Professor Kate Poirier | OL67 | Fall 2020

# Category: Project #1(Page 1 of 6)

## General form of a Differential Equation Involving Growth and Decay

$Q’=\frac{dQ}{dt}$

Growth and decay problems are commonly generalized under the exponential model,

$Q’=aQ$

$a$ would be the constant of proportionality.

Upon quick inspection, we can treat this model as a separable equation. Thus, the solution for this differential equation will be:

$Q=ce^{at}$

For IVPs, the solution would be,

$Q=Q_0e^{a(t-t_0)}$

where $Q_0$ is the value/function of $Q$ at a given time, and $t_0$ is a given value of time. We use $(t-t_0)$ because we solve for the value of $Q(t)$ at a given time period.

#### Quick Review

With,

$Q’=aQ$

We separate the variable $Q$ into one side of the equation,

$\frac{Q’}{Q}=a$

Then we integrate the entire equation with respect to $t$

$\int\frac{Q’}{Q} dt=\int (a) dt$

Integrating gives us:

$\ln(Q)=at+c$

Now we must solve for $Q$. To solve for $Q$, we use our natural logarithm and exponent rules,

$e^{\ln(Q)}=e^{at+c}$

Thus,

$Q=ce^{at}$

## Applications

What is radioactive decay? Radioactive decay is a natural phenomenon of certain materials “losing” ( i.e. decaying) energy and matter over time due to their unstable atomic nucleus. Recall that atoms are made of particles called protons, neutrons, and electrons; under radioactive decay, these three particles are ejected out of the atomic nucleus, thus, a radioactive material will lose mass over time.

The general exponential decay function is defined as:

$Q(t) = Q_0e^{-k(t-t_0)}$

$Q_0$  is the initial quantity, $-k$ is the is treated as the â€śdecay constantâ€ť, $t_0$ is the initial time (essentially zero in most cases), and $t$ would be any time duration.

Since $\tau$, or the â€śhalf-life,â€ť is the amount of time at which a radioactive materialâ€™s quantity is reduced to half, we can turn $Q(t)$ into,

$Q(\tau) = \frac{Q_0}{2} = Q_0e^{-k(\tau-t_0)}$

$\frac{Q_0}{2} = Q_0e^{-k\tau}$

where $t_0 = 0$ and $t = \tau$.

We then solve for $k$ by canceling like terms and taking the natural logarithm of the equation:

$\ln(\frac{1}{2}) = \ln(e^{-k\tau})$

((Recall that $\ln(\frac{a}{b}) = \ln(a)-\ln(b)$ and $\ln(1) = 0$))

$-\ln(2) = -k\tau$

$k = \frac{ln(2)}{\tau}$

If a given problem indicates that $Q_0$ lost a certain percentage, $P$, of its mass at a certain time, then we can set $Q(t) = PQ_0$.

For an IVP where $Q’=-kQ$ and $Q(t_0)=Q_0$, the general solution is the same as the exponential decay function:

$Q(t) = Q_0e^{-k(t-t_0)}$

### 2.Carbon Dating

Carbon dating is a process based on the concept of radioactive decay. The unstable isotope, Carbon-14, decays into the stable isotope, Carbon-12. By determining the ratio of Carbon-14 and Carbon-12 in deceased organisms, scientists can determine the age of an organism– higher levels of Carbon-14 in a sample means that an organism died at a more recent time period.

Carbon dating uses the same function for radioactive decay problems, $Q(t) = Q_0e^{-k(t-t_0)}$. Most problems, however, have us solve for $t$. Also, unless otherwise stated, $t_0=0$.

