Thread: Need help with a AB Problem!

1. Need help with a AB Problem!

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

Thank you very much!

2. 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 $\displaystyle R' = kR$

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

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 $\displaystyle R = 10$ and solve for $\displaystyle t$

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

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

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

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

solveing for R

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

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

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

You should be able to finish from here.

Good luck.

4. Originally Posted by TheEmptySet
$\displaystyle \frac{dR}{dt}=kR \iff \frac{dR}{r}=kdt$ so integrate both sides

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

solveing for R

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

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

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

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?

5. 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

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

using the value of k you got from part A.

Good luck.

6. Originally Posted by TheEmptySet
Solve the equation

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

using the value of k you got from part A.

Good luck.
Thank you very much to you all! I think I got it!