Solve Related Rates: Water Tank Filled/Drained at 5m^3/min and 7m^3/min

[Draggu]

Homework Statement

A water tank has the shape of an inverted circular cone with a base diameter of 8m and a height of 12m.
a) If the tank is being filled with water at the rate of 5m^3/min, at what rate is the water level increasing when the water is 5m deep?
b) If the tank is full of water and being drained at the rate of 7m^3/min, at what rate is the water level decreasing when the water is 7m deep?

Homework Equations

The Attempt at a Solution

a)

V'(t) = 5
r = (1/3)h
V = (1/3)(pi)((1/3h)^2)h
=(1/27)(pi)(h^3)


V' = (1/9)(pi)(h^2)h'
sub in known values~

5 = (1/9)(pi)(144)h'
h' = 5 / (50.24)

=.09

I have a feeling I have to sub in the "water is 5m deep" part somewhere, but I don't know where.

b) Same problem as above, basically.

 

[HallsofIvy]

Homework Statement

A water tank has the shape of an inverted circular cone with a base diameter of 8m and a height of 12m.
a) If the tank is being filled with water at the rate of 5m^3/min, at what rate is the water level increasing when the water is 5m deep?
b) If the tank is full of water and being drained at the rate of 7m^3/min, at what rate is the water level decreasing when the water is 7m deep?

Homework Equations

The Attempt at a Solution

a)

V'(t) = 5
r = (1/3)h
V = (1/3)(pi)((1/3h)^2)h
=(1/27)(pi)(h^3)


V' = (1/9)(pi)(h^2)h'
sub in known values~

5 = (1/9)(pi)(144)h'
h' = 5 / (50.24)

=.09

I have a feeling I have to sub in the "water is 5m deep" part somewhere, but I don't know where.

b) Same problem as above, basically.

Well, yes. I notice you have "144" above. That is from 12' height of the entire tank isn't it? But if the water is only 5' deep, it isn't filling the entire tank. Use the 5' depth instead of 12'.

 

[Draggu]

Well, yes. I notice you have "144" above. That is from 12' height of the entire tank isn't it? But if the water is only 5' deep, it isn't filling the entire tank. Use the 5' depth instead of 12'.

a) 0.57~

b)

V'(t) = -7
r = (1/3)h
V = (1/3)(pi)((1/3h)^2)h
=(1/27)(pi)(h^3)


V' = (1/9)(pi)(h^2)h'
sub in known values~

-7 = (1/9)(pi)(49)h'
h' = -7 / (17)

-0.41

 

[Draggu]

a) 0.57~

b)

V'(t) = -7
r = (1/3)h
V = (1/3)(pi)((1/3h)^2)h
=(1/27)(pi)(h^3)


V' = (1/9)(pi)(h^2)h'
sub in known values~

-7 = (1/9)(pi)(49)h'
h' = -7 / (17)

-0.41

Can someone please confirm this?^^ :(