Rate of Change in a circular cone

[xsoon2bprox]

I need help in finding the rate of change in a circular cone.

The V= (Pi/3)h(r^2)
were h =height and r= radios

the radios is 5cm and the height is 12cm

Water is pouring out of the bottom of the circular cone at 10 cm/sec. What is the change in rate when the height is 8 cm.

 

[xsoon2bprox]

I know that you have to take the derivative of the volume in respect to time.

V ' = Pi/3 [h(2r)(r ' )+r^2(h ' )]

and that is as far as I was able to get.

 

[fasteddie65]

I need help in finding the rate of change in a circular cone.

The V= (Pi/3)h(r^2)
were h =height and r= radios

the radios is 5cm and the height is 12cm

Water is pouring out of the bottom of the circular cone at 10 cm/sec. What is the change in rate when the height is 8 cm.

First of all, note that r/h = 5/12, so 5h = 12r and r = (5/12)h and dr/dt = (5/12)dh/dt.

h = 8, so r = (5/12)(8) = 10/3.

V = (?/3) r^2 h

dV/dt = (?/3) r^2 dh/dt + (2?/3) r h dr/dt

-10 cm^3/sec = (?/3) [ (5 cm)^2 (10/3 cm) dh/dt + 2(5 cm) (10/3 cm) (5/12) dh/dt)]

-10 = (?/3) (250/3 + 125/9) dh/dt

You can finish that up and find dh/dt.

 

[Deleted member 4993]

I need help in finding the rate of change [of what? Volume? Height? diameter? - whose change and with respect to what (time? height?..)] in a circular cone.

The V= (Pi/3)h(r^2)
were h =height and r= radios

the radios is 5cm and the height is 12cm

Water is pouring out of the bottom of the circular cone at 10 cm/sec. What is the change in rate when the height is 8 cm.<<< Is that statement correct?