Related rates volume pouring in and out question

[r3dxP]

At a rate of 10ft^3/min, water is pouring into a conical cistern that is 16ft deep and 8ft in diameter at the top. But the cistern has developed a small leak. At the same time the water is 12ft deep, the water level is observed to be rising at 1/3 ft/min. How fast is the water leaking out?

i got .. Vout = 1.46ft^3/min
is this answer correct? if not, what is the correct answer and why?

 

[r3dxP]

help needed fast!~~~ thanks anybody/.? :P

 

[lurflurf]

i got .. Vout = 1.46ft^3/min
is this answer correct? if not, what is the correct answer and why?

That is wrong, how did you arrive at that?

vnet=vin-vout
vi=10
the height of water at time t can be
h(t)
the radius of the water cone can be expressed in terms of height
r(h(t))
you know
r(0)=0
r(16)=4
and
r(t)=a+b*h(t) for real numbers a and b
(it is a linear function)
V(t)=pi*r(h(t))^2*h(t)/3
you know for some time t
h(t)=12
h'(t)=1/3
find
V'(t)
use h(t) and h'(t) to write V(t) as a number then
vout=vin-V'(t)

 

[r3dxP]

That is wrong, how did you arrive at that?

vnet=vin-vout
vi=10
the height of water at time t can be
h(t)
the radius of the water cone can be expressed in terms of height
r(h(t))
you know
r(0)=0
r(16)=4
and
r(t)=a+b*h(t) for real numbers a and b
(it is a linear function)
V(t)=pi*r(h(t))^2*h(t)/3
you know for some time t
h(t)=12
h'(t)=1/3
find
V'(t)
use h(t) and h'(t) to write V(t) as a number then
vout=vin-V'(t)

oh thanks a lot.
so as for the answer, i got Vout = 10-9*pi = 10-9*3.14 = -18.26
so conclusion : water is pouring out at a rate of 18.26 ft^3/s when water is 12ft deep.
is that correct?

 

[lurflurf]

oh thanks a lot.
so as for the answer, i got Vout = 10-9*pi = 10-9*3.14 = -18.26
so conclusion : water is pouring out at a rate of 18.26 ft^3/s when water is 12ft deep.
is that correct?

That should be 10-3*pi
a negative vout would imply water is coming in