## anonymous one year ago The rate of decay in the mass, M, of a radioactive substance is given by the differential equation dM /dt = -kM, where k is a positive constant. If the initial mass was 200g, then find the expression for the mass, M, at any time t.

1. anonymous

@Astrophysics @ganeshie8 @nincompoop

2. anonymous

$\frac{ 1 }{ -k }$$\int\limits \frac{ dM }{ M } = \int\limits dt$

3. anonymous

is that right?

4. anonymous

Yea you mean $\frac{ -1 }{ k} \times \int\limits \frac{ dM }{ M}=\int\limits dt$ R Right?

5. anonymous

integrate both side and it is $\frac{ -1 }{ k}( \ln M)=t$

6. anonymous

okay how did you integrate the left side

7. anonymous

okay nvm i see what you did

8. anonymous

Cross multiply so you can get M by it self. $-1(\ln M)=kt$

9. anonymous

now i just solve for M right?

10. anonymous

ln M=-kt

11. anonymous

M = e ^-kt

12. anonymous

e both sides and it should look $M=e ^{-kt}$

13. Astrophysics

Don't forget the constant you will need it for your initial amount

14. anonymous

Yea integrate both side and solve for M because he want to find an expression for M

15. anonymous

200 = e^-k(0) ...?

16. anonymous

Sorry had trouble with equation table.

17. anonymous

these are my answer choices by the way M = 200ln(kt) M = 2e−kt M = 200 ekt M = 200 e−kt

18. anonymous

is it D

19. Astrophysics

yes

20. anonymous

Yes, because if it is M=200e^-kt

21. anonymous

okay thanks for the help guys :D

22. anonymous

if it is M=200e^-kt*

23. anonymous

you multiply it by 200 since it is decaying and it is the initial principal.

24. Astrophysics

$\frac{ dM }{ dt } = -kM \implies \int\limits \frac{ d M}{ M } = \int\limits - k dt$ $\ln(M) = -k t+C$ $\large e^{\ln(M)} = e^{-kt+C}$ $M = e^{-kt}e^C$ let $e^c = C$ so we have $M = Ce^{-kt}$ after putting our initial conditions we get $M = 200e^{-kt}$

25. Astrophysics

Just to note initial conditions is t = 0 meaning M = 200, so C = 200

26. anonymous

This method works too, but I just multiply the initial mass amount by the expression.

27. anonymous

Since he probably is taking a calculus class with some basic differential equation problems. From my experience, it is similar to when I took Calculus. Doubt this is a Differential Equation class questions. I never took DE.