Wednesday

March 4, 2015

March 4, 2015

Posted by **Chaz** on Wednesday, March 24, 2010 at 4:51pm.

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?

- calculus -
**bobpursley**, Wednesday, March 24, 2010 at 5:01pmMost 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

- calculus -
**Damon**, Wednesday, March 24, 2010 at 5:05pmA)

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

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