#### Step-by-step solution for ‘t’

(See “Radioactive Decay” section for finding $k$)

From $Q(t) = Q_0e^{-k(t-t_0)}$, we have,

$\frac{Q(t)}{Q_0}=e^{-k(t)}$

Take the natural logarithm of the equation to have,

$\ln(\frac{Q(t)}{Q_0})=\ln(e^{-k(t)})$

$\ln(\frac{Q(t)}{Q_0})=-k(t)$

Finally, solve for $t$,

$t=\frac{-\ln(\frac{Q}{Q_0})}{k}$

### 3.Continuously Compounding Interest

The concept of compounding interest applies to savings accounts, loans, credit cards, or most financial services that involves “interest.” Essentially, a certain amount grows in value at a specified percentage at specified intervals of time.

The general formula would be:

$Q(t) = Q_0(1+\frac{r}{n})^{nt}$

$Q_0$ is the initial amount deposited or owed, $r$ is the annual rate or interest rate, $n$ is the number of times per year interest is compounded, and “t” would be the number of years.

Semi-annually would mean $n=2$, quarterly is $n=4$, monthly is $n=12$, and daily is $n=365$. Also note that $r$ should be written in it’s decimal form.

Recall that,

$\lim_{n \to \infty} (1+\frac{r}{n})^{n}=e^{r}$

Therefore, if we infinitely compound interest, then we have this function,

$Q(t)=\lim_{n \to \infty} Q_0(1+\frac{r}{n})^{nt}=Q_0 e^{rt}$

The solution then to an IVP regarding continuously compounding interest where $Q’=rQ$, $Q(t)=Q_0$, and $t=0$ is

$Q(t)=Q_0e^{rt}$

### 4.Mixed Growth and Decay

Mixed growth and decay is a general term for problems that involve rate of changes that increase and decrease a value simultaneously.

The general forms would be:

Form A

$Q’ =$ rate of increase of $Q$ – rate of decrease of $Q$

Form B

$Q’ =$ rate of increase of $Q$ + rate of increase of $Q$

Form C

$Q’ =$ rate of decrease of $Q$ – rate of decrease of $Q$

If one of the rate of changes are constant then, we’ll have

$Q’ = \pm a \pm cQ$

where $a$ would be a constant rate of change (negative or positive), and $cQ$ would be $Q(t)$ multiplied to a proportionality constant, which could also be negative or positive depending on the problem (recall the exponential model at the beginning of this post).

Notice that we now have a first order linear differential equation. To solve this type of problem, we can either use variation of parameters, separation of variables, integrating factors, transformations from Bernoulli equations and homogeneous equations, or substitution.

#### Example Situations

An example of general form A would be the production of a radioactive material (while it’s being produced (constant rate increase), the material radioactively decays (decrease)). Therefore, we have,

$Q’ = a – kQ$

$k$ is the decay constant, $a$ is the production rate, and $Q$ is the decay rate.

An example of general form B would be a savings program. Say a constant amount is periodically deposited (constant rate increase) and the entire balance of the savings account accrues interest(increase). Therefore, we have,

$Q’ = a + rQ$

$r$ is the annual interest, $a$ is the periodic deposit of a constant amount, and $Q$ would be rate of increase due to interest.

An example of general form C be a tank of water is leaking (decrease) while a person scoops out a constant amount at a steady rate (constant rate decrease). Therefore we have,

$Q’ = -a – cQ$

$-c$ would be proportionality constant, $-a$ would be the rate of water scooped out the tank, and $Q$ would be the rate of leakage.

## Sample Problems

#### Example 1 (Denny/Sheyla)

A radioactive material will lose 34% of its mass in 55 minutes. What will itâ€™s half life be?

#### Example 2 (Jennifer/Ariel)

A process creates a radioactive substance at a rate of 3 g/hr and the substance decays at a rate proportional to its mass, with a constant of proportionality, k = .05(hr)^-1. If $Q(t)$ is the mass of the substance at time, $t$, find limit of $Q(t)$ as $t$ approaches infinity.

