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.

1 answer

decreasing rate dP/dt= k sqrt (P)

dP/(sqrtP)= k dt
integrate both sides.

-1/2 sqrtP=kT+ C
square both sides
1/4 P= (C+kT)^2
P= 4 (C+kT)^2 and by choosing the constansts C, k
P= (a-bt)^2

400=(a-b*o)^2
a= 20

100=(20-b30)^2
10=20-30b
b=1/3

P=(20-1/3*50)^2=(20-17.7)^2=...