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
Your derivative is almost correct. 100 should appear only once, as it is a common factor of both terms.
ohh so its 100e^-0.027725t lim h--->0 (e^-0.027725h - 1)/h?
Yes, that's correct.
okay but how do i find the final derivative? its suppose to be -2.77e^-0.0277t
nvm i found it! thank you! now all i need is d)
Plug the given value of t into the derivative, remembering that it's the number of years since 2005.
To solve part (a), we need to find an expression for the amount of the isotope, A(t), that is still active using time t measured in years since 2005. We know that the half-life of the isotope is 25 years, which means that after each half-life, the amount of the isotope remaining is halved.
Let's start by considering the general form of the decay equation: A(t) = P(1/2)^(t/h), where P is the initial amount of the isotope, t is the time elapsed since 2005, and h is the half-life.
We know that P = 100 mg, t is the time elapsed since 2005, and h = 25 years. So, plugging in the values, we have A(t) = 100(1/2)^(t/25).
To solve part (b), we need to find the amount of the isotope we have at the present time, which is 2018. Since 2005 to 2018 is a span of 13 years, we can substitute t = 13 into the equation A(t) = 100(1/2)^(t/25) and calculate the value to one decimal place.
A(13) = 100(1/2)^(13/25) ≈ 44.4 mg
Therefore, we have approximately 44.4 mg of the isotope at the present time (2018) to one decimal place.
To solve part (c), we need to determine an expression for the instantaneous rate of change or the derivative of A(t) with respect to t, denoted as A'(t). We will use the definition of a derivative:
A'(t) = lim(h → 0) (A(t + h) - A(t))/h
Let's plug in the expression for A(t):
A'(t) = lim(h → 0) [100e^(-0.027725(t+h)) - 100e^(-0.027725t)]/h
To simplify further, we can use the properties of exponential functions:
A'(t) = lim(h → 0) [100e^(-0.027725t) * (e^(-0.027725h) - 1)]/h
Now we can evaluate this limit by plugging in h = 0 and using the fact that e^0 = 1:
A'(t) = 100e^(-0.027725t) * lim(h → 0) (e^(-0.027725h) - 1)/h
To evaluate the limit, you can use a spreadsheet by plugging in different small values of h (e.g., 0.001, 0.0001, etc.), calculating the expression (e^(-0.027725h) - 1)/h for each value, and observing how it approaches a certain value. As h approaches 0, you should see the expression converge to a constant value.
Once you have obtained the constant value from the spreadsheet, you can use it as the c value in part (a) to compare it against your c value. If they match, it confirms the correctness of your derivative calculation.
To solve part (d), we need to find the instantaneous rate of decay in 2018, which can be obtained by evaluating A'(t) at t = 13. Plugging t = 13 into your derivative expression or using the computed derivative value from part (c), we can find the instantaneous rate of decay in 2018.
A'(13) = 100e^(-0.027725 * 13)
Use a calculator to calculate this value to one decimal place. The result will give you the instantaneous rate of decay in 2018. Don't forget to include proper units, which in this case would be mg/year. Additionally, you can interpret the rate as the rate at which the isotope is decaying at the present time.