} = e^{2 l n (50 + t)} = (50 + t)^{2}$ .
Integrate. $((50 + t)^{2} x)^{'} = 1.2 (50 + t)^{2} ⟹ (50 + t)^{2} x = 0.4 (50 + t)^{3} + C .$
Initial condition. At $t = 0$ : $2500 \cdot 5 = 0.4 \cdot 125000 + C$ , so $C = 12500 - 50000 = - 37500$ .
Solution. $x (t) = 0.4 (50 + t) - \frac{37 500}{( 50 + t ) ^{2}} .$
At overflow ( $t = 100$ ). $x = 0.4 \cdot 150 - \frac{37 500}{22 500} = 60 - 1.67 \approx 58.33$ kg, and the concentration is $\frac{58.33}{300} \approx 0.194$ kg/L.

The concentration is close to, but still below, the inflow value $0.2$ kg/L, as expected. After $t = 100$ min the model no longer applies: the outflow rate changes when the tank overflows.

Use the mixing simulator to explore. With $r_{in} = r_{out}$ the volume is constant; otherwise it shows $V (t)$ and the linear equation with its integrating factor.

Logistic population growth

Exponential growth cannot last. Food, space or nutrients run short, and the growth rate per individual falls as the population grows. The simplest model of this lets the per-capita growth rate $P^{'} / P$ decrease linearly, from a value $r$ when the population is tiny to zero when it reaches a level $K$ :

\frac{1}{P} \frac{d P}{d t} = r (1 - \frac{P}{K}) ⟺ \frac{d P}{d t} = r P (1 - \frac{P}{K}) .

This is the logistic equation. The constant $r > 0$ is the intrinsic growth rate and $K > 0$ is the carrying capacity, the largest population the environment can sustain. The same equation describes the spread of a new technology through a market, or of an infection through a closed group.

Solving the logistic equation

Before solving, look at the right side. It is zero at $P = 0$ and at $P = K$ , so these constants are equilibrium solutions. For $0 < P < K$ the right side is positive (growth); for $P > K$ it is negative (decline). Now let us solve with $P (0) = P_{0} > 0$ , $P_{0} \neq = K$ .

Rewrite and separate. $\frac{d P}{d t} = \frac{r}{K} P (K - P)$ , so $\frac{K d P}{P ( K - P )} = r d t .$
Partial fractions. $\frac{K}{P ( K - P )} = \frac{1}{P} + \frac{1}{K - P} .$
Integrate. $ln ∣ P ∣ - ln ∣ K - P ∣ = r t + C ⟹ \frac{P}{K - P} = A e^{r t},$ where $A = \frac{P _{0}}{K - P _{0}}$ from the initial condition (this keeps the sign right in both cases $P_{0} < K$ and $P_{0} > K$ ).
Solve for $P$ . $P = A e^{r t} (K - P)$ , so $P (1 + A e^{r t}) = A K e^{r t}$ and $P = \frac{A K e ^{r t}}{1 + A e ^{r t}} = \frac{K}{1 + A ^{- 1} e ^{- r t}} .$

Theorem 3.3.1 (Solution of the logistic equation)

For $r > 0$ , $K > 0$ and $P_{0} > 0$ , the solution of $P^{'} = r P (1 - P / K)$ , $P (0) = P_{0}$ , is

P (t) = \frac{K}{1 + ( \frac{K - P _{0}}{P _{0}} ) e ^{- r t}}, t \geq 0.

It satisfies $P (t) \to K$ as $t \to \infty$ . If $0 < P_{0} < K /2$ , the graph is S-shaped, with an inflection point where $P = K /2$ ; there the growth rate takes its largest value, $rK /4$ .

Where does the inflection come from? Differentiate the equation once more with the chain rule: $P^{''} = r P^{'} (1 - \frac{2 P}{K})$ . While $0 < P < K$ we have $P^{'} > 0$ , so $P^{''}$ changes sign exactly where $P = K /2$ . Growth speeds up below half the capacity and slows down above it.

Figure 3.3.5. Logistic curves with

r = 0.5

per year and

K = 8000

. Starting below

K /2

, the curve is S-shaped with its steepest point at

P = K /2

. Starting above

K

, the population decreases to

K

Example 3.3.7

A lake is stocked with $500$ fish. Ecologists estimate that the lake can support $8000$ fish and that the intrinsic growth rate is $r = 0.5$ per year. Using the logistic model, find $P (t)$ , the population after $10$ years, the time when the population reaches $4000$ and when it reaches $7000$ , and the largest yearly growth rate.

