Solving half-life problems with exponential decay
Formulas for half-life
Growth and decay problems are another common application of derivatives.
We actually don’t need to use derivatives in order to solve these problems, but derivatives are used to build the basic growth and decay formulas, which is why we study these applications in this part of calculus.
We won’t work through how to prove these formulas, because in addition to derivatives, we also use integrals to build them, and we won’t learn about integrals until later in calculus.
So, for now, we’ll just state that the basic equation for exponential decay is
???y=Ce^{kt}???
where ???C??? is the amount of a substance that we’re starting with, ???k??? is the decay constant, and ???y??? is the amount of the substance we have remaining after time ???t???. Since substances decay at different rates, ???k??? will vary depending on the substance.
Half life equation
Every decaying substance has its own half life, because half life is the amount of time required for exactly half of our original substance to decay, leaving exactly half of what we started with. Because every substance decays at a different rate, each substance will have a different half life. But regardless of the substance, when we’re looking at half life, we know that
???y=\frac{C}{2}???
Because ???y??? is the amount of substance that remains as the substance decays, and because ???C??? is the amount of substance we started with originally, when the substance has decayed to half of its original amount, ???y??? will be equivalent to ???C/2???. So we can substitute this value in for ???y???, and then simplify the decay formula.
???\frac{C}{2}=Ce^{kt}???
???\frac{1}{2}=e^{kt}???
So, when we’re dealing with half life specifically, instead of exponential decay in general, we can use this formula we got from substituting ???y=C/2???.
How to solve for the half-life of any substance
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How much of a substance will be left after a certain amount of time?
Example
Fermium-???253??? has a half life of ???3??? days. If we start with ???1,200\ \text{mg}???, how much Fermium-???253??? will be left after ???10??? days?
First we need to find ???k???. Since we know that ???C=1,200??? and that ???t=3???, we can use
???\frac{C}{2}=Ce^{kt}???
???\frac{1,200}{2}=1,200e^{k(3)}???
Solving the equation for ???k???, we get
???\frac{1}{2}=e^{3k}???
???\ln{\frac{1}{2}}=\ln{e^{3k}}???
???\ln{0.5}=3k???
???k=\frac{\ln{0.5}}{3}???
Now that we have a value for ???k???, we can solve for ???y??? using ???y=Ce^{kt}???, where ???C=1,200???, ???t=10??? and ???k=\frac{\ln{0.5}}{3}???.
???y=1,200e^{\frac{\ln{0.5}}{3}(10)}???
???y=1,200e^{\frac{10}{3}\ln{0.5}}???
???y=119.06\ \text{mg}???
There would be ???119.06\ \text{mg}??? of Fermium-???253??? left after ???10??? days.
Example
The half life of Americium-???243??? is ???7,370??? years. How long will it take a mass of Americium-???243??? to decay to ???73\%??? of its original size?
First we need to find ???k???. We can assume that the original mass is ???100\%??? of its original size and say that ???C=1???, and we already know that ???t=7,370???, so we can use
???\frac{C}{2}=Ce^{kt}???
to find ???k???.
???\frac{1}{2}=1e^{k(7,370)}???
???\frac{1}{2}=e^{7,370k}???
???\ln{\frac{1}{2}}=\ln{e^{7,370k}}???
???\ln{\frac{1}{2}}=7,370k???
???k=\frac{\ln{0.5}}{7,370}???
Now that we have ???k???, we can find ???t??? using ???y=Ce^{kt}???, where ???C=1??? and ???y=0.73???.
???0.73=1e^{\frac{\ln{0.5}}{7,370}t}???
???\ln{0.73}=\ln{e^{\frac{\ln{0.5}}{7,370}t}}???
???\ln{0.73}=\frac{\ln{0.5}}{7,370}t???
???t=\frac{7,370\ln{0.73}}{\ln{0.5}}???
???t=3,346.21??? years
It would take ???3,346.21??? years for our sample to decay to ???73\%??? of its original size.