-->

Sunday 24 February 2013

Simple O.D.E.'s Part 2

In part 1 I The simple form are those where dx/dt = kx. We can integrate these with a result with of form X0e^(-kt). The next thing to deal with is the relativity in a system. In the first question the ambient temperature of the water is given as 32 degrees and so this should be treated as zero. Questions are from the Calculus: Single Variable course on Coursera.



On a cold day you want to brew a nice hot cup of tea. You pour boiling water (at a temperature of 212oF) into a mug and drop a tea bag in it. The water cools down in contact with the cold air according to Newton's law of cooling:
dTdt=κ(AT)
where T is the temperature of the water, A=32oF the ambient temperature, and κ=0.36 min1.
The threshold for human beings to feel pain when entering in contact with something hot is around 107oF. How many seconds do you have to wait until you can safely take a sip? Round your answer to the nearest integer.

Instead of 212 and 107 degrees we should be dealing with 180 and 75 respectively. This gives us:
75 = 180 e^(-0.36t) or 5/12 = e^(-0.36t)

Now take the logs of both sides

ln (5/12) = -0.36t giving us t = 2.432 which when multiplied by 60 (converting from minutes to seconds) and rounding gives us 146 seconds



On the night of April 14, 1912, the British passenger liner RMS Titanic collided with an iceberg and sank in the North Atlantic Ocean. The ship lacked enough lifeboats to accommodate all of the passengers, and many of them died from hypothermia in the cold sea waters. Hypothermia is the condition in which the temperature of a human body drops below normal operating levels (around 36oC). When the core body temperature drops below 28oC, the hypothermia is said to have become severe: major organs shut down and eventually the heart stops.
If the water temperature that night was 2oC, how long did it take for passengers of the Titanic to enter severe hypothermia? Recall from lecture that heat transfer is described by Newton's law of cooling:
dTdt=κ(AT)
where T is the body temperature of a passenger, A the water temperature, and κ=0.016 min1. Give your answer in minutes and round it to the nearest integer.


This is very much like the last question except it uses a negative ambient temperature so we must increase our figures by two degrees. This gives us:


30 = 38 e^(-0.016t) 

Now take the logs of both sides

ln (30/38) = -0.016t giving us t = 14.77  rounding gives us 15 minutes.


The next question is on the Malthusian Trap. Part 1 is given a population of 6bn in 2002 and a growth rate of 1.1% what will the population be in 2030.

P(2030) = P(2002)e^(growth rate * (2030-2002)) = 6bne^(0.011*28) = 8.16bn

Part 2 attempts to estiamte the Malthusian Catastrophe but that probably needs a post of its own.





Tuesday 19 February 2013

Simple O.D.E.'s Part 1


Differential equations result when the quantity being changed ends up in the differential. The simple form are those where dx/dt = kx. We can integrate these with a result with of form X0e^(-kt). Questions are from the Calculus: Single Variable course on Coursera.

1) The amount I of a radioactive substance in a given sample will decay in time according to the following equation:
dIdt=λI
Nuclear engineers and scientists tend to be concerned with the half-life of a substance, that is, the time it takes for the amount of radioactive material to be halved.
Find the half-life of a substance in terms of its decay constant λ.
Let I = Ce^(-λt) so half-life means Ce^(-λt0) = Ce^(-λt1) and take the log of both sides.
ln C -λt0 = ln 2C -λt1
λ(t1-t0)  = ln 2 + ln C - ln C 
t1-t0        = ln 2/λ

2) After drinking a cup of coffee, the amount C of caffeine in a person's body obeys the differential equation
dCdt=αC
where the constant α has an approximate value of 0.14 hours1.
How many hours will it take a human body to metabolize half of the initial amount of caffeine? Round your answer to the nearest integer.
C(t) = C0e^(-αt) and we are interested in when halft the coffe is gone so

1/2C0  = C0e^(-αt) or 1/2 = e^(-αt) if we take the log of both sides

- ln 2   = -at which works out at 0.69/0.14 using the figures above


3) In a highly viscous fluid, a falling spherical object of radius r decelerates right before reaching the bottom of the container. A simple model for this behavior is provided by the equation
dhdt=αrh
where h is the height of the object measured from the bottom, and α is a constant that depends on the viscosity of the fluid.
Find the time it would take the object to drop from h=6r to h=2r in terms of α and r.

H(t) = Ce^(-ta/r)

2r = 6re^(-ta/r)) => 1 = 3e^(-ta/r)) and again take the logs

ln 3 + (-ta/r) =       0
               t    =  (r*ln 3)/a 








Arrow Key Nav