Investigating Exponential Models

Problem 1. Given $k \neq 0$ and any real constant $y_0$ , solve the initial value problem

$\displaystyle \frac{\mathrm dy}{\mathrm dx} = ky,\quad y(0) = y_0.$

This is known as the exponential model. If $k > 0$ , this problem describes an exponential growth model. If $k < 0$ , this problem describes an exponential decay model.

(Click for Solution)

Solution. By the method of separable variables,

$\begin{aligned}\int \frac 1y\, \mathrm dy &= \int k\, \mathrm dx \\ \ln|y| &= kx + C\\ |y| &= e^{kx+C}\\ y &= \pm e^C \cdot e^{kx} \\ y &= Ae^{kx},\end{aligned}$

where $A = \pm e^C$ is an arbitrary constant. Plugging in the initial value $y(0) = y_0$ ,

$y_0 = Ae^{k \cdot 0} \quad \Rightarrow \quad A = y_0 \quad \Rightarrow \quad y = y_0 e^{kx}.$

Problem 2. Let $T \equiv T(t)$ denote the temperature of an object at time $t$ and $T_{\text{env}}$ denote the (constant) temperature of the environment. Denote $\Delta T := T - T_{\text{env}}$ and assume $\Delta T > 0$ .

Newton’s law of cooling states that $T$ decreases at a rate proportional to $\Delta T$ with proportionality constant $\lambda > 0$ . Given that $T_0$ is the initial temperature of $T$ , find an expression for $T$ in terms of $T_0$ , $T_{\text{env}}$ , $\lambda$ , $t$ .

(Click for Solution)

Solution. By Newton’s law of cooling,

$\displaystyle \begin{aligned} \frac{\mathrm dT}{\mathrm dt} &= -\lambda \Delta T. \end{aligned}$

On the left-hand side,

$\displaystyle \begin{aligned} \frac{\mathrm dT}{\mathrm dt} &= \frac{\mathrm d}{\mathrm dt}(T_{\text{env}} + \Delta T) \\ &= 0 + \frac{\mathrm d}{\mathrm dt}(\Delta T) \\ &= \frac{\mathrm d(\Delta T)}{\mathrm dt}.\end{aligned}$

Hence,

$\displaystyle \frac{\mathrm d(\Delta T)}{\mathrm dt} = -\lambda \Delta T.$

By Problem 1, $(\Delta T)(t) = (\Delta T)(0) e^{-\lambda t}$ . By the definition of $\Delta T$ ,

$(\Delta T)(0) = (T - T_{\text{env}})(0) = T_0 - T_{\text{env}}.$

Therefore, $T = T_{\text{env}} + \Delta T = T_{\text{env}} + (T_0 - T_{\text{env}}) e^{-\lambda t}$ . Furthermore, as $t \to \infty$ , since $e^{-\lambda t} \to 0$ , we see that the long run temperature tends to $T_{\text{env}}$ since taking $t \to \infty$ yields

$T \to T_{\text{env}} + (T_0 - T_{\text{env}}) \cdot 0 = T_{\text{env}}.$

Problem 3. A radioactive substance decays with a decay rate of $\lambda > 0$ . Let $x \equiv x(t)$ denote the amount of a radioactive substance at time $t$ . The radioactive decay model states that $x$ decreases at a rate proportional to $x$ with proportionality constant $\lambda$ .

Compute the half-life of the substance (i.e. the time taken for the substance to decay to half of its quantity) in terms of $\lambda$ .

(Click for Solution)

Solution. By the radioactive decay model,

$\displaystyle \frac{\mathrm d x}{\mathrm dt} = -\lambda x.$

By Problem 1, letting $x_0 := x(0)$ , $x = x_0 e^{-\lambda t}$ . Fix $t \geq 0$ and denote the half-life by $t_{1/2}$ . Then

$\displaystyle \frac x2 = x_0 e^{-\lambda(t + t_{1/2})} = x_0 e^{-\lambda t} e^{-\lambda t_{1/2}} = x e^{-\lambda t_{1/2}}.$

Dividing by $x > 0$ , and applying logarithmic properties (exercise),

$\displaystyle e^{-\lambda t_{1/2}} = \frac 12 \quad \Rightarrow \quad t_{1/2} = \frac{\ln 2}{\lambda}.$

This proof also highlights that $t_{1/2}$ is constant no matter how much material $x$ of the radioactive substance remains; after $(\ln 2)/{\lambda}$ units of time, $x$ decreases by $50\%$ .