#### Example 3 (Richard/Raisa)

Wonka’s factory makes 1000 pounds of chocolate per week, while the workers eat the candy at a rate equal to Q(t)/25 pounds per week, where $Q(t)$ is the amount of chocolate present at time $t$.
a. Find $Q(t)$  for t > 0 if Wonka’s factory has 350 pounds of chocolate at t = 0b. lim t —> infinity of $Q(t)$

#### Example 4 (Tushar /Shisir)

Students tend to use google chrome at an alarming increased exponential rate, with the usage on the browser doubling every 100 days. What is the exponential growth rate?

#### Example 5 (Brian/ Jian Hui)

A young Warren Buffet initially deposits 50 dollars into his savings account. He also tries to make monthly recurring deposits of 10 dollars every month. If his savings account has an interest rate of 0.05% APY, make a mathematical model to predict the value of his account at any point in time, where $t>0$

## Links to Solutions for Select Section 4.1 Exercises

Exercise 1

Exercise 3

Exercise 5

Exercise 7

Exercise 11

Exercise 13

Exercise 17

Objective:

Learn how to apply differential equations to solve problems dealing with position, velocity and acceleration.

Introduction:

Mechanics is the general study of the relationships between motion, forces, and energy. In the past we have solved problems dealing with motion in one and two dimensions and even incorporated inclined planes and friction.

The incline and friction properties in previous courses were generally uniform in nature and did not change with respect to any other parameters. We solved these problems by analyzing the derivative of motion equations, that is to say the derivative of a function is the rate of change of the output value with respect to its input value. For example the velocity of an object is the change of its position with respect to time.

In this lesson we will be analyzing the relationship (differential equation) between motion parameters (functions) and their rates of change (derivatives). We will be using first order ordinary differential equations (ODE).

The parameters that we are going to be using will be standard units for length, mass, force, and time and are generally quantified in three standard systems: Centimeter-Gram-Seconds (CGS); Meter-Kilogram- Seconds (MKS); and Foot-Pound-Slug (British).

Parameters and Relationships:

Time (t)

Displacement/ Position (y): y = y(t)

Velocity (v): v = v(t) v = yâ€™

Terminal Velocity (vt)

Acceleration (a)(): a = a(t) a = vâ€™ a = yâ€™â€™

Force (F): F = m a

Force of Gravity/ Weight (Fg): Fg = m g

Resistive Force (F1): F1= k|v| = -kv

Acceleration due to Gravity (g): 9.8 m/s2 (mks); 980 cm/s2 (cgs); 32 f/s2(British)

We will refer to an object’s displacement as the length measured from its original position to its final position, while we will say that distance is the total length traveled to its final position.

Velocity is the rate of change of position with respect to time. An example of this is a car’s speedometer which measures forward speed (velocity) in either miles per hour, or kilometers per hour. Velocity is the first derivative of position, the rate of change in position with respect to time.

Acceleration is the rate of change of velocity with respect to time. It is the first derivative of velocity, or the second derivative of position. We will constrain ourselves to only using First Order equations in this lesson.

Newtonâ€™s second law of motion asserts that for a constant mass (m), the forceÂ (F)Â and the accelerationÂ (a)Â are related by the equation:

$F=ma$

This equation can be further defined by the form:

$F(t, y, y’) = my’$

Here we see that the force F may depend uponÂ t,Â y, andÂ y’. Remembering our constraint to use first order differential equations we will omit the use of y to attain the form:

$F(t, y’) = my’$

Then recast the form in terms of v:

$F(t, v) = mv’$

Let’s also recall that weight is:

$W=F=mg$

Motion Through a Resisting Medium Under Constant Gravitational Force:

In previous 2D motion physics problems our projectile motions were not inhibited by a resisting medium. Among the list of assumptions in these problems was that there was no effect due to water or air resistance (drag). This led to possible inaccuracies in the solutions in comparison to real world applications.

In order to attain a more accurate solution we will introduce resistive force as a parameter in our motion equation.

