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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.
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.

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anonymous
 one year ago
Best ResponseYou've already chosen the best response.0@Astrophysics @ganeshie8 @nincompoop

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0\[\frac{ 1 }{ k }\]\[\int\limits \frac{ dM }{ M } = \int\limits dt\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Yea you mean \[\frac{ 1 }{ k} \times \int\limits \frac{ dM }{ M}=\int\limits dt\] R Right?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0integrate both side and it is \[\frac{ 1 }{ k}( \ln M)=t\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0okay how did you integrate the left side

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0okay nvm i see what you did

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Cross multiply so you can get M by it self. \[1(\ln M)=kt\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0now i just solve for M right?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0e both sides and it should look \[M=e ^{kt}\]

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.0Don't forget the constant you will need it for your initial amount

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Yea integrate both side and solve for M because he want to find an expression for M

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Sorry had trouble with equation table.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0these are my answer choices by the way M = 200ln(kt) M = 2e−kt M = 200 ekt M = 200 e−kt

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Yes, because if it is M=200e^kt

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0okay thanks for the help guys :D

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0if it is M=200e^kt*

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0you multiply it by 200 since it is decaying and it is the initial principal.

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.0\[\frac{ dM }{ dt } = kM \implies \int\limits \frac{ d M}{ M } = \int\limits  k dt \] \[\ln(M) = k t+C\] \[\large e^{\ln(M)} = e^{kt+C}\] \[M = e^{kt}e^C\] let \[e^c = C\] so we have \[M = Ce^{kt}\] after putting our initial conditions we get \[M = 200e^{kt}\]

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.0Just to note initial conditions is t = 0 meaning M = 200, so C = 200

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0This method works too, but I just multiply the initial mass amount by the expression.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Since 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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