Problem 4. During my tutorial class with $27$ students, I assigned students to solve Problem 3. Once a student has solved the problem, I task said student to help other students solve the problem.

Let $x \equiv x(t)$ denote the number of students who successfully solved Problem 3 at time $t$ minutes. It is given that $x$ follows an exponential growth model with proportionality constant $\lambda$ , called the growth rate. Assume $x$ can take non-integer values for simplicity (i.e. so that $0 < x(0) < 1$ ).

Given that the first person solved the problem at $5$ minutes and the time taken for all students to complete solving Problem 3 is $25$ minutes, estimate the value of $\lambda$ . Hence, if the class instead has $N$ students, estimate the time taken for the whole class to solve Problem 3, giving your answer in terms of $N$ .

(Click for Solution)

Solution. By Problem 1, $x = x_0 e^{\lambda t}$ , where $x(0) = x_0$ .

Since $x(5) = 1$ and $x(25) = 27$ ,

$1 = x_0 e^{5 \lambda},\quad 27 = x_0 e^{25 \lambda}.$

Dividing the second term by the first,

$\displaystyle 27 = \frac{27}{1} = \frac{e^{25 \lambda}}{e^{5 \lambda}} = e^{20\lambda}.$

Hence, $\lambda = \frac 1{20} \ln 27 \approx 0.165$ .

For a class of $N$ students, we need to find $t$ such that

$\displaystyle N = x_0 e^{\lambda t} \quad \Rightarrow \quad t = \frac 1{\lambda} \ln \frac N{x_0}.$

Since $x(5) = 1$ ,

$1 = x_0 e^{5 \lambda}\quad \Rightarrow \quad x_0 = e^{-5\lambda}.$

Hence,

$\displaystyle t = \frac 1{\lambda} \ln \frac N{e^{-5\lambda}} = 5 + \frac 1{\lambda} \ln N = 5 + \frac{20}{\ln 27} \ln N.$

Problem 5. Generalise the result in Problem 4.

(Click for Solution)

Solution. Use the exponential model $x = x_0 e^{\lambda t}$ . For a given class with $M$ students, suppose that the first student solves at time $t_1$ and the whole class solves at time $t_2$ . Then

$1 = x_0 e^{\lambda t_1},\quad M = x_0 e^{\lambda t_2}.$

Dividing by one another and performing algebruh,

$\displaystyle \lambda = \frac{\ln (M)}{t_2 - t_1} .$

Finally, consider any class with $N$ students with the same growth rate $\lambda$ , and suppose the first student solves Problem 3 in time $t_3$ . This means

$\displaystyle 1 = x_0 e^{\lambda t_3} \quad \Rightarrow \quad x = e^{\lambda (t-t_3)} \quad \Rightarrow \quad t - t_3 = \frac {\ln(x)}{\lambda}.$

Thus, the class of $x = N$ students will take a total time of $t$ given by

$\displaystyle \frac{t - t_3}{t_2 - t_1} = \frac{\ln(N)}{\ln(M)} \quad \Rightarrow \quad t = t_3 + \frac{\ln (N)}{\ln (M)} \cdot (t_2 - t_1).$

Setting $t_1 = t_3 = 5$ , $t_2 = 25$ , and $M = 27$ , we obtain the result in Problem 4:

$\displaystyle t = 5 + \frac{\ln(N)}{\ln(27)} \cdot (25 - 5) = 5 + \frac{20}{\ln 27} \ln N.$

—Joel Kindiak, 25 Apr 25, 1256H

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Investigating Exponential Models

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