Post

Created by @nathanedwards
 at November 4th 2023, 11:31:56 pm.

AP Calculus AB Exam Question - Exponential Growth and Decay

A radioactive substance has a decay rate of 0.05 per year. At time t=0t=0, there are 200 grams of the substance.

(a) Write a differential equation to represent the rate of change of the mass of the radioactive substance, M(t)M(t), with respect to time.

(b) Solve the differential equation from part (a) to find an expression for M(t)M(t).

(c) Determine the amount of the radioactive substance remaining after 10 years, to the nearest gram.


Answer:

(a) To write a differential equation for the rate of change of the mass with respect to time, we can use the fact that the rate of decay is given as 0.05 per year. This means that the rate of decrease of mass is proportional to the current mass.

Let M(t)M(t) be the mass of the radioactive substance at time tt. Then, the differential equation is:

dMdt=0.05M\frac{{dM}}{{dt}} = -0.05M(b)
dMM=0.05dt\frac{{dM}}{{M}} = -0.05 dt

Integrating both sides, we get:

lnM=0.05t+C\ln|M| = -0.05t + C

where CC is the constant of integration.

Applying the exponential function to both sides, we have:

M=e0.05t+C=e0.05teC|M| = e^{-0.05t + C} = e^{-0.05t} \cdot e^C

Multiplying both sides by a suitable constant KK to represent the absolute value, we obtain:

M=Ke0.05tM = Ke^{-0.05t}

where K=±eCK = \pm e^C is another constant of integration. We are only concerned with the positive mass, so we can drop the absolute value sign.

Therefore, the expression for M(t)M(t) is:

M(t)=Ke0.05tM(t) = Ke^{-0.05t}(c)
200=Ke0.05(0)=K200 = Ke^{-0.05(0)} = K

Thus, K=200K = 200.

To find the amount of the radioactive substance remaining after 10 years, evaluate M(10)M(10):

M(10)=200e0.05(10)121 gramsM(10) = 200e^{-0.05(10)} \approx 121\text{ grams}

Therefore, to the nearest gram, there will be approximately 121 grams of the radioactive substance remaining after 10 years.