Math Help

>A tissue culture grows until it has an area of 9cm^2.Let A(t) be the area of the tissue at time t. On model for the growth rate is A'(t)=k*sqrt(A(t))*(9-A(t))

>Assume k=6, find the solution to A(0)=1 and graph it.

Can anyone show me how to start solving it...

no

Which part confuses you: subtracting, multiplying or the square root?

OP, you can't integrate this easily, i.e. finding A(t) would require higher level math. See below:

dA/dt = k (sqrtA)(9-A)

dA/[(sqrtA)(9-A)] = k dt

Now see this: http://integrals.wolfram.com/index.jsp?expr=1%2F((x^.5)*(9-x))&random=false

You have an integral on the LHS that you can't solve easily. Check to see if you've written down the problem correctly.


HOWEVER, you can just plug in the values for k and A(0) to get what A'(0) is. This is what >>3 was getting at, I assume.

So now you know your initial slope (48 cm^2/time unit). From A(0), you know the initial value. And you know that lim A as t -> infinity is 9. So assuming a standard model for tissue culture growth, you'll have a concave upward graph that starts from the point (t, A) = (0,1) with an initial slope of 48 cm^2/time, that reaches a horizontal asymptote of A = 9.



tl;dr: Graph will look like this: http://cs.gmu.edu/cne/modules/dau/calculus/limits/gif/ffnewfig6.gif
(don't pay attention to the actual values, just the overall behavior of the graph)

Thanks anyway for the help, but I got the answer. It's A(t)=(6e^(18t)-3)^2/(2e^(18t)+1)^2

>"Now see this: http://integrals.wolfram.com/index.jsp?expr=1%2F((x^.5)*(9-x))&random=false"
>"You have an integral on the LHS that you can't solve easily. Check to see if you've written down the problem correctly."

A "Gauss hypergometric function" as part of the solution??? What a piece of crap software. But the graph looks right, I solved it by just using a u substitution with u=sqrt(A(t)) and then used partial factions....