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?
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)
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
To show that the maximum rate of growth occurs at any time when A(t) = 3 cm^2, we need to find the value of t for which A'(t) is maximum.
Given, A'(t) = k * sqrt(A(t)) * (9 - A(t)).
To find the maximum, we need to find the critical points of A'(t), where the derivative of A'(t) is equal to zero. So, let's differentiate A'(t) with respect to t.
dA'(t)/dt = k * (1/2) * (A(t))^(-1/2) * (9 - A(t)) + k * sqrt(A(t)) * (-1).
Setting the derivative equal to zero and simplifying:
0 = k * (1/2) * (A(t))^(-1/2) * (9 - A(t)) - k * sqrt(A(t)).
Now, let's simplify further:
(1/2) * (A(t))^(-1/2) * (9 - A(t)) = sqrt(A(t)).
Squaring both sides:
(1/4) * (9 - A(t))^2 = A(t).
Expanding:
(1/4) * (81 - 18A(t) + A(t)^2) = A(t).
Rearranging terms:
A(t)^2 - 20A(t) + 81 = 0.
Now we have a quadratic equation. To solve for A(t), we can factor or apply the quadratic formula.
Factoring, we get:
(A(t) - 9)(A(t) - 9) = 0.
Hence, A(t) = 9.
So, when A(t) = 9 cm^2, the growth rate A'(t) is maximum.
Moving on to the second part of the question:
A) Assuming k = 6, to find the solution corresponding to A(0) = 1, we can use a differential equation solver or solve it analytically.
Given, A'(t) = k * sqrt(A(t)) * (9 - A(t)).
Substituting k = 6 into the equation, we have:
A'(t) = 6 * sqrt(A(t)) * (9 - A(t)).
To solve this differential equation, we can separate the variables and integrate both sides:
(1/sqrt(A(t))) * dA(t)/dt = 6 * (9 - A(t)).
Integrating both sides:
∫(1/sqrt(A(t))) dA(t) = ∫6 * (9 - A(t)) dt.
Integrating the left side gives:
2√A(t) = 54t - 6∫A(t) dt.
We can find the integral of A(t) with respect to t using any integration technique suitable to solve the equation.
Simplifying:
2√A(t) = 54t - 6A(t) + C.
Now we can use the initial condition A(0) = 1 to solve for the constant C.
When A(0) = 1:
2√1 = 54*0 - 6*1 + C,
2 = -6 + C,
C = 8.
Substituting C back into the equation:
2√A(t) = 54t - 6A(t) + 8.
Squaring both sides to get rid of the square root:
4A(t) = 2916t^2 - 648t + 36A(t) - 48 + 3A(t).
Combining like terms:
37A(t) = 2916t^2 - 648t - 48.
Dividing by 37:
A(t) = (2916t^2 - 648t - 48)/37.
This is the solution corresponding to A(0) = 1.
For A(0) = 4, the process is the same. We substitute A(0) = 4 into the equation and solve for the constant C using the same method.