Question

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.

Answer 1

@Astrophysics @ganeshie8 @nincompoop

Answer 2

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

Answer 3

is that right?

Answer 4

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

Answer 5

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

Answer 6

okay how did you integrate the left side

Answer 7

okay nvm i see what you did

Answer 8

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

Answer 9

now i just solve for M right?

Answer 10

ln M=-kt

Answer 11

M = e ^-kt

Answer 12

e both sides and it should look
\[M=e ^{-kt}\]

Answer 13

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

Answer 14

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

Answer 15

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

Answer 16

Sorry had trouble with equation table.

Answer 17

these are my answer choices by the way

M = 200ln(kt)
M = 2e−kt
M = 200 ekt
M = 200 e−kt

Answer 18

is it D

Answer 19

yes

Answer 20

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

Answer 21

okay thanks for the help guys :D

Answer 22

if it is M=200e^-kt*

Answer 23

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

Answer 24

\[\frac{ dM }{ dt } = -kM \implies \int\limits \frac{ d M}{ M } = \int\limits - k dt \]

\[\ln(M) = -k t+C\]

\[\large e^{\ln(M)} = e^{-kt+C}\]

\[M = e^{-kt}e^C\]

let \[e^c = C\]

so we have \[M = Ce^{-kt}\] after putting our initial conditions we get \[M = 200e^{-kt}\]

Answer 25

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

Answer 26

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

Answer 27

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.