Answer to Question #112056 in Calculus for Andreas Cjay

Question

2.1 Fragments of a dinosaur skeleton are discovered and dated using radioactive decay with
potassium-40. Suppose the half-life of the 14K is 1.5 billion years.
(a) If the bones are known to have 70 million years what is the remaining percentage of
14K in those bones today? Round up to 2 decimal places. [5]
(b) Suppose you have a 500 million year old skeleton, what percentage of 14K will remain
today? Round up to the nearest integer. [5]
(c) If you cannot detect amounts below 1%, what is the maximum age you can date
using 14K? Round up to the next integer.

Answer 1

let time be measured in millions. then given that

half life =1.5 billion years

=1500 million years

then N(1500)=(total amount of radioactive)/2

this implies

N(1500)=N₀/2

let N(t) be the remaining amount after time t

and N₀ be the initial amount of radioactive substance.

then

by decaying model $N(t)=N\omicron e^{\phantom{i}-kt}$

implying

$N\omicron/2=N\omicron e^{\phantom{i}-1500k}\\1/2=e^{\phantom{i}-1500k}\\\implies (2)^{\phantom{i}-1/1500}=e^{\phantom{i}-k}$

$\therefore N(t)=N\omicron( ( e)^{\phantom{}-k})^{\phantom{}t}\\\implies N(t)=N\omicron( ( 2)^{\phantom{}-1/1500})^{\phantom{}t}\\\implies N(t)=N\omicron ( 2)^{\phantom{}-t/1500}$

• If the bones are known to have 70 million years what is the remaining percentage of 14K in those bones today? Round up to 2 decimal places

t=70

$N(70)=N\omicron ( 2)^{\phantom{}-70/1500}\implies N(70)=N\omicron\phantom{i}0.968171\\ =96.82\phantom{k}percent \phantom{k}of \phantom{k}initial\phantom{k}quantity$

• Suppose you have a 500 million year old skeleton, what percentage of 14K will remain today? Round up to the nearest integer.

t=500

$N(500)=N\omicron ( 2)^{\phantom{}-500/1500}\implies N(500)=N\omicron\phantom{i}0.793701\\ =79.37\phantom{k}percent \phantom{k}of \phantom{k}initial\phantom{k}quantity$

• If you cannot detect amounts below 1%, what is the maximum age you can date using 14K? Round up to the next integer

given that $N(t)\geqslant N\omicron\ 0.01 =N\omicron/100$

$N(t)=N\omicron ( 2)^{\phantom{}-t/1500}\geqslant N\omicron/100\\\implies N\omicron ( 2)^{\phantom{}-t/1500}\geqslant N\omicron/100\\ ( 2)^{\phantom{}-t/1500}\geqslant 1/100\\\implies( 2)^{\phantom{}t/1500}\geqslant 100$

make t the subject from the inequality

$t/1500 ln(2)\geqslant ln(100)\\t\geqslant 1500(ln(100)/ln(2))\\\implies t\geqslant 9965.784285\phantom{i}million\phantom{i}years\\\implies t\simeq 9965784285 \phantom{i} years$