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 hours−1.

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