Respuesta :
Answer:
845758 [tex]cm^{3} / min[/tex]
Step-by-step explanation:
Let V be the volume of water in the tank, in c m 3 ; let h be the depth/height of the water, in cm; and let r be the radius of the surface of the water (on top), in cm. Since the tank is an inverted cone, so is the mass of water. Since the tank has a height of 6 m and a radius at the top of 2 m, similar triangles implies that h/ r = 6 /2 = 3
so that h = 3 r
The volume of the inverted cone of water is then
[tex]V= \frac{1}{3 } \pi r^{2} h[/tex] = [tex]\pi r^{3}[/tex]
Now differentiate both sides with respect to time t (in minutes) to get
[tex]\frac{dv}{dt}=3\pi r^{2} * \frac{dr}{dt}[/tex] (the Chain Rule is used in this step).
If [tex]V_{i}[/tex] is the volume of water that has been pumped in, then
[tex]\frac{dV}{dt}= \frac{dV_{i} }{dt} -8000 =3\pi (\frac{200}{3} )^{2} *20[/tex] (when the height/depth of water is 2 meters, the radius of the water is [tex]\frac{200}{3} cm[/tex]
Therefore [tex]\frac{dV_{i} }{dt} =\frac{800000\pi }{3} + 8000[/tex] ≈ 845758 [tex]cm^{3} / min[/tex].
