The rate at which an amount of a radioactive substance decays is modeled by the differential equation dA/dt = kA, where A is the mass in grams, t is the time in years, and k is a constant. Answer the following.
a) If a 100-gram sample of the radioactive substance decays to 95 grams after 1 year, find an equation that can model the mass of the of the sample after t years.
b) Find the mass of the sample after 50 years.
c) The half-life of a substance is the amount of time it takes for a sample to decay to half its original size. Find the half-life of the radioactive substance.
a) To find an equation that models the mass of the sample after t years, we can solve the differential equation.
Given the initial condition that a 100-gram sample decays to 95 grams after 1 year, we can substitute these values into the equation:
dA/dt = kA
Plugging in A = 100 grams and t = 1 year gives us:
dA/dt = k(100)
Since dA = (dA/dt)dt, we can rewrite the equation as:
dA = k(100)dt
Let's integrate both sides to solve for A:
∫dA = ∫k(100)dt
A = 100kt + C
Where C is the constant of integration.
Now we can use the initial condition that A = 95 grams when t = 1 year to find the value of C:
95 = 100k(1) + C
C = 95 - 100k
Therefore, the equation that models the mass of the sample after t years is:
A = 100kt + (95 - 100k)
b) To find the mass of the sample after 50 years, we substitute t = 50 into the equation:
A = 100k(50) + (95 - 100k)
Simplifying, we get:
A = 5000k + (95 - 100k)
A = 5000k - 100k + 95
A = 4900k + 95
c) The half-life of a substance is the amount of time it takes for a sample to decay to half its original size.
In this case, we need to find the value of t when A = 50 grams.
Substituting A = 50 into the equation, we have:
50 = 100kt + (95 - 100k)
Simplifying, we get:
50 = 100kt -100k + 95
-45 = 100kt - 100k
Dividing both sides by -100, we get:
0.45 = -kt + k
Rearranging the equation, we have:
0.45 = k(1 - t)
Dividing by (1 - t), we have:
k = 0.45 / (1 - t)
Since the half-life is the time it takes for A to decay to half its original size (50 grams), we set A = 50 and solve for t:
50 = 100kt + (95 - 100k)
Substituting k = 0.45 / (1 - t), we have:
50 = 100(0.45 / (1 - t))t + (95 - 100(0.45 / (1 - t)))
Solving this equation will give us the half-life of the radioactive substance. However, the equation is quite complex and requires numerical methods or approximation techniques to solve. So, unfortunately, I can't provide an exact answer without using more advanced mathematical tools. But hey, maybe the half-life will be shorter than waiting for me to solve it!
a) To find an equation that can model the mass of the sample after t years, we use the differential equation dA/dt = kA.
Given that the mass decreases from 100 grams to 95 grams in 1 year, we can use this information to find the value of k.
We have:
dA/dt = kA
=> (dA/A) = k(dt)
Integrating on both sides:
∫ (dA/A) = ∫ k(dt)
ln|A| = kt + C
Since we know that A = 100 grams when t = 0, we can substitute these values into the equation:
ln|100| = (0)k + C
ln|100| = C
Therefore, our equation becomes:
ln|A| = kt + ln|100|
Simplifying, we get:
ln|A/100| = kt
Rearranging, we find the equation that models the mass of the sample after t years:
A = 100e^(kt)
b) To find the mass of the sample after 50 years, we substitute t = 50 into the equation:
A = 100e^(kt)
A = 100e^(k * 50)
We don't have the value of k, so we can't find the mass without this information.
c) The half-life of a radioactive substance is the amount of time it takes for a sample to decay to half its original size. In this case, half of the original mass is 100 grams / 2 = 50 grams.
Setting up the equation:
A = 100e^(kt)
50 = 100e^(kt)
Dividing both sides by 100:
0.5 = e^(kt)
Taking the natural logarithm of both sides:
ln(0.5) = kt
We can solve for t by knowing the value of k. However, without more information about k, we cannot determine the half-life of the radioactive substance.
a) To find an equation that models the mass of the sample after t years, we can use the given differential equation.
The differential equation dA/dt = kA represents the rate of change of mass with respect to time. We can solve this equation by separating variables.
Start by dividing both sides by A to get dA/A = kdt.
Next, integrate both sides of the equation. The integral of dA/A is ln|A|, and the integral of kdt is kt. So we have ln|A| = kt + C, where C is the constant of integration.
To find C, we can use the given information that a 100-gram sample decays to 95 grams after 1 year. Substituting A = 100 and t = 1 into the equation, we get ln|100| = k(1) + C, which simplifies to ln(100) = k + C.
Using the property of logarithms, we can rewrite this as ln(100) = ln(e^k * e^C), where e is the base of the natural logarithm. ln(e^k * e^C) = k + C, so we have k + C = ln(100).
Rearranging the equation, we have C = ln(100) - k.
Substituting this value of C back into the equation ln|A| = kt + C, we have:
ln|A| = kt + ln(100) - k
Simplifying further, we get:
ln|A| = k(t-1) + ln(100)
This is the equation that can model the mass of the sample after t years.
b) To find the mass of the sample after 50 years, we can substitute t = 50 into the equation obtained in part a).
ln|A| = k(50-1) + ln(100)
Simplifying further, we get:
ln|A| = 49k + ln(100)
Exponentiating both sides, we have:
|A| = e^(49k + ln(100))
Since mass cannot be negative, we can drop the absolute value signs, giving us:
A = e^(49k + ln(100))
This equation gives the mass of the sample after 50 years.
c) The half-life of a substance is the amount of time it takes for a sample to decay to half its original size. Mathematically, this means A = A_0/2, where A_0 is the initial mass.
Using the equation obtained in part a), we can substitute A_0/2 for A:
ln(A_0/2) = k(t-1) + ln(100)
Simplifying further, we have:
ln(A_0) - ln(2) = k(t-1) + ln(100)
Since ln(2) and ln(100) are constants, we can combine them into a single constant:
ln(A_0/2) = k(t-1) + ln(100)
ln(A_0/2) = k(t-1) + ln(2*50)
ln(A_0/2) = k(t-1) + ln(2) + ln(50)
ln(A_0/2) = k(t-1) + ln(2) + ln(A_0/100)
Rearranging the equation, we have:
ln(A_0/2) - ln(A_0/100) = k(t-1) + ln(2)
Simplifying further, we get:
ln(100) = k(t - 1) + ln(2)
Since the half-life is the time it takes for the sample to decay to half its original size, we can substitute A_0/2 for A in the equation obtained in part a). The half-life occurs when A = A_0/2, so we have:
ln(A_0/2) = k(t-1) + ln(100)
Comparing this equation with the equation for the time t in part a), we can see that k(t-1) is equal to ln(100). Therefore, the half-life occurs when t - 1 = 1 (since ln(100) = 1) or t = 2.
Hence, the half-life of the radioactive substance is 2 years.
let the equation be
A = 100 e^(kt)
a) if amount = 95
95 = 100 e^(1k)
.95 = e^k
k = ln .95
so A = 100 e^(ln.95 t)
when t = 50
A = 100 e^(50ln.95) = 7.69 g are left
for half-life time, only 50 g of the original 100g would remain
50 = 100 e^(ln.95 t)
.5 = e^(ln.95 t)
ln.95t = ln.5
t = ln.5/ln.95 = appr13.5 years