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 = (a-bt)^2

if

a = 2c

b = 2k

Now, we are told that P(0) = 400

(a-0)^2 = 400

a = 20 or -20

P(30) = 100

If a=20

(20-30b)^2 = 100

so b = 1 or 1/3

If a = -20

(-20-30b)^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.

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