Answer:
The water is being pumped at a speed of
[tex] \boxed{\bf \frac{10592000*\pi}{27}\;cm^3/min}[/tex]
Step-by-step explanation:
By congruence of triangles, the radius r of the cone base when its height is 200 cm satisfies the relation
r/200 = 400/600
So, r = 400/3 cm when the water is 200 cm high.
The volume of a cone with radius of the base = R is given by
[tex] \bf V=\frac{\pi R^2h}{3}[/tex]
So, the volume of water when it is 200 cm high is
[tex] \bf V_1=\frac{\pi* (400/3)^2*200}{3}=\frac{32000000*\pi}{27}\;cm^3[/tex]
One minute later, the height of the water is 200 cm + 20 cm = 220 cm
The radius now satisfies
r/220 = 400/600
and now the radius of the base is
r = 440/3
and the new volume of water is
[tex] \bf V_2=\frac{\pi* (440/3)^2*220}{3}=\frac{42592000*\pi}{27}\;cm^3[/tex]
So, the water is raising (being pumped) at a rate (speed) of
[tex] \bf V_2-V_1=(\frac{42592000*\pi}{27}-\frac{32000000*\pi}{27})=\frac{10592000*\pi}{27}\;cm^3/min[/tex]
Using implicit differentiation, it is found that water is being pumped at a rate of 837758 cm³/min.
--------------------------
The volume of a conical tank is given by:
[tex]V = \frac{\pi r^2h}{3}[/tex]
In which
Applying the implicit differentiation, we have that:
[tex]\frac{dV}{dt} = \frac{\pi}{3}\left(2rh \frac{dr}{dt} + r^2\frac{dh}{dt}\right)[/tex]
We have that:
The rate at which water is being pumped into the tank is [tex]\frac{dV}{dt}[/tex], thus:
[tex]\frac{dV}{dt} = \frac{\pi}{3}\left(2rh \frac{dr}{dt} + r^2\frac{dh}{dt}\right)[/tex]
[tex]\frac{dV}{dt} = \frac{\pi}{3}\left(200^2(20)\right)[/tex]
[tex]\frac{dV}{dt} = 837758[/tex]
Rate of 837758 cm³/min.
A similar problem is given at https://brainly.com/question/5889603