The total force acting on our object with the addition of a resistive force would be:

$F= -mg+F_{1}$

where $F_{1}$ is equal to a positive constant * velocity:

$F_{1}= k \lvert v \rvert =- kv$

this yields:

$mv’= -mg-kv$

rewrite in standard form for a first order linear equation and solve:

$v’+\frac{k}{m}v=-g$

Strategies:

When attempting to solve the differential equation consider:

• Are all units in the same system (CGS, MKS, British)?
• Recall that weight and mass are not the same.
• Mass = ?
• Acceleration due to gravity = ?
• k = ?
• Determine the general solution to the differential equation.
• Verify what the question is asking.
• Apply any additional parameters from the problem and solve the Initial Value Problem (IVP).

Escape Velocity:

Escape velocity is the speed that an object needs to be traveling to break free of a planet or moon’s gravity well and leave it without further propulsion (northwestern.edu).

• Time (t)
• Altitude (y, h), Maximum Altitude (ym)
• Escape Velocity (ve)
• Radius of Earth (R): 6378000m
• Gravitational Constant (G): Â 6.67 Ă— 10-11 Newtons kg-2 m2
• Mass of the Earth (M)

Let $GM=K$

The gravitational force on the vehicle at an altitudeÂ $y$Â above Earth is:

$F=-\frac{K}{(y+R)^2}$

Since $F=-mg$ when $y=0$ setting $y=0$ yields:

$-mg=-\frac{K}{R^2}$

and thus $K=mgR^2$:

$F=-\frac{mgR^2}{(y+R)^2}$

incorporate Newton’s second law of Motion formula:

$F=my”$ or $F=mv’$ or $F=ma$

therefore:

$a=\frac{gR^2}{(y+R)^2}$

or

$v’=\frac{gR^2}{(y+R)^2}$

Â When $t=0$ the velocity is $v_{0}$ and the altitude is $h$,

Therefore we can obtainÂ $v$Â as a function ofÂ $y$Â by solving the initial value problem when $v(h) = v_{0}$. That is: when the altitude of the spacecraft is 0, it is on the surface (ie :its position is equal to the radius of the earth).

Note: $y+R$ is essentially distance from the center of the Earth. Where $y$ in this case is more accurately described as height above the surface, or altitude from the surface. Therefore $y+R$ yield $h$ which is position.

Integrate both sides:

$\frac{1}{2}v^2= \frac{gR^2}{h}+c$

Since $v(h) = v_{0}$, c equals:

$\frac{1}{2}v_{0}^2= \frac{gR^2}{0+R}+c$

simplify:

$\frac{1}{2}v_{0}^2= gR}+c$

solve for the constant of integration:

$c=\frac{1}{2}v_{0}^2-gR$

plug in $c$ to IVP yields:

$\frac{1}{2}v^2= \frac{gR^2}{h}+\frac{1}{2}v_{0}^2}-gR$

in terms of $v(h)$:

$v(h)=\sqrt{\frac{2gR^2}{h}+v_{0}^2-2gR$

When $\frac{1}{2}v_{0}^2-2gR>0$ the space vehicle has enough velocity to escape Earth.

in terms of $v_{0}$:

$v_{0}>\sqrt{2gR}$

Escape Velocity of Earth:

$v_{e}=\sqrt{2(9.8)(6378000)}=11.18\: m/s$

Escape Velocity Practical Exercise:

4.3.18: A space vehicle is to be launched from the moon, which has a radius of about 1080 miles. TheÂ acceleration due to gravity at the surface of the moon is aboutÂ 5.31Â ft/s . Find the escape velocity in miles/s.

Solution:

Radius of moon: $1080\:miles\rightarrow \frac{1\: mile}{5280\: feet}=5702400\:ft$

Gravity of moon: 5.31ft/s

$v_{e}=\sqrt{2(5.31)(5702400)}=7782\: ft/s=1.47 mi/s$

Examples of Applications:

Review the following examples in preparation for your practical exercise question:

__________________________________________________________________________

4.3.3: A boat weighs 64,000 lb. Its propellor produces a constant thrust of 50,000 lb and the water exerts a resistive force with magnitude proportional to the speed, with k D 2000 lb-s/ft. Assuming that the boat starts from rest, find its velocity as a function of time, and find its terminal velocity.

