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 limh→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.
Ok, let's start with A(t) = P e^(ct)
when t = 0, A(0) = 100 , assuming a time of zero <---> 2005
100 = P(e^0) or P = 100
so we have A(t) = 100 e^(ct)
when t = 25 , A(25) = 50
50 = 100 e^(25c)
.5 = e^(25c)
take ln of both sides and use log rules ...
25c = ln .5
c = ln .5 /25
A(t) = 100 e^(ln .5/25 t)
b) 2018 ---> t = 13
A(18) = 100 e^(13 ln .5/25) =appr 69.7 mg
c) you are on your own, since it asked for a spreadsheet.
d) Using your result from c) you should have had
A'(t) = (ln .5/25)(100) e^(ln .5/25 t) = 4ln .5 e^(ln .5/25 t)
so at t = 18 (.... 2018)
A'(13) = 4ln .5 e^(ln .5/25 (13) ) = ..... mg/year
(a) To find an expression for the amount of the isotope, A(t), that is still active using time t, we can use the formula A(t) = Pe^ct, where P is the initial amount of the isotope.
Given that there was 100 mg of the isotope in 2005, we have A(0) = 100 mg. Since the half-life of the isotope is 25 years, we know that after 25 years, the amount of the isotope will be halved. Therefore, we have A(25) = 50 mg. Substituting these values into the general formula, we can solve for P and c.
From A(t) = Pe^ct, we substitute A(0) = 100 and t = 0: 100 = Pe^0, which simplifies to P = 100.
Next, substitute A(25) = 50 and t = 25: 50 = 100e^(25c), which can be simplified to 0.5 = e^(25c).
To solve for c, we take the natural logarithm (ln) of both sides: ln(0.5) = ln(e^(25c)). By the property of logarithms, we can move the exponent down: ln(0.5) = 25c.
Finally, we solve for c: c = ln(0.5)/25.
Thus, the expression for the amount of the isotope, A(t), still active at time t, measured in years since 2005, is A(t) = 100e^(ln(0.5)/25)t.
(b) To find the amount of the isotope at the present time (2018), we need to find A(2018-2005). A(13) ≈ 100e^(ln(0.5)/25 * 13). Using a calculator, the value of A(13) is approximately 44.3333 mg (rounded to 1 decimal place).
Therefore, at the present time (2018), there is approximately 44.3 mg of the isotope remaining.
(c) To determine the instantaneous rate of change, A'(t), we can use the definition of the derivative: A'(t) = lim(h->0) (A(t+h) - A(t))/h.
Considering A(t) = 100e^(ln(0.5)/25)t, we substitute this expression into the definition of the derivative, while keeping A(t) as a variable: A'(t) = lim(h->0) (100e^(ln(0.5)/25)(t+h) - 100e^(ln(0.5)/25)t)/h.
Expanding the expression, we get A'(t) = lim(h->0) (100e^(ln(0.5)/25)te^(ln(0.5)/25)h - 100e^(ln(0.5)/25)t)/h.
Simplifying further, we have A'(t) = lim(h->0) (100e^(ln(0.5)/25)t(e^(ln(0.5)/25)h - 1))/h.
Now, we evaluate the limit as h approaches 0: A'(t) = 100e^(ln(0.5)/25)t * lim(h->0) (e^(ln(0.5)/25)h - 1)/h.
The limit (e^(ln(0.5)/25)h - 1)/h is equal to ln(0.5)/25. Therefore, the expression for the instantaneous rate of change, A'(t), is A'(t) = (ln(0.5)/25) * 100e^(ln(0.5)/25)t.
(d) The instantaneous rate of decay in 2018 can be found by substituting t = 13 into the expression above.
A'(13) = (ln(0.5)/25) * 100e^(ln(0.5)/25 * 13). Evaluating this expression with a calculator, we find that A'(13) ≈ -3.6969 mg/year.
The negative sign indicates that the isotope is decaying, and the value 3.6969 mg/year represents the rate at which the isotope is decaying in 2018.