anonymous
  • anonymous
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.
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
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anonymous
  • anonymous
@Astrophysics @ganeshie8 @nincompoop
anonymous
  • anonymous
\[\frac{ 1 }{ -k }\]\[\int\limits \frac{ dM }{ M } = \int\limits dt\]
anonymous
  • anonymous
is that right?

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anonymous
  • anonymous
Yea you mean \[\frac{ -1 }{ k} \times \int\limits \frac{ dM }{ M}=\int\limits dt\] R Right?
anonymous
  • anonymous
integrate both side and it is \[\frac{ -1 }{ k}( \ln M)=t\]
anonymous
  • anonymous
okay how did you integrate the left side
anonymous
  • anonymous
okay nvm i see what you did
anonymous
  • anonymous
Cross multiply so you can get M by it self. \[-1(\ln M)=kt\]
anonymous
  • anonymous
now i just solve for M right?
anonymous
  • anonymous
ln M=-kt
anonymous
  • anonymous
M = e ^-kt
anonymous
  • anonymous
e both sides and it should look \[M=e ^{-kt}\]
Astrophysics
  • Astrophysics
Don't forget the constant you will need it for your initial amount
anonymous
  • anonymous
Yea integrate both side and solve for M because he want to find an expression for M
anonymous
  • anonymous
200 = e^-k(0) ...?
anonymous
  • anonymous
Sorry had trouble with equation table.
anonymous
  • anonymous
these are my answer choices by the way M = 200ln(kt) M = 2e−kt M = 200 ekt M = 200 e−kt
anonymous
  • anonymous
is it D
Astrophysics
  • Astrophysics
yes
anonymous
  • anonymous
Yes, because if it is M=200e^-kt
anonymous
  • anonymous
okay thanks for the help guys :D
anonymous
  • anonymous
if it is M=200e^-kt*
anonymous
  • anonymous
you multiply it by 200 since it is decaying and it is the initial principal.
Astrophysics
  • 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 \[M = 200e^{-kt}\]
Astrophysics
  • Astrophysics
Just to note initial conditions is t = 0 meaning M = 200, so C = 200
anonymous
  • anonymous
This method works too, but I just multiply the initial mass amount by the expression.
anonymous
  • 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.

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