Respuesta :
Explanation:
As the given data is as follows.
h = [tex]\frac{3}{5}d[/tex]
= [tex]\frac{3}{5} \times (2r)[/tex]
Also, we know that r = [tex]\frac{4}{3}h[/tex]
and Volume (V) = [tex]\frac{1}{3} \pir^{2}h[/tex]
= [tex]\frac{1}{3} \pi (\frac{4}{3}h)^{2} h[/tex]
= [tex]\frac{16}{27} \pi h^{3}[/tex]
And, [tex]\frac{dV}{dt} = \frac{3 \times 16}{27} \pi h^{2} \frac{dh}{dt}[/tex]
[tex]\frac{dV}{dt} = \frac{16}{9} \pi h^{2} \frac{dh}{dt}[/tex]
Putting the given values into the above formula as follows.
[tex]\frac{dV}{dt} = \frac{16}{9} \pi h^{2} \frac{dh}{dt}[/tex]
[tex]30 m^{3}/min = \frac{16}{9} \pi (2)^{2} \frac{dh}{dt}[/tex]
[tex]\frac{dh}{dt} = 1.343 m/min
or, = 134.3 cm/min (as 1 m = 100 cm)
thus, we can conclude that the height changing at 134.3 cm/min when the pile is 2 m high.
