first part is trivial, just put t = 0 into the function
c'(t) = t^3(-1/2)ln2(2^(-t/2) + 3t^2(2^(-t/2)
= 2^(-t/2)((-1/2)ln2(t^3) + 3t^2)
= 0 for max
2^(-t/2) = 0 ---> no solution
or
(t^2)(ln2(t) - 6)
t = 0, or t = 6/ln2
clearly t = 0 gives zero concentration, so
t = 6/ln2 = 8.656
C(8.656) = 32.29
c(t) = t^3 * 2^(-t/2)
Find the initial concentration c(0) at time t=0 min.
Finds the maximum concentration of the dye sand the time when this occurs.
4 answers
Thanks a lot.
Can you tell me where you got:
(t^2)(ln2(t) - 6)
Can you tell me where you got:
(t^2)(ln2(t) - 6)
from
(-1/2)ln2(t^3) + 3t^2
we are setting that equal to 0
(-1/2)ln2(t^3) + 3t^2 = 0
multiply by -2
ln2(t^3) - 6t^2 = 0
t^2( ln2(t) - 6) = 0 etc.
ok?
(-1/2)ln2(t^3) + 3t^2
we are setting that equal to 0
(-1/2)ln2(t^3) + 3t^2 = 0
multiply by -2
ln2(t^3) - 6t^2 = 0
t^2( ln2(t) - 6) = 0 etc.
ok?
are the answers t=0 or t=6/ln2