Solution

Solution. $\frac{K - P _{0}}{P _{0}} = \frac{7500}{500} = 15$ , so $P (t) = \frac{8000}{1 + 15 e ^{- 0.5 t}} .$
After 10 years. $P (10) = \frac{8000}{1 + 15 e ^{- 5}} \approx \frac{8000}{1.1011} \approx 7266$ fish.
4000 fish. $1 + 15 e^{- 0.5 t} = 2$ , so $e^{- 0.5 t} = \frac{1}{15}$ and $t = 2 ln 15 \approx 5.42$ years.
7000 fish. $1 + 15 e^{- 0.5 t} = \frac{8}{7}$ , so $e^{- 0.5 t} = \frac{1}{105}$ and $t = 2 ln 105 \approx 9.31$ years.
Largest growth rate. At $P = K /2 = 4000$ : $P^{'} = rK /4 = 0.5 \cdot 8000/4 = 1000$ fish per year.

Fisheries use exactly this fact: harvesting is most productive when the stock is kept near $K /2$ , where the population replaces itself fastest.

The logistic simulator draws $P (t)$ for your own $r$ , $K$ and $P_{0}$ and marks the inflection point.

Chemical reaction rates

In chemical kinetics, the rate of a reaction is the rate at which reactants are used up (or products form). Two simple cases lead to equations we can solve.

First-order reactions. If a substance A converts to products at a rate proportional to its own concentration, $[A]^{'} = - k [A]$ . This is the decay law again: $[A] = [A]_{0} e^{- k t}$ , with a half-life $ln 2/ k$ that does not depend on the starting concentration. The decomposition of many compounds, and the decay of nuclei, behave this way.

Second-order reactions. In a reaction $A + B \to C$ , one molecule of A meets one molecule of B. The rate is then proportional to the product of the two concentrations. Let the initial concentrations be $a$ and $b$ (mol/L), and let $x (t)$ be the concentration of C formed. Each mole of C uses one mole of A and one of B, so the concentrations left are $a - x$ and $b - x$ , and

\frac{d x}{d t} = k (a - x) (b - x), x (0) = 0.

The equation is separable. For $a \neq = b$ , the partial-fraction identity

\frac{1}{( a - x ) ( b - x )} = \frac{1}{a - b} (\frac{1}{b - x} - \frac{1}{a - x})

gives, after integration,

\frac{1}{a - b} ln \frac{a - x}{b - x} = k t + C .

As $t \to \infty$ , $x$ approaches the smaller of $a$ and $b$ : the reaction stops when the limiting reactant is used up.

Example 3.3.8

In the reaction $A + B \to C$ , the initial concentrations are $[A]_{0} = 0.5$ mol/L and $[B]_{0} = 0.3$ mol/L. After $10$ min, $0.1$ mol/L of C has formed. Find the rate constant, $x (t)$ , the concentration of C after $30$ min, and its final value.

Solution

Equation. $x^{'} = k (0.5 - x) (0.3 - x)$ , $x (0) = 0$ . Here $a - b = 0.2$ , so $\frac{1}{a - b} = 5$ .
Integrate. $5 ln \frac{0.5 - x}{0.3 - x} = k t + C$ . At $t = 0$ : $C = 5 ln \frac{5}{3}$ . Subtract and divide by 5: $ln (\frac{3}{5} \cdot \frac{0.5 - x}{0.3 - x}) = 0.2 k t .$
Rate constant. At $t = 10$ , $x = 0.1$ : $\frac{3}{5} \cdot \frac{0.4}{0.2} = 1.2$ , so $ln 1.2 = 2 k$ and $k = \frac{ln 1.2}{2} \approx 0.0912 L/(mol \cdotp min) .$
Solve for $x$ . Write $E = e^{0.2 k t} = 1. 2^{t /10}$ . Then $0.5 - x = \frac{5}{3} E (0.3 - x)$ , which rearranges to $x (t) = \frac{1.5 ( E - 1 )}{5 E - 3} .$
After 30 min. $E = 1. 2^{3} = 1.728$ , so $x (30) = \frac{1.5 \times 0.728}{8.64 - 3} \approx 0.194$ mol/L.
Final value. As $t \to \infty$ , $E \to \infty$ and $x \to \frac{1.5}{5} = 0.3$ mol/L: all of B, the limiting reactant, is used up.

