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 November 24th, 2015, 09:12 PM #1 Newbie   Joined: Oct 2015 From: Canada Posts: 5 Thanks: 0 Derivatives, trignometric functions and exponential functions Hey guys I need help with few questions they are: 1- A dinosaur bone from an archaeological site has one-fifth of the amount of Carbon 14 than it originally contained. Determine the approximate age of the bone given that Carbon 14 has a half life of 5770 years. 2- The number of people in a small town became zombies after t days during a zombie apocalypse is given by N(t) and is approximated by the exponential model N(t) =10000/1 + 12.5e^-kt If 80 people are zombies by the day 10 , find how many people become zombies by day 15. Thanks
 November 24th, 2015, 10:27 PM #2 Math Team   Joined: Dec 2013 From: Colombia Posts: 7,655 Thanks: 2633 Math Focus: Mainly analysis and algebra $c(t) = A\mathrm e^{-kt}$ and $c(5770)=\frac12c(0)$. So \begin{aligned} A\mathrm e^{-5770k} &= \frac12A \\-5770k &= \log \frac12 \\ k &= {\log 2 \over 5770} \end{aligned}Then we have $c(0) = \frac15 c(-T)$ so \begin{aligned} A &= \frac15A\mathrm e^{{\log 2 \over 5770}T} \\ \log 5 &= {\log 2 \over 5770}T \\ T &= {5770\log 5 \over \log 2} \\ &= 13397.5 = 13400 \text{ years} \end{aligned} I think there's something wrong with question 2.
November 25th, 2015, 06:20 AM   #3
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Quote:
 Originally Posted by Nij Hey guys I need help with few questions they are: 1- A dinosaur bone from an archaeological site has one-fifth of the amount of Carbon 14 than it originally contained. Determine the approximate age of the bone given that Carbon 14 has a half life of 5770 years.
So, after t years, the fraction of Carbon 14 left would be $\displaystyle \left(\frac{1}{2}\right)^{-\frac{t}{5770}}$. You want to solve $\displaystyle \left(\frac{1}{2}\right)^{-\frac{t}{5770}}= 0.20$. I recommend taking the logarithm of both sides to begin.

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 2- The number of people in a small town became zombies after t days during a zombie apocalypse is given by N(t) and is approximated by the exponential model N(t) =10000/1 + 12.5e^-kt
Do you mean 10000/(1+ 12.5e^(-kt))? That would be a much more reasonable problem than what you wrote.

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 If 80 people are zombies by the day 10
So $\displaystyle \frac{10000}{1+ 12.5e^{-10k}}= 80$. That is the same as $\displaystyle 1+ 12.5e^{-10k}= 800000$ so $\displaystyle 12.5e^{-10k}= 79999$.
Solve that for k. Start by dividing both sides by 12.5 then take the logarithm of both sides.
(Since the number of zombies is increasing, k will be negative.)

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 , find how many people become zombies by day 15.
Using the value of k you just found and t= 15 evaluate
$\displaystyle \frac{10000}{1+ 12.5e^{-kt}}$

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