Karl's Calculus Tutor - Solution to Exercise 6.3-5

Solution to Exercise 6.3-5

You are looking for a function of time, t, that reduces itself by half every 72 hours. If t is in hours, then t/72 increases by 1 every 72 hours. So the function

   f(t)  =  2^-t/72

does that. Make sure you see why.

Trouble is, at t = 0, this function is not equal to 6 × 10¹⁷. What is it equal to? Well, whenever you raise a positive number to the zero power, you get 1.

What's more, if you multiply f(x) by any constant, it still reduces itself by half every 72 hours. So the function

   g(t)  =  6 × 10¹⁷ × 2^-t/72

meets all the requirements for part a of the problem.

Part b asks you to express this in terms of an exponential based upon e^x. In this section we discussed the identity

   b^x  =  e^{x ln(b)}

So simply substitute 2 for b and -t/72 for x and you get

   g(t)  =  6 × 10¹⁷ × 2^-t/72  =  6 × 10¹⁷ × e^{(-t/72) ln(2)}

Using a calculator, you can compute that this is the same as

   g(t)  =  6 × 10¹⁷ × e^{-t/103.8740429}

Part c wants to know the rate at which the material decays. For each scintillation there will be the loss of one atom of radioactive material. And the term, rate, suggests that what we need to do here is take the derivative with respect to time, t. Using the chain rule you get

   g'(t)  =  (-6 × 10¹⁷ / 103.8740429) × e^{-t/103.8740429}

and using a calculator, you can see that this is the same as

   g'(t)  =  -5.776226505 × 10¹⁵ × e^{-t/103.8740429}

Note that this number is negative. That indicates that the number of radioactive atoms is decreasing. The rate of scintillations per hour, then, must be the opposite of that:

   scintillations/hour  =  5.776226505 × 10¹⁵ × e^{-t/103.8740429}

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