Exercises

Carbon-14 has a half-life of about $5730$ years. A wooden beam from an archaeological site contains $30%$ of the carbon-14 found in living wood. Estimate its age.
Answer
$t = 5730 \frac{ln ( 1/0.3 )}{ln 2} \approx 9950$ years.
The concentration of a drug in the blood decays exponentially. It is $16$ mg/L two hours after an injection and $9$ mg/L six hours after it. Find the initial concentration and the half-life.
Answer
$c (0) = 16 \cdot \frac{4}{3} \approx 21.3$ mg/L; $T_{1/2} = \frac{4 ln 2}{ln ( 16/9 )} \approx 4.82$ h.
A sample of transformer oil cools from $9 0^{\circ}$ C to $7 0^{\circ}$ C in $15$ min in a room at $2 2^{\circ}$ C. Find its temperature after $45$ min and the time when it reaches $3 0^{\circ}$ C.
Answer
$T = 22 + 68 (\frac{48}{68})^{t /15}$ ; $T (45) \approx 45. 9^{\circ}$ C; $t = 15 \frac{ln ( 68/8 )}{ln ( 68/48 )} \approx 92.2$ min.
A thermometer at $2 0^{\circ}$ C is placed in a bath at $8 0^{\circ}$ C and reads $5 0^{\circ}$ C after $6$ s. When does it read $7 9^{\circ}$ C?
Answer
$T = 80 - 60 (\frac{1}{2})^{t /6}$ ; $t = 6 lo g_{2} 60 \approx 35.4$ s.
A body cooling in a room of unknown temperature reads $70$ , $50$ and $3 8^{\circ}$ C at $t = 0$ , $10$ and $20$ min. Assuming Newton's law, find the room temperature and $k$ . (Hint: with $T_{j} = T_{m} + A q^{j}$ , show $T_{m} = \frac{T _{0} T _{2} - T _{1}^{2}}{T _{0} + T _{2} - 2 T _{1}}$ .)
Answer
$T_{m} = \frac{2660 - 2500}{8} = 2 0^{\circ}$ C; $q = e^{- 10 k} = 0.6$ , $k = \frac{1}{10} ln \frac{5}{3} \approx 0.0511 min^{- 1}$ .
A $1000$ L tank is full of pure water. Brine with $0.05$ kg/L enters at $20$ L/min and the mixture leaves at the same rate. Find the salt after $30$ min and in the long run.
Answer
$x = 50 (1 - e^{- t /50})$ ; $x (30) \approx 22.6$ kg; $x \to 50$ kg.
A $400$ L tank holds water with $60$ kg of dissolved salt. Pure water flows in at $8$ L/min and the mixture flows out at $8$ L/min. When will only $6$ kg of salt remain?
Answer
$x = 60 e^{- t /50}$ ; $t = 50 ln 10 \approx 115$ min.
A $200$ L tank holds water with $30$ kg of salt. Pure water flows in at $3$ L/min and the mixture is drained at $5$ L/min. Find $x (t)$ , and the amount and concentration of salt after $60$ min.
Answer
$V = 200 - 2 t$ ; $x = 30 (1 - \frac{t}{100})^{5/2}$ for $0 \leq t < 100$ ; $x (60) \approx 3.04$ kg, concentration $\approx 0.038$ kg/L.
A population obeys the logistic law with $r = 0.3$ per year, $K = 1200$ and $P (0) = 100$ . Find $P (t)$ and the time when $P = 900$ .
Answer
$P = \frac{1200}{1 + 11 e ^{- 0.3 t}}$ ; $t = \frac{ln 33}{0.3} \approx 11.7$ years.
An invasive water plant covers $200$ m $^{2}$ of a $10000$ m $^{2}$ lake. Five weeks later it covers $1000$ m $^{2}$ . Assuming logistic growth with $K = 10000$ m $^{2}$ , find $r$ and the time when half the lake is covered.
Answer
$A = \frac{10 000}{1 + 49 e ^{- r t}}$ with $r = \frac{1}{5} ln \frac{49}{9} \approx 0.339$ per week; half covered at $t = ln 49/ r \approx 11.5$ weeks.
In $A + B \to C$ with $[A]_{0} = [B]_{0} = 0.2$ mol/L and $k = 0.5$ L/(mol·min), the product concentration satisfies $x^{'} = k (0.2 - x)^{2}$ . Find $x (t)$ and the time when half of A has reacted.
Answer
$\frac{1}{0.2 - x} = 0.5 t + 5$ , so $x = 0.2 - \frac{1}{0.5 t + 5}$ ; half reacted ( $x = 0.1$ ) at $t = 10$ min.
In a first-order reaction, $25%$ of the reactant is converted in $8$ min. When will $90%$ be converted?
Answer
$e^{- 8 k} = 0.75$ ; $t = 8 \frac{ln 10}{ln ( 4/3 )} \approx 64.0$ min.