Posted by PPP on Wednesday, November 23, 2011 at 1:10pm.
A quantity has the value P at time t seconds and is decreasing at a rate proportional to sqrt(P).
a) By forming and solving a suitable differential equation, show that P= (a  bt)^2 , where a and b are constants.
Given that when t= 0, P = 400,
b) find the value of a.
Given also that when t= 30, P = 100,
c) find the value of P when t = 50.

calculus  Steve, Wednesday, November 23, 2011 at 5:01pm
we are told that
dP/dt = k * P^(1/2)
P^(1/2) dP = k dt
2P^(1/2) = kt + c
P = (2c  2kt)^2 = (abt)^2
if
a = 2c
b = 2k
Now, we are told that P(0) = 400
(a0)^2 = 400
a = 20 or 20
P(30) = 100
If a=20
(2030b)^2 = 100
so b = 1 or 1/3
If a = 20
(2030b)^2 = 100
so b = 1 or 1/3
So far, we have 4 combinations of values for a and b
I'll let you figure out P(50). Maybe you have more info that eliminates some of the choices.