# calculus

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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 -

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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