Let y(t) denote the mass of a radioactive substance at time t.t. Suppose this substance obeys the equation
y′(t)=−14y(t).
Assume that, initially, the mass of the substance is y(0)=M>0. At what time does half of the mass remain? (Round your answer to 3 decimal places).
PLease explain
this is just like the other one. You already know about radioactive decay, and how it is an exponential function, right?
Yes but I'm not sure what the m>0 is for
surely the initial mass is positive. Kind of makes things more real.
Looking at the original equation, if y(0) < 0, we have exponential growth, not decay.
To solve this problem, we need to find the time at which half of the initial mass remains.
First, let's find the general solution to the given differential equation:
The differential equation is y'(t) = -14y(t). This is a separable first-order linear differential equation, which means we can separate the variables and solve for y(t).
Divide both sides of the equation by y(t):
y'(t)/y(t) = -14
Integrate both sides with respect to t:
∫(y'(t)/y(t)) dt = ∫-14 dt
Simplifying the integrals:
ln|y(t)| = -14t + C
where C is the constant of integration.
Now, let's use the initial condition y(0) = M to find the particular solution. When t = 0, y(t) = M:
ln|M| = -14(0) + C
ln|M| = C
Substituting this value back into our general solution:
ln|y(t)| = -14t + ln|M|
Now, we can solve for y(t) to find an equation that describes the mass of the substance at any given time:
|y(t)| = e^(-14t + ln|M|)
|y(t)| = e^(-14t) * |M|
Since the mass of the substance cannot be negative, we can remove the absolute value:
y(t) = e^(-14t) * M
Now, we need to find the time at which half of the initial mass remains.
Let's set y(t) = M/2 and solve for t:
M/2 = e^(-14t) * M
Divide both sides by M:
1/2 = e^(-14t)
Taking the natural logarithm of both sides:
ln(1/2) = -14t
Simplifying:
-14t = ln(1/2)
Divide both sides by -14:
t = ln(1/2) / -14
Calculating this value:
t ≈ 0.049 (rounded to 3 decimal places)
Therefore, at approximately t = 0.049 (or 0.049 units of time), half of the initial mass remains.