Carbon has a half life of approximately 5715 years and a given amount decreases according to the law of exponential decay.

Question

anonymous · Answer

Starting with $$dy/dt= ky$$, show all steps in solving this separable differential equation and use the result to determine what percent of a given amount of carbon remains after 250 years.

whpalmer4 · Answer

$$\frac{dy}{dt} = k y$$rearrange so that all of the $y$ variables are on one side
$$\frac{dy}{y} = k\,dt$$Now integrate both sides to get your solution.

whpalmer4 · Answer

Next, you have to use your knowledge of the half life to find the value of the decay constant.  I'm not sure how to describe how to do that without giving away the answer, so you'll just have to try it and ask questions if you can't suss it out.

Finally, once you have the decay constant, you just plug in 250 years for $t$ to find the amount left.

anonymous · Answer

Thats the problem. I'm not sure what to do next. I think I integrate the equation into y=kt?

anonymous · Answer

integral of 1/y is ln|y|

integral k dt is kt+c

set them equal
solve for y

anonymous · Answer

jeez what's the matter with just $$\left(\frac{1}{2}\right)^{\frac{t}{5700}}$$

anonymous · Answer

Do you see how bad I am at this! ok so when I solve for y I get $$y= \pm e ^{kt+c}$$

whpalmer4 · Answer

@satellite73 Nothing's wrong with it, except that the problem requests the student solve it a different way...and that should be 5715 in the denominator of the fraction :-)

whpalmer4 · Answer

$$y = e^{kt+c}$$can be written as $$y(t) = Ce^{kt}$$because $$e^{kt+c} = e^c*e^{kt}$$

At $t=0$, $$y(0) = Ce^{k0} = C$$If we want to deal in just the fraction of the amount left, rather than an actual number, we can set $C=1$ and $y(t)$ will just give the fraction left.  At $t=5715$, after one half-life, $$y(5715) = Ce^{k5715} = \frac{C}{2}$$$$e^{k5715} = 0.5$$Take the ln of both sides and solve for $k$.  Plug $k$ back into your equation and calculate $y(250)$.  It should be pretty close to 1, given that you've only gone about 4% of the way toward cutting it in half!

anonymous · Answer

So I'm not sure I did this right but I got 8.749(10^-5)=K?

whpalmer4 · Answer

No, that is incorrect. Can you show me how you arrived at that value?