CALCULUS - Exponential Growth and Decay? Please help FAST!!?

Hi,

I閳ユ獡 doing home schooling, and there is no one to help me with CALCULUS right now. I don閳ユ獩 clearly understand how to do this. Please help me fast! I need to submit my answers by tomorrow. I submitted this earlier , but people are giving different answers and if you can tell me of some website that can help solve this, I will appreciate it.

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Exponential Growth and Decay------

1. Solve the given differential equation subject to the given condition. Note that y(a) denotes the value of y at t = a.

dy/dt = 6y, y (0) = 1

2. The population of US was 3.9 million in 1790 and 178 million in 1960. if the rate of growth is assumed proportional to the number present, what estimate would you give for the population in 2000? (Compare your answer with the actual 2000 population, which is 275 million?)

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3. if a radioactive substance loses 15% of its radioactivity in 2 days, what is its half life?

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4. Human hair from a grave in Africa proved to have only 51% of the carbon 14 of living tissue. When was the body buried? See problem 5)

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5. (carbon Dating) all living things contain carbon 12, which is stable, and carbon 14, which is radioactive. While a plant or animals is alive, the ratio of these two isotopes of carbon remains unchanged, since the carbon 14 is constantly renewed: after death no more carbon 14 is absorbed. The half-life of carbon 14 is 5730 years. If charred logs of an old fort show only 70% of the carbon 14 expected in living matter, when did the fort burn down? Assume that the fort burned soon after it was built of freshly cut logs.

CALCULUS - Exponential Growth and Decay? Please help FAST!!?

#1: dy/dt = 6y

There's an analytic solutimethod for solving this, but in this case it's easiest to note that there is only one type of function whose derivative is an exact multiple of the function itself, and that's ke^(ct), where c is the multiplier (in this case, 6), and k corresponds to an arbitrary constant of integration (which is 1, since y(0)=1). Ergo, y(t)=e^(6t)

#2: Since the rate of growth is proportional to the population, we are (again) looking for a function of the form ke^(ct), that passes through 3.9 million in 1790 and 178 million in 1960. Thus:

ke^(1790c)=3.9 * 10^6 and

ke^(1960c)=1.78 * 10^8

Therefore 1.78 *10^8/e^(1960c) = k = 3.9 *10^6/e^(1790c)

e^(1790c)/e^(1960c) = 3.9*10^6 / (1.78*10^8)

1790c - 1960c = ln (3.9 * 10^6 / (1.78 * 10^8)

-170c = ln (39/1780)

c = ln (1780/39) / 170 閳?0.0224753352

ke^(1790 * ln (1780/39)/170) = 3.9*10^6

k (1780/39)^(1790/170) = 3.9*10^6

k=3.9*10^6 / (1780/39)^(179/17) 閳?1.3153108000 * 10^(-11)

Looking at the population for 2000:

3.9*10^6 / (1780/39)^(179/17) * e^(2000 * ln (1780/39)/170)

3.9*10^6 * (1780/39)^(200/17 - 179/17)

3.9*10^6 * (1780/39)^(21/17)

閳?437,377,629

This is somewhat higher than the actual U.S. population in 2000. We conclude that the rate of growth of the population has slowed significantly since 1790.

#3 Knowing e^(2c)=17/20, we solve for c and then find t such that e^(ct)=1/2. Thus:

2c=ln (17/20)

c=ln(17/20)/2

e^(ln (17/20)/2 t)=1/2

ln(17/20)/2 t = ln (1/2)

t=2 ln (1/2)/ln (17/20) 閳?8.53 days

The other problems are based on the same principle: use the given information (in this case, the half-life) to find c, and then plug that into e^(ct) to find t.