A radioactive isotope has a half life of 25 years. There was 100 mg of the isotope in 2005.
(a) Find an expression for the amount of the isotope, A(t), that is still active using time t, measured in years since 2005. Your expression must be of the form A(t)=Pect (base "e").
(b) How much of the isotope do we have at the present time (2018) to 1 decimal place?
(c) Determine an expression for the instantaneous rate of change, A'(t), using the limh→0A(t+h)−A(t)h. You will need to use a spreadsheet again and compare the limit value to your c value from part (a).
(d) What is the instantaneous rate of decay in 2018 to 1 decimal place? Be sure to include proper units and an interpretation of this rate in the context of the problem.
so ive gotten all the way up to c but the spread sheet doesnt come out right and i dont know how to do d can someone help?
i finished c) can some one help me with d?
(a) A(t) = 100*(1/2)^(t/25)
Now 1/2 = e^(-ln2), so
A(t) = 100* e^(-ln2*t/25) = 100e^(-0.0277t)
(b) A(13)
(c) A'(t) = -0.0277 A(t)
I assume you've seen the proof that d/dt e^t = e^t
(d) A'(13)
The snowplough problem will help you solve this.
Sure, I can help you with parts (c) and (d).
To determine the expression for the instantaneous rate of change, A'(t), we can use the definition of the derivative. The derivative of A(t) with respect to time t gives us the rate at which the amount of the isotope is changing at any given time t.
We can express A(t) as A(t) = P * e^(ct), where P is the initial amount of the isotope at t = 0 (in this case, 100 mg), c is the decay constant, and e is the base of natural logarithms.
Using the definition of the derivative, A'(t) can be calculated as follows:
A'(t) = lim(h→0) [A(t+h) - A(t)] / h
Substituting the expression for A(t), we get:
A'(t) = lim(h→0) [P * e^(c(t+h)) - P * e^(ct)] / h
Expanding the equation, we have:
A'(t) = lim(h→0) P * [e^(ct) * e^(ch) - e^(ct)] / h
Using the properties of exponents, we can rewrite this as:
A'(t) = lim(h→0) P * e^(ct) * [e^(ch) - 1] / h
Now, to evaluate the limit as h approaches 0, we can approximate it using a spreadsheet by plugging in small values of h (e.g., 0.0001) and observing the trend as it approaches 0. By calculating this expression for smaller and smaller values of h, we can estimate the value of A'(t).
For part (d), to find the instantaneous rate of decay in 2018, you will evaluate A'(t) at t = 2018 - 2005 (since we're measuring time since 2005) using the value of c obtained in part (c). This will give you the rate of decay at that specific time.
Remember to include the proper units (e.g., mg/year) and interpret the rate of decay as the rate at which the amount of the isotope is decreasing at that particular time.
I hope this helps! Let me know if you have any further questions.