calculus
posted by PPP on .
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.

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= (abt)^2
400=(ab*o)^2
a= 20
100=(20b30)^2
10=2030b
b=1/3
P=(201/3*50)^2=(2017.7)^2=...