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

Answer:

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

$V = \frac{1}{3}\pi r^2h$    ......[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

$\frac{r}{h} = \frac{16}{8} = 2$

or

r =2h

Substitute in [1]; we have

$V = \frac{1}{3}\pi (2h)^2h = \frac{1}{3}\pi 4h^2 \cdot h = \frac{4}{3} \pi h^3$                          ......[2]

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

$\frac{dV}{dt} = 4 cm^3/min$

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

$\frac{dV}{dt} = \frac{4}{3} \pi (3h^2) \frac{dh}{dt}= 4 \pi h^2\frac{dh}{dt}$  

{$\frac{d x^n}{dx} = nx^{n-1}$}

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

$4 = 4 \pi h^2\frac{dh}{dt}$

Simplify:

$\frac{dh}{dt} = \frac{1}{\pi h^2}$

Substitute h = 6 cm we have;

$\frac{dh}{dt} = \frac{1}{\pi 6^2}$

or

$\frac{dh}{dt} = \frac{1}{36\pi}$     [Use $\pi =3.14$ ]

Simplify:

$\frac{dh}{dt} \approx 0.00885 cm/min$

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