I tried an exponential and that did not work so try a polynomial:
change in n per hour 5 7 9 11
change in change in n per hour = 2
constant second derivative --> parabola
n = a + k t + c t^2
dn/dt = k + 2 c t
d^2n/dt^2 = 2 c
so
2c = 2
c = 1
and
n = a + k t + t^2
when t = 1, n = 3
3 = a + k + 1
a + k = 2
when t = 2, n = 8
8 = a + 2 k + 4
a + 2 k = 4
subtract those 2 equations
-k = - 2
k = 2
so
n = a + 2 t + t^2
when t = 3, n = 15
15 = a + 2(3) + 9
a = 0
so
n = + 2 t + t^2
check when t = 5
n = 10 + 25 = 35 check
so I claim that
n = 2 t + t^2
The table below shows the number of molecules in experiment B.
Time(h) : 1st , 2nd , 3rd , 4th , 5th
No. of molecules : 3 , 8 , 15 , 24, 35
(i) how many molecules will there be in nth hour?
(ii) from which hour onwards would the no. of molecules be in experiment B?
1 answer
1 answer