4.3.3 Solution

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4.3.5: A stone weighing 1/2 lb is thrown upward from an initial height of 5 ft with an initial speed of 32 ft/s. Air resistance is proportional to speed, with k = 1/128 lb-s/ft. Find the maximum height attained by the stone.

4.3.5 Solution

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4.3.7: A 96 lb weight is dropped from rest in a medium that exerts a resistive force with magnitude proportional to the speed. Find its velocity as a function of time if its terminal velocity is -128 ft/s.

4.3.7 Solution

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4.3.10: An object weighing 256 lb is dropped from rest in a medium that exerts a resistive force with magnitude proportional to the square of the speed. The magnitude of the resisting force is 1 lb when |v| = 4 ft/s. Find v for t > 0, and find its terminal velocity.

4.3.10 Solution

Â __________________________________________________________________________

Practical Exercise:

An object with massÂ mÂ is given an initial velocityÂ $v\_{0}\leq 0$Â in a medium that exerts a resistive force with magnitude proportional to the square of the speed. Find the velocity of the object forÂ $t > 0$, and find its terminal velocity.

Practical Exercise Solution:

$m\frac{dv}{dt}=mg-kv^2$

$\frac{dv}{dt}=g-\frac{k}{m}v^2$

$\int\frac{dv}{g-\frac{k}{m}v^2}=\int\ dt$

$\frac{-ln[\frac{\frac{k}{m}v-\sqrt{\frac{gk}{m}}}{\frac{k}{m}v+\sqrt{\frac{gk}{m}}}]}{2\sqrt{\frac{gk}{m}}}=t+c$

$t=0; v=v_{0}; c=$

$\frac{-ln[\frac{\frac{k}{m}v_{0}-\sqrt{\frac{gk}{m}}}{\frac{k}{m}v_{0}+\sqrt{\frac{gk}{m}}}]}{2\sqrt{\frac{gk}{m}}}$

$\frac{-ln[\frac{\frac{k}{m}v-\sqrt{\frac{gk}{m}}}{\frac{k}{m}v+\sqrt{\frac{gk}{m}}}]}{2\sqrt{\frac{gk}{m}}}=t-\frac{-ln[\frac{\frac{k}{m}v_{0}-\sqrt{\frac{gk}{m}}}{\frac{k}{m}v_{0}+\sqrt{\frac{gk}{m}}}]}{2\sqrt{\frac{gk}{m}}}$

$ln[\frac{\frac{k}{m}v-\sqrt{\frac{gk}{m}}}{\frac{k}{m}v+\sqrt{\frac{gk}{m}}}]=-\beta t+ln[\frac{\frac{k}{m}v_{0}-\sqrt{\frac{gk}{m}}}{\frac{k}{m}v_{0}+\sqrt{\frac{gk}{m}}}]$

$ln[\frac{\frac{k}{m}v-\sqrt{\frac{gk}{m}}}{\frac{k}{m}v+\sqrt{\frac{gk}{m}}}*\frac{\frac{k}{m}v_{0}-\sqrt{\frac{gk}{m}}}{\frac{k}{m}v_{0}+\sqrt{\frac{gk}{m}}}]=e-\beta t$

$v=\alpha \frac{v_{0}(1+e^-\beta t)-\alpha(1-e^-\beta t)}{\alpha(1+e^-\beta t)-v_{0}(1-e^-\beta t)}$

where:

$\alpha = \sqrt{\frac{mg}{k}$ and $\beta = 2\sqrt{\frac{kg}{m}}$

$v_{t}={t\to\infty}; {e^-\beta t\to}0$

$v_{t}=\alpha\frac{v_{0}-\alpha}{\alpha-v_{0}}=-\alpha(\frac{v_{0}-\alpha}{v{0}-\alpha})=-\alpha$

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