Q. 62 - Calculus

Question

We may use a recursively defined sequence to approximate the current amount of a radioactive element. For example, radioactive radium changes into lead over time. The rate of decay is proportional to the amount of radium present. Experimental data suggests that a gram of radium decays into lead at a rate of $\frac{1}{2337}$gram per year. Let ak be the amount of radium at the end of year k. Since the decay rate is constant, if we use a linear model to approximate the amount that remains after one year has passes, we have

$a1 = a0 -\frac{ 1}{2337} =\frac{2336}{2337}a0.$ 

More generally, we obtain the recursion formula

$ak+1 = \frac{2336}{2337}ak.$

Use this formula to estimate how much radium remains after 100 years if we start off with a 0 = 10 grams of radium.

Step-by-Step Solution

Answer

If k=0, $a_1=10\left(\frac{2336}{2337}\right)$

If k=1, $a_2=10{\left(\frac{2336}{2337}\right)}^2$

Step 1. Given information

$a1 = a0 -\frac{ 1}{2337} =\frac{2336}{2337}a0$

Step 2. Calculate a k + 1 for k=0

$\begin{array}{r}{a}_1=\frac{2336}{2337}{a}_0\\ =\frac{2336}{2337}\left(10\right)\\ =10\left(\frac{2336}{2337}\right)\end{array}$

Step 3. Calculate a k + 1 for k=1

$\begin{array}{r}{a}_1=\frac{2336}{2337}\left(\frac{2336\times 10}{2337}\right)\\ =10{\left(\frac{2336}{2337}\right)}^2\end{array}$

Step 3. Calculate for 100 years

$\begin{array}{r}{a}_{99+1}=\frac{2336}{2337}{a}_{99}\\ a_{100}=\frac{2336}{2337}\left(10{\left(\frac{2336}{2337}\right)}^{100}\right)\\ =10{\left(\frac{2336}{2337}\right)}^{101}\end{array}$