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.

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

   ......[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

or

r =2h

Substitute in [1]; we have

                         ......[2]

It is also, given the rate i.e,

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

{}

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

Simplify:

Substitute h = 6 cm we have;

or

    [Use ]

Simplify:

Therefore, the rate of depth of the water changing is,

b

2

−

b

b

3

+

7

b

2

+

14

b

+

6

%

−

3

b

3

+

7

b

2

+

14

b

+

6

%

step-by-step explanation: