Question

An inverted cone has a depth of 40 cm and a base of radius 5 cm. Water is poured into it at a rate of 1.5 cubic centimetres per minute. Find the rate at
which the level of water in the cone is rising when the depth is 4 cm.​

Answer 1

The rate at  which the level of water in the cone is rising when the depth is 4 cm is $\frac{21}{11} cm/s$

Step-by-step explanation:

Let the volume in the inverted cone = V

Volume (V) = $\frac{1}{3}$π$r^{2} h$

By similar triangles,

5/40 = r/h

⇒ r = h/8

∴ V = 1/3π(h/8) $(\frac{h}{8})^{2} }$ h

⇒ V = $\frac{1}{192}$π$h^{3}$

Water is being poured in the funnel @1.5$cm^{3}$/sec

$\frac{dV}{dt} = 1.5$

⇒ $\frac{d}{dt} \frac{1}{192}$π$h^{3}$) = 3/2

⇒ $\frac{1}{64}$ π$h^{2} \frac{dh}{dt} = \frac{3}{2}$

⇒ $\frac{dh}{dt} =$ 96/π$h^{2}$

∴ The rate at which water level changes

= dh/dt = 96/π$h^{2}$

water is at 4cm, i.e., when h=4,

∴ $\frac{dh}{dt}$ = 96/π$4^{2}$

⇒ $\frac{dh}{dt} = \frac{6}{(\frac{22}{7}) }$

= 6 * 7/22

= $\frac{21}{11}$cm/s.

Learn more: inverted Cone

brainly.in/question/5373225