Physics: A conical tank has height 3 m and radius 2 m at the top. Water flows in at a rate of 1.5 m3/min. How fast is the water level rising when it is 1.1 m from the bottom of the tank? (Round your answer to three decimal places.)

The answer is 4.8 I know yes because I did the question and Good luck on edunity...

A conical tank has height 3 m and radius 2 m at the

bryn2433

20.01.2022

Explanation:

It is known that general formula to calculate the volume of a cone is as follows.

Volume of the cone = $\frac{1}{3} \pi r^{2}h$

and here, r = $\frac{2}{3}h$

So, V = $\frac{1}{3} \pi \times (\frac{4}{9}) \times (h^{3})$

On putting the given values into the above formula we will calculate the value f h as follows.

V = $\frac{1}{3} \pi \times (\frac{4}{9}) \times (h^{3})$

= $\frac{1}{3} \times \pi \times (\frac{4}{9}) \times (h^{3})$

= $\frac{4 \pi}{27} h^{3}$

Now, we will differentiate w.r.t t as follows.

$\frac{dV}{dt} = \frac{d}{dt}(\frac{4 \pi}{27}h^{3})$

= $\frac{4\pi}{27}(3h^{2}) \frac{dh}{dt}$

It is given that water here flows at a rate of 1.5 $m^{3}/min$ and h = 1.1 m.

$\frac{dV}{dt} = \frac{4 \pi}{27}(3h^{2}) \frac{dh}{dt}$

1.5 = $\frac{4 \pi}{27}(3h^{2})\frac{dh}{dt}$

1.5 = $\frac{4 \pi}{27}(3(1.1)^{2})\frac{dh}{dt}$

$\frac{dh}{dt}$ = 0.888

= 0.9 m/min (approx)

Thus, we can conclude that the water level is rising at a rate of 0.9 m/min.

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