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- Thread starter cemtu
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- #3

cemtu

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dN0/dt = 2 is given thus N0 is 2t. There is no problem there sir.

- #4

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Your original post makes no sense. If you want any help you'll have to organise it into something meaningful.dN0/dt = 2 is given thus N0 is 2t. There is no problem there sir.

- #5

cemtu

- 52

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Okay right away.

- #6

cemtu

- 52

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I put the whole question and its solution.Your original post makes no sense. If you want any help you'll have to organise it into something meaningful.

- #7

Of course, start off with the equation you wrote of ##N = N_{0}e^{-\lambda t}##. Then you can derive both sides with respect to ##t## to obtain a relation pertaining to the activities.

How would you derive the RHS wrt ##t##? Or perhaps more specifically, what can we do about the constant out the front?

- #8

cemtu

- 52

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I know how to derive however our professor just replaced N with dN/dt (LHS) and N0 with dN0/dt (RHS). I don't know how. He didnt took the derivative as the way you and I want.

Of course, start off with the equation you wrote of ##N = N_{0}e^{-\lambda t}##. Then you can derive both sides with respect to ##t## to obtain a relation pertaining to the activities.

How would you derive the RHS wrt ##t##? Or perhaps more specifically, what can we do about the constant out the front?

- #9

I know how to derive however our professor just replaced N with dN/dt (LHS) and N0 with dN0/dt (RHS). I don't know how. He didnt took the derivative as the way you and I want.

I can't speak for what your professor was trying to imply however I can say that ##\frac{dN_{0}}{dt}## is zero, as is the case for all constants. It's essentially a meaningless statement.

What do you get if you derive both sides of ##N = N_0 e^{-\lambda t}## with respect to ##t##?

- #10

cemtu

- 52

- 5

HereI can't speak for what your professor was trying to imply however I can say that ##\frac{dN_{0}}{dt}## is zero, as is the case for all constants. It's essentially a meaningless statement.

What do you get if you derive both sides of ##N = N_0 e^{-\lambda t}## with respect to ##t##?

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- #11

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I suspect this might be an initial value problem. But, you've misunderstood what's been given as initial values.Here

Without seeing the problem stated clearly, we are all just guessing.

- #12

Here

You've done the differentiation right, ##\frac{dN}{dt} = -\lambda N_{0} e^{-\lambda t} \implies A = A_{0}e^{-\lambda t}## since ##A = \lambda N##.

You must stop using ##\frac{dN_{0}}{dt}##. It is wrong. I strongly suspect the intended meaning is the rate of change of ##N## at ##t=0##, however we might just denote this (negative) ##A_{0}## (or that more clunky expression in post #7).

You should be able to just plug in the ratio of the activities.

- #13

cemtu

- 52

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Sir, thank you.You've done the differentiation right, ##\frac{dN}{dt} = -\lambda N_{0} e^{-\lambda t} \implies A = A_{0}e^{-\lambda t}## since ##A = \lambda N##.

You must stop using ##\frac{dN_{0}}{dt}##. It is wrong. I strongly suspect the intended meaning is the rate of change of ##N## at ##t=0##, however we might just denote this (negative) ##A_{0}## (or that more clunky expression in post #7).

You should be able to just plug in the ratio of the activities.

- #14

Sir, thank you.

You're welcome, though no need to call me 'sir', I haven't met the Queen yet!

- #15

Mark44

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This...Without seeing the problem stated clearly, we are all just guessing.

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