Question

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)=Pe^ct (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 lim h→0 A(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.

WHAT I HAVE FINISHED:
i finished up till c) and for c my derivative is 100e^-0.027725t lim h--->0 (100e^-0.027725h - 1)/h

however my spread sheet is not right and i dont know how to do part d

Answer 1

Your derivative is almost correct. 100 should appear only once, as it is a common factor of both terms.

Answer 2

ohh so its 100e^-0.027725t lim h--->0 (e^-0.027725h - 1)/h?

Answer 3

Yes, that's correct.

Answer 4

okay but how do i find the final derivative? its suppose to be -2.77e^-0.0277t

Answer 5

nvm i found it! thank you! now all i need is d)

Answer 6

Plug the given value of t into the derivative, remembering that it's the number of years since 2005.