P3 rates of change question

can anyone help me with this specimen question??

a cylindrical container has a height of 200cm. the container was initially full of a chemical but there is a leak from a hole in the base. when the leak is noticed, the container is half-full and the level of the chemical is dropping at a rate of 1cm per minute, it is required to find for how many minutes the container has been leaking. to model the situation, it is assumed that when the depth of the chemical remaining is x cm, the rate at which the level is dropping is proportional to x ^1/2. set up and solve an appropriate differential equation and hence show that the container has been leaking for about 80 minutes.

thanks :smile:

Reply 1

C4>O7

dx/dt=-kx^½
dx/dt=1 @ x=100:
=> 1=100^½k
k=-0.1

=>dx/dt=-0.1x^½
∫ x^-½ dx = ∫ -0.1 dt
=> 2x^½ = -0.1t + C
Initially full, at x=200, t=0:
=> 2(200)^½=0+C
C=20√2

=> 2x^½ = -0.1t+20√2
x=100:
20√2-20=0.1t
t = 82.84...≈80m

Reply 2

Jonny W

dx/dt = -k sqrt(x)

When t = 0 (the instant at which the leak is noticed), x = 100 and dx/dt = -1. So -1 = -k sqrt(100) and k = 1/10.

dx/dt = -(1/10) sqrt(x)
1/sqrt(x) dx/dt = -(1/10) . . . . . separating the variables
2sqrt(x) = -(1/10)t + c . . . . . integrating wrt t
2sqrt(x) = -(1/10)t + 20 . . . . . since x = 100 when t = 0

When the container is full, x = 200 and so

2sqrt(200) = -(1/10)t + 20
t = 10(20 - 2sqrt(200)) = -82.8427