The Student Room Group

A rock contains a radioactive substance which is decaying


The mass of the rock, m grams, at time t years after initial observation is given by

m = 400 + 80e^(-kt),

where k is a positive constant.

Given that the mass of the rock decreases by 0.2% in the first 10 years, find

(i) the value of k,

(ii) the value of t when m = 475,

(iii) the rate at which the mass of the rock is decreasing when t = 100.


I'm clueless from the start; how can the mass of the rock decrease when "m = 400 + 80e^(-kt)"? :confused:
Reply 1
The derivative of m = 400 + 80e^(-kt) with respect to t = -k80e^(-kt) < 0.
Reply 2
Ah, thanks.

Then would I work out k from -k80e^(-10k) = -0.2?
Reply 3
Okay I think this is it, I may be wrong i)
after 0 years t=0
so if you substitute 0 for t in the equation you get the original mass of the rock

m=400+80e^0
so m=480

so when t=10 the mass= 99.8% of 480
= 479.04

so the equation is 479.04= 400+80e^(-k10)
79.04=80e^(-k10)
0.988=e^(-k10)
ln0.988=-k10
(ln0.988)/(-10)=k
(0.001) 3dp=k

this gives the equation m=400+ 80e^(-0.001t)

use this to answer 2 and 3 by substitution
Reply 4
tomtheone
Okay I think this is it, I may be wrong i)
after 0 years t=0
so if you substitute 0 for t in the equation you get the original mass of the rock

m=400+80e^0
so m=480

so when t=10 the mass= 99.8% of 480
= 479.04

so the equation is 479.04= 400+80e^(-k10)
79.04=80e^(-k10)
0.988=e^(-k10)
ln0.988=-k10
(ln0.988)/(-10)=k
(0.001) 3dp=k

this gives the equation m=400+ 80e^(-0.001t)

use this to answer 2 and 3 by substitution


So you don't need to differentiate it? :confused:
Reply 5
I would use logs
Reply 6
The mass of the rock, m grams, at time t years after initial observation is given by
m = 400 + 80e^(-kt),
where k is a positive constant.
Given that the mass of the rock decreases by 0.2% in the first 10 years, find

(i) the value of k,

First find mass of rock when t=0 -> m = 400 + 80 = 480g

After 10 years mass => 480 x (100-0.2)/100 = 479.04g

sub (t=10, m=479.04) into equation:
=> 479.04 = 400 + 80e^(-10k)
=> -(1/10)ln[(479.04 - 400)/80] = k
=> k = 0.001

equation: m = 400 + 80e^(-t/1000)
(ii) the value of t when m = 475,

sub (m=475) into m
=> 475 = 400 + 80e^(-t/1000)
=> -1000ln(475-400)/80] = t
=> t = 64.5 years
(iii) the rate at which the mass of the rock is decreasing when t = 100

dm/dt = -80/1000e^(-t/1000)
dm/dt = -8/100e^(-t/1000)

sub (t = 100) into dm/dt => -8/100e^(-100/1000) = -0.072g per year.