Need help with a AB Problem!

**mak15** · March 19th 2008, 05:53 PM

Dear all, I am a high school student & I have the following problem which to me it is very confusing.

Thank you very much!

**Jhevon** · March 19th 2008, 05:58 PM

Originally Posted by mak15

Dear all, I am a high school student & I have the following problem which to me it is very confusing.

Thank you very much!

(a) this is a separable differential equation.

you want to solve $R' = kR$

rewrite as $\frac {R'}R = k$

now integrate both sides and continue (don't forget the arbitrary constant). use the clues given in the question to solve for the unknown.

(b) once you answered part (a), plug in $R = 10$ and solve for $t$

(c) $kt_h = \ln 2$ , always

here $t_h$ is the half-life, $k$ is the rate of decay.

**TheEmptySet** · March 19th 2008, 06:00 PM

$\frac{dR}{dt}=kR \iff \frac{dR}{r}=kdt$ so integrate both sides

$\int \frac{dR}{R}= \int kdt \iff ln(R)=kt+c$

solveing for R

$R=e^{kt+c}=e^{kt} \cdot e^c=Ae^{kt}$

$R(t)=Ae^{kt}$

$R(0)=10^6=Ae^{0}$

You should be able to finish from here.

Good luck.

**mak15** · March 19th 2008, 07:11 PM

Originally Posted by TheEmptySet

$\frac{dR}{dt}=kR \iff \frac{dR}{r}=kdt$ so integrate both sides

$\int \frac{dR}{R}= \int kdt \iff ln(R)=kt+c$

solveing for R

$R=e^{kt+c}=e^{kt} \cdot e^c=Ae^{kt}$

$R(t)=Ae^{kt}$

$R(0)=10^6=Ae^{0}$

You should be able to finish from here.

Good luck.

Thank you so much for ALL of your replies!

I understand the A is a constant and I got A = 1,000,000
But I am not sure about Part C, Half-Life?

**TheEmptySet** · March 19th 2008, 07:35 PM

Originally Posted by mak15

Thank you so much for ALL of your replies!

I understand the A is a constant and I got A = 1,000,000
But I am not sure about Part C, Half-Life?

Solve the equation

$\frac{1000000}{2}=1000000e^{kt}$

using the value of k you got from part A.

Good luck.

**mak15** · March 20th 2008, 07:00 AM

Originally Posted by TheEmptySet

Solve the equation

$\frac{1000000}{2}=1000000e^{kt}$

using the value of k you got from part A.

Good luck.

Thank you very much to you all! I think I got it!

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