Instantaneous rates of change. Exponential growth and Decay

[josh_123]

The equation I used for these are:
If N=N^oe^kt then DN/Dt=N₀ke^kt

so the two problems I have trouble with is
A radioactive substance has a decay constant (k) of -.0539 per year. If 371 grams of the material is initially present, what is the instaneous rate of change of the substance at times t=4 weeks and 18 months?

so what I did is 371(-.0539)e^{(-.0539)(4/48)=-19.907 gram/week}
and 371(-.0539)e^{(-.0539)(1.5)}=-28.421 gram/month

Is this right?

2. A radioactive substance has a half life of 23.7 days. If 4983 grams of the material is initially present, what is the instantaeous rate of change of the substance at times t=1 day

I find the k constant by -2491.5/23.7 which is -105.127 gram/day which mean -.288 gram/year? (by dividing 365). Is this right?

then for 1 day 4983(-.288)e^(-.288)(1/365)=-2.948 gram/day?
Is this right?

 

[Hammie]

In your first equation, where did you get the 48 from? Maybe it should be 52. Then at t=4weeks, you would have t=4/52 years. I think your answer is in terms of years, not weeks.

Same for the second equation. You converted 18 months into 1.5 years. The solution you have is in terms of grams per year, not per month. You may want to check your work, I got a different result for this part.

For the second problem, why are you converting to years? Everything is given in days. You are doing more work than needed, unless the question is asking for an answer in terms of gram/year.

And you need to reconsider how you are calculating the k value. The half life is 23.7 days, so half of the material is present after t=23.7 days. Try setting the exponent equal to 23.7 k. e^(23.7k) = 1/2. Then solve for the k value. That would be a good first step.