1. Some time in the future a human colony is started on Mars. The colony begins

Question

1. Some time in the future a human colony is started on Mars. The colony begins with 50000 people and grows exponentially to 150000 in 250 years.

Assuming the population continues to grow exponentially, how long will it take to reach a size of 450000?

What is the rate of change of the size of the population 250 years after the founding of the original colony?

2. Solve the differential equation with given initial condition

What is the temperature of the coffee after 14 minutes?

After how many minutes will the coffee be 100 degrees?

Some time in the future a human colony is started on Mars. The colony begins with 50000 people and grows exponentially to 150000 in 250 years. Give a formula for the size of the human population on Mars as a function of t = time (in years) since the founding of the original colony Assuming the population continues to grow exponentially, how long will it take to reach a size of 450000? What is the rate of change of the size of the population 250 years after the founding of the original colony? Solve the differential equation with given initial condition dy/dt = 0.5y;y(2) = 100 A cup of coffee at 183 degrees is poured into a mug and left in a room at 72 degrees. After 4 minutes, the coffee is 136 degrees. Assume that the differential equation describing Newton's Law of Cooling is (in this case) dT/dt = k(T - 72). What is the temperature of the coffee after 14 minutes? After how many minutes will the coffee be 100 degrees?

Explanation / Answer

1.) Since the colony grows exponentialy then the

size of the human population = H(t) = Ho * e^(k*t) (t is in years)

Ho = Initial population = 50000

after 250 years H = 150000

=> H(250) = 50000 *e^(k*250) = 150000

=> k = (1/250) * ln (150000/50000) = 0.004394   

equation becomes

H(t) = 50000 * e^(0.004394*t) (t is in years)

for population to reach 450000

= t = (1/k) * ln (450000/50000) = 500 years

rate of change = dH/dt = Ho * k * e^(k*t)

so at t=0

rate of change = Ho*k = 50000*0.004394 = 220

2.) solution is ln(y) = 0.5*t + c => ( c is constan)

given

at t=2 , y(2) = 100

=> ln(100) = 0.5*2 + c

=> c = ln(2) - 1

=> y = (100/e) * e^(0.5*t)

3.) eqution for Temperature would be

T - 72 = c*e^(k*t) (where c is constant)

you can solve 3) using the answer in 2)

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