Respuesta :
Explanation:
It is known that general formula to calculate the volume of a cone is as follows.
Volume of the cone = [tex]\frac{1}{3} \pi r^{2}h[/tex]
and here, r = [tex]\frac{2}{3}h[/tex]
So, V = [tex]\frac{1}{3} \pi \times (\frac{4}{9}) \times (h^{3})[/tex]
On putting the given values into the above formula we will calculate the value f h as follows.
V = [tex]\frac{1}{3} \pi \times (\frac{4}{9}) \times (h^{3})[/tex]
= [tex]\frac{1}{3} \times \pi \times (\frac{4}{9}) \times (h^{3})[/tex]
= [tex]\frac{4 \pi}{27} h^{3}[/tex]
Now, we will differentiate w.r.t t as follows.
[tex]\frac{dV}{dt} = \frac{d}{dt}(\frac{4 \pi}{27}h^{3})[/tex]
= [tex]\frac{4\pi}{27}(3h^{2}) \frac{dh}{dt}[/tex]
It is given that water here flows at a rate of 1.5 [tex]m^{3}/min[/tex] and h = 1.1 m.
[tex]\frac{dV}{dt} = \frac{4 \pi}{27}(3h^{2}) \frac{dh}{dt}[/tex]
1.5 = [tex]\frac{4 \pi}{27}(3h^{2})\frac{dh}{dt}[/tex]
1.5 = [tex]\frac{4 \pi}{27}(3(1.1)^{2})\frac{dh}{dt}[/tex]
[tex]\frac{dh}{dt}[/tex] = 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.
