1) In the year 1970, the population of the planet Earth was estimated to be 3.5

Question

1) In the year 1970, the population of the planet Earth was estimated to be 3.5 Billion people and it was growing at a rate of about 2% per year. When does this simple population model, assuming population growth is directly proportional to population size, predict we will reach 10 billion people (by year)?

2) A Turkey comes out of the fridge at a temperature of 35 degrees Fahrenheit. It must be cooked to an internal temperature of 180 degrees in order to be safe to eat. After 1 hour, you check the turkey and it measures 100 degrees. If you put it in an oven set to 200 degrees, using Newton's Law of Heating, how long will it take to cook the Turkey in a pre-heated oven?

3) A 200-Volt electromotive force is applied to an RC series circuit in which the resistance is 1000 ohms, the capacitance 5x10^-6 farad. Find the charge, q(t) on the capacitor if i(0) = .4. Recall that i = dq/dt.

Explanation / Answer

d n /d t = - k * n
where:
n= amount of substance remaining
k = constant
(the minus sign is because the amount of substance is decreasing)

This is a separable equation having as answer:

n = A* e ^ (-kt) (where A is yet another constant)

Initially (at t= 0) n = A so from the given data we know that A=100mg
After 6 hours (t=6) the mass decreased by 3%, so the mass left now is 97mg n= 100e^(-6k) = 97
so e^(-6k) = 97/100= 0.97

Take the natural logarithm of both sides:

-6k = ln (0.97)
k =~ 5 * 10 ^-3

We now have the full equation: n = 100 e ^(- 5*10^-3 * t)
To find the amount remaining after 24 hours simply put t=24 in the formula.
Als, to get the half life, put n=50 (half of 100mg) and solve for t.


dT/dt = k ( T - Tm )

What this mean is a body at a temp of 20C is drop in the container of hot water whose temp. is kept
at 100 C. If after .583333 min (35sec) the temp of the metal is 90C.

here the temp of the surrounding medium is the temp of the water, which is held constant at Tm=100 , dT/dt +kT=100k , So, T=100+ce^(kt). At t=0, we have T=20; hence 20=100+ce^(k0)=

c=-80 and the solution becomes T=100-80e^(kt)

at t=.583333 min, we have T=98 hence 98=100-80e^(k*.35), k=-.059413

98=100-80e^(-059413*t), t=62.1 sec


V = 500 gallons
A-in = 10 pounds/min

5 gal/min is pumped out from the tank. 5 gal is 1% of the tank's volume
A-out = 1% x S

(Where A-in is the rate of salt being pumped into the tank, and A-out for the rate of that being pumped out of the tank)
Always use A-in and A-out if you are confused on how to write down the first order differential ;) .

dA/dt = A-in - A-out = 10 - (A/100)

Solve the equation by separating variables.
dA / {10 - (A/100)} = dt
100 dA / (1000-A) = dt

integrate both sides
-100 ln|1000-A| = t

ln|1000-A| = -t/100
1000 - A = e^(-t/100)
A = 1000 - e^(-t/100)

b). c(t) = A(t) : V = {1000 - e^(-t/100)} : 500

The algebra should look much easier when done on paper I guess, but the main idea is to state the rate at which salt goes in (A-in) and the rate at which it goes out (A-out).
Then, the rate at which the mass of the salt changes can be stated as A-in - A-out.
Finally, use separating variable method to solve the first order differential equation.

the last one for you to do hint: RI+L(dI/dt)+(1/C)q-E(t)=0

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