CALCULUS, Exponential Growth and Decay? Please help!!?

Question

Hi,
I’m doing homeschooling, and there is no one to help me with CALCULUS right now. I don’t clearly understand how to do this. Please help me fast!
Thanks
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?)

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

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 4)

4. (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.

Thanks a lot James L!!!!!

For #5, I have the correct answer.

It is:
½ = e^(5730)
k= ln (1/2)/ 5730=-1.210x10^-4
0.7 yo = yo e^((-1.210x10^-4)t)
t = ln (.07)/-1.210x10^-4=2950
Fort burned down about 2950 years ago.

Answer 1

1. Divide both sides by y:

(1/y) dy/dt = 6

Integrate both sides from 0 to s:

integral from 0 to s 1/y dy/dt dt = integral from 0 to s of 6 dt

In the left integral, use the substitution u = y(t), so du = dy/dt dt. Change the limits accordingly by plugging t=0 and t=s into u=y(t).

integral from y(0) to y(s) 1/u du = integral from 0 to s of 6 dt

Perform the integration:

ln|u| = 6t

plug in the limits y(0) and y(s) on the left side, and the limits 0 and s on the right:

ln|y(s)| - ln|y(0)| = 6s

Rearrange:

ln|y(s)| = 6s + ln|y(0)|

Exponentiate:

y(s) = e^[6s + ln|y(0)|] = y(0)e^(6s).

Plug in y(0) = 1, and you get y(s) = e^(6s). Or you can use any letter in place of s, so y(t) = e^(6t). I only used s earlier because I was already using t inside the integrals.

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?)

Let P(t) be the population in millions at time t, t is in years since 1790. Then P(t) = 3.9e^(kt), where k is chosen to satisfy P(170) = 78.

You have 78 = 3.9e^(170k), or 20 = e^(170k). Take ln of both sides: ln 20 = 170k. So k = (ln 20)/170. Now plug in t=210, and you get

P(210) = 3.9e^(210*(ln 20)/170)
= 157.84 million, much less than the actual number

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

Mass at time t is M(t) = M0*exp(-kt), where k is a positive constant and M0 is the original mass. t is time in days.

You have M(2) = 0.15M0, so 0.15 = exp(-2k). Therefore k = -(ln 0.15)/2.

To get the half-life, solve the equation

M0/2 = M0*exp(t(ln 0.15)/2)

so 1/2 = exp(t(ln 0.15)/2)
=> ln 0.5 = t (ln 0.15)/2
=> t = 2(ln 0.5)/(ln 0.15) = 0.73 days

4. Let t be the time since the body was buried. Then exp(-kt) = 0.51, where k is chosen so that exp(-5730k) = 0.5. to get k, take ln of both sides:

5730k = -ln(0.5), so k = -(ln 0.5)/5730.

Now solve exp(t(ln 0.5)/5730) = 0.51 for t.

ln of both sides: t(ln 0.5)/5730 = ln(0.51), so
t = 5730*ln(0.51)/ln(0.5) = 5566.3 years

4. (carbon Dating) same as previous problem, but use 0.7 in place of 0.51