Professor Kate Poirier | OL67 | Fall 2020

Category: Project #1 groups

Project 1 Part 2 – Chapter 4.2: Cooling and Mixing

Group Members: Ryjll Morris, Jason Zhu, Aron Singh


Newtons law of cooling states that the rate of heat loss of an object is directly proportional to the difference in the temperatures between the object and it’s medium. In other words, according to the textbook, for the temperature T(t) of an object at time (t), in a medium with a temperature Tm(t), the rate of change of T at any time (t) is proportional to T(t) – Tm(t).

This is ultimately expressed as

T = Tm + (To - Tm){e^-kt}

This formula is derived from the first order linear differential equation:

{T^'} = -k(T - Tm)

We solved this equation in the following steps:

Distributing K yields

{T^'} + kT = kTm

The complementary equation solution is in the form of T= ue^{-kt}, and {u^'}e^{-kt} = kTm

Therefore, solving for {u^'} we get

{u^'} = kTme^{-kt}

Integrating this equation to get u gives us

u = Tme^{kt} +c

Now, here we can recall that T = ue^{-kt}so substituting what we got for u into this equation we get

T = Tm +ce^{-kt}

Because c represents any constant, if the time is set to T(0) = To for the initial temperature, c would be (To – Tm), which is the inital temperature of the object minus the temperature of the medium.

So now substituting the value for c into the equation above we get

T = Tm + (To - Tm)e^{-kt}


The mixing section deals with the same first order linear differential equations and how a solution changes over time. The application is a salt water solution with a given concentration being added at a specific rate to a tank that initially contains saltwater with a different concentration. the objective with these problems is to determine the quantity of salt in the tank as a function of time. So, lets say we have a tank that initially contains 40 lbs of salt dissolved in 600 gallons of water. At to = 0, the initial time zero, water that contains a \frac{1}{2} pound of salt per gallon is poured into the tank at the rate of 4 gallons per minute and the mixture is drained from the tank at the same rate.

Here we have to find a differential equation for the quantity Q(t) of salt in the tank at time t>0, solve the equation to determine Q(t) and then find the lim t \rightarrow \infty Q(t).

So here, Q , which is the rate of change of the quantity of salt in the tank, changes with respect to time. Therefore, if rate in represents th rate at which salt enters the tank and rate out represents the rate at which salt leaves the tank then

Q^' = rate in – rate out.

Here, rate in is (concentration) * (rate of flow)

=\frac{1}{2} (lb/ gal) * 4 (gal/min) = 2 lb/min.

Now to determine rate out, we have to do a little more work. Think about the fact that we are removing 4 gallons of mixture per minute but there is always 600 gallons in the tank. Using this as a ratio we can see that

6 : 4 = \frac{6}{4} = 1.5

1.5 : 1

so we are actually removing 1/150 of the mixture per mintue. Now because the salt is evenly distrubuted, we are also removing 1/150 of the salt per minute.

Therefore, if at time (t) there are Q(t) pounds present in the tank the rate out at any time (t) = \frac{Q(t)} {150}

So this will look like

rate out =\frac {Q(t)}{600} *4 = \frac{Q(t)} {150}

Recalling our previous formula for Q^'

{Q^'} = 2lb/min – \frac{Q(t)}{150}

Rearranging this in the Linear differential format gives us

{Q^'} + \frac{Q}{150} = 2

Now we know thate^{-t/150}is a solution of the complementary equation for this differential, which is in the form of

Q = ue^{-t/150}, where {u^'}e^{-t/150} = 2

So once again, making u^' the subject we get

{u^'} = 2e^{t/150}

Integrating this we get

u = 300e^{-t/150} +c

Recalling now that Q =ue^{-t/150} and substituting the value of u into this equation we obtain

Q = 300 +ce^{-t/150}

From the previous section, we learnt that c = (To -Tm) so, applying value to this case would make c = (Q(0) – Q(t))

From our example we know that Q(0) the inital quantity of salt is 40 lbs, and we just found that Q(t) is 300

So c = 40 – 300 = -260

Therefore Q =300 -260e^{-t/150}

Now from this we can see that the limt \rightarrow \infty Q(t) = 300 for any value of Q(0).


Cooling Problem 1

Cooling Problem 3

Cooling Problem 5 and 7

Mixing Problems


In textbook: Section 4.2 Exercises

#8 – A tank initially contains 40 gallons of pure water. A solution with 1 gram of salt per gallon of water is added to the tank at 3 gal/min, and the resulting solution drains out at the same rate. Find the quantity Q.t / of salt in the tank at time t > 0.

#4(a) – A thermometer initially reading 212 F is placed in a room where the temperature is 70 F. After 2 minutes the thermometer reads 125 F.
(a) What does the thermometer read after 4 minutes

Project #1: Section 4.1: Growth and Decay

General form of a Differential Equation Involving Growth and Decay


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


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:


For IVPs, the solution would be,


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



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


Then we integrate the entire equation with respect to t

\int\frac{Q'}{Q} dt=\int (a) dt

Integrating gives us:


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





1.Radioactive Decay

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 “decay constant, t_0 is the initial time (essentially zero in most cases), and t would be any time duration.

Since \tau, or the “half-life,” is the amount of time at which a radioactive material’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,


Take the natural logarithm of the equation to have,



Finally, solve for t,


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


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 it’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

Example 6

Example 7

Links to Solutions for Select Section 4.1 Exercises

Exercise 1

Exercise 3

Exercise 5

Exercise 7

Exercise 11

Exercise 13

Exercise 17