The rate of decay is proportional to the mass for radioactive material. For a certain radioactive isotope, this rate of decay is given by the differential equation dm/dt = -.022m, where m is the mass of the isotope in mg and t is the time in years.

A. If m(0)=20, write a function m(t) for the radioactive decay of the isotope. Show the steps in separating variables of the given differential equation and solving the equation for m.

B. The half-life of a radioactive substance is the time required for half of the substance to decay. What is the half-life of this radioactive isotope to the nearest tenth of a year?

Steve ur a savage

c'mon, you can do this.

dm/dt = -0.22m
dm/m = -0.22 dt
ln(m) = -0.22t + c
m = c e^(-0.22t)

c is the initial amount, so

m(t) = 20 e^(-0.22t)

I'm sure you can find the half-life now, ok?

A. To separate variables in the given differential equation, we need to express dm and dt separately and rearrange the equation. Let's start:

Given differential equation: dm/dt = -0.022m

Step 1: Separate the variables by multiplying both sides by dt:
dm = -0.022m dt

Step 2: Divide both sides by m:
dm/m = -0.022 dt

Step 3: Integrate both sides with respect to their respective variables:
∫(dm/m) = ∫(-0.022 dt)

Step 4: Evaluate the integrals:
ln|m| = -0.022t + C

Here, C is the constant of integration.

Step 5: Solve for m by exponentiating both sides:
e^(ln|m|) = e^(-0.022t + C)

Simplifying this gives:
|m| = e^C * e^(-0.022t)

Step 6: Since e^C is a positive constant, let's denote it as k:
|m| = k * e^(-0.022t)

Step 7: To find the function m(t), we consider the initial condition m(0) = 20:
|m(0)| = k * e^(-0.022 * 0)
|20| = k * e^0
20 = k

Therefore, our final function m(t) is:
m(t) = 20 * e^(-0.022t)

B. To find the half-life of the radioactive isotope, we need to find the time at which the mass has decreased to half its initial value (m(0) = 20/2 = 10 mg).

Substituting this into the function m(t), we have:
10 = 20 * e^(-0.022t)

Divide both sides by 20:
0.5 = e^(-0.022t)

Take the natural logarithm of both sides to isolate t:
ln(0.5) = -0.022t

Solve for t:
t = ln(0.5) / -0.022

Using a calculator, we find t ≈ 31.5548 years.

Therefore, the half-life of this radioactive isotope is approximately 31.6 years (rounded to the nearest tenth of a year).

Thank you

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