Water i draining from a conical tank at the rate of 2 m3

/min. The tank i 16 meter high
and the top radiu i 4 meter. How fat i the water level falling when the water’ level
i 12 meter high?



Answer :

By using the concept of rate of change, it is obtained that

Water level is falling at the rate of [tex]\frac{2}{9\pi}[/tex] m per minute

What is rate of change?

Suppose there is a function and there are two quantities. If one quantity of a function changes, the rate at which other quantity of the function changes is called rate of change of a function.

Here the concept of rate of change has been used

Let the radius of cone at any point be r m and height be h m

Volume of cone = [tex]\frac{1}{3}\pi\times r^2\times h[/tex]

Now,

By the concept of similarity,

[tex]\frac{r}{h} = \frac{4}{16}\\\\\frac{r}{h} = \frac{1}{4}\\\\r = \frac{1}{4}h[/tex]

Volume of cone (V) =

[tex]\frac{1}{3}\times \pi \times (\frac{1}{4} h)^2\timesh\\\\\frac{1}{48}\times \pi \times h^3\\[/tex]

[tex]\frac{dV}{dt} = \frac{1}{48} \times \pi \times 3h^2\times \frac{dh}{dt}\\\\\frac{dV}{dt} = \frac{1}{16} \times \pi \times h^2\times \frac{dh}{dt}\\\\[/tex]

Now,

[tex]\frac{dV}{dt} = -2,\ h = 12[/tex]

So,

[tex]-2 = \frac{1}{16}\times \pi \times (12)^2 \times \frac{dh}{dt}\\\\\frac{dh}{dt} = -\frac{2 \times 16}{\pi \times 12 \times 12}\\\\\frac{dh}{dt} =- \frac{2}{9\pi}[/tex]

So water level is falling at the rate of [tex]\frac{2}{9\pi}[/tex] m per minute

To learn more about rate of change, refer to the link-

brainly.com/question/24313700

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