Most of the cells are on the edge,so A'=k(sqrt(A)) which implies that growth rate is proportional to diameter.
growth rate= k(sqrt(A) * (9-A)
growth rate ' = -ksqrtA+ k/2sqrtA * (9-A)
set growth rate '=0
2A=(9-A) or A=3cm^3
A tissue culture grows until it has an area of 9 cm^2. A(t) is the area of the tissue at time t. One model for the growth rate is A'(t)=k*sqrt(A(t))*(9-A(t)) for some constant k. The number of cells on the edge is proportional to sqrt(A(t)) and most of the growth occurs on the edge.
A) How do you show that the maximum rate of growth occurs at any time when A(t)=3 cm^2?
B)Assuming that k=6, then how do you find the solution corresponding to A(0)=1?
and for A(0)=4?
2 answers
A)
dA/dt = k A^.5 (9-A) = 9 k A^.5 -k A^1.5
d^2A/dt^2 = .5(9)k A^-.5 - 1.5 k A^.5
when that is 0 we have a max or min
4.5 /A^.5 = 1.5 A^.5
A = 4.5/1.5 = 3 done
dA/dt = k A^.5 (9-A) = 9 k A^.5 -k A^1.5
d^2A/dt^2 = .5(9)k A^-.5 - 1.5 k A^.5
when that is 0 we have a max or min
4.5 /A^.5 = 1.5 A^.5
A = 4.5/1.5 = 3 done
1 answer