MATH SOLVE

7 months ago

Q:
# A paper cup has the shape of a cone with height 10 cm and radius 3 cm (at the top. if water is poured into the cup at a rate of 2cm3/s, how fast is the water level rising when the water is 5 cm deep?

Accepted Solution

A:

Let

h: height of the water

r: radius of the circular top of the water

V: the volume of water in the cup.

We have:

r/h = 3/10

So,

r = (3/10)*h

the volume of a cone is:

V = (1/3)*π*r^2*h

Rewriting:

V (t) = (1/3)*π*((3/10)*h(t))^2*h(t)

V (t) =(3π/100)*h(t)^3

Using implicit differentiation:

V'(t) = (9π/100)*h(t)^2*h'(t)

Clearing h'(t)

h'(t)=V'(t)/((9π/100)*h(t)^2)

the rate of change of volume is V'(t) = 2 cm3/s when h(t) = 5 cm.

substituting:

h'(t) = 8/(9π) cm/s

Answer:

the water level is rising at a rate of:

h'(t) = 8/(9π) cm/s

h: height of the water

r: radius of the circular top of the water

V: the volume of water in the cup.

We have:

r/h = 3/10

So,

r = (3/10)*h

the volume of a cone is:

V = (1/3)*π*r^2*h

Rewriting:

V (t) = (1/3)*π*((3/10)*h(t))^2*h(t)

V (t) =(3π/100)*h(t)^3

Using implicit differentiation:

V'(t) = (9π/100)*h(t)^2*h'(t)

Clearing h'(t)

h'(t)=V'(t)/((9π/100)*h(t)^2)

the rate of change of volume is V'(t) = 2 cm3/s when h(t) = 5 cm.

substituting:

h'(t) = 8/(9π) cm/s

Answer:

the water level is rising at a rate of:

h'(t) = 8/(9π) cm/s