Question

Water is drained out of tank, shaped as an inverted right circular cone that has a radius of 16cm and a height of 8cm, at the rate of 4 cm3/min. At what rate is the depth of the water changing at the instant when the water in the tank is 6 cm deep? Show all work and include units in your answer. Give an exact answer showing all work and include units in your answer.

Answer 1

Let h=height of water
Let r=radius of water surface
r/h=16/8 =2, so r=2h.
The volume of water is:
v=(1/3)×π×r²×h
=(1/3)×π×(2h)²×h
=4/3πh³
dv/dh=4πh^2
By chain rule:
dv/dt=dv/dh×dh/dt
but
dv/dt=4
thus:
4=(4πh)×dh/dt
dh/dt=4/(4πh²)
when h=6cm we have:
dh/dt=4/(4π6²)
=0.00884 cm³/min

Answer 2

Volume(V) of a circular cone is given by :

`V = (1)/(3)\pi r^2h` ......[1] where r is the radius and h is height of the cone respectively.

Given: Radius of cone(r) = 16 cm and height of cone(h) = 8 cm

`(r)/(h) = (16)/(8) = 2`

or

r =2h

Substitute in [1]; we have

`V = (1)/(3)\pi (2h)^2h = (1)/(3)\pi 4h^2 \cdot h = (4)/(3) \pi h^3` ......[2]

It is also, given the rate
`4 cm^3/min` i.e,

`(dV)/(dt) = 4 cm^3/min`

Differentiate V with respect to t in equation [2] we get;

`(dV)/(dt) = (4)/(3) \pi (3h^2) (dh)/(dt)= 4 \pi h^2(dh)/(dt)`

{
`(d x^n)/(dx) = nx^(n-1)`}

To find the rate of depth of the water when tank is 6 cm deep.

`4 = 4 \pi h^2(dh)/(dt)`

Simplify:

`(dh)/(dt) = (1)/(\pi h^2)`

Substitute h = 6 cm we have;

`(dh)/(dt) = (1)/(\pi 6^2)`

or

`(dh)/(dt) = (1)/(36\pi)` [Use
`\pi =3.14` ]

Simplify:

`(dh)/(dt) \approx 0.00885 cm/min`

Therefore, the rate of depth of the water changing is,
`0.00885 cm/min`