Question

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?

Answer 1

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

Answer 2

Using implicit differentiation, it is found that the water level is rising at a rate of 0.0764 centimetres per second.

The volume of a cone of radius r and height h is given by:

$V = \frac{\pi r^2h}{3}$

Applying implicit differentiation, the rate of change is given by:

$\frac{dV}{dt} = \frac{2\pi r}{3}\frac{dr}{dt} + \frac{\pi r^2}{3}\frac{dh}{dt}$

In this problem:

• The radius is constant, thus $\frac{dr}{dt} = 0$.
• Height of 5 cm and radius of 3 cm, thus $h = 5, r = 3$.
• Water poured at a rate of 2 cm³/s, thus $\frac{dV}{dt} = 2$

Then

$\frac{dV}{dt} = \frac{2\pi r}{3}\frac{dr}{dt} + \frac{\pi r^2}{3}\frac{dh}{dt}$

$2 = \frac{25\pi}{3}\frac{dh}{dt}$

$\frac{dh}{dt} = \frac{6}{25\pi}$

$\frac{dh}{dt} = 0.0764$

The water level is rising at a rate of 0.0764 centimetres per second.

A similar problem is given at https://brainly.com/question/13461339