water is running out a conical funnel at the rate of 1cu in per sec. if the radius of the base of the funnel is 4an and the altitude is 8in. find the rate at which the water level is dropping when it is 2 in from the top.

Question

water is running out  a conical funnel at the rate of 1cu in per sec. if the radius of the base of the funnel is 4an and the altitude is 8in. find the rate at which the water level is dropping when it is 2 in from the top.

amistre64 · Answer

change of rates eh....

amistre64 · Answer

v' = 1 ft^3 per sec; or simply 1

the rate of h is what we wanna determine

amistre64 · Answer

volume of a cone = (1/3)(base area)(height)

amistre64 · Answer

the relation of height to raduis is given by the cross section of the cone

amistre64 · Answer

h = -2r+8 right?

anonymous · Answer

can you show the solution?

amistre64 · Answer

v = (1/3)(pi 2^2)(-2r +8)

amistre64 · Answer

im working on it; havent got to the end yet lol

amistre64 · Answer

pi 2^2 was meant to be pi r^2

amistre64 · Answer

lets re work that with h as the variabe so we can see the rate of change with respect to h and not r

amistre64 · Answer

r = (h-8)/-2 or (8-h)/2

amistre64 · Answer

v = (1/3)(pi ((8-h)/2)^2)(h)  perhaps?

amistre64 · Answer

$$v = \frac{h}{3}*\frac{\pi.(8-h)^2}{4}$$

amistre64 · Answer

64hpi +h^3pi -16h^2pi
--------------------- = v  derive now
               12

amistre64 · Answer

64pi +3h^2pi -32hpi
--------------------- = v'  derive now
               12

amistre64 · Answer

pi(64 +3h^2 -32h) = 12

64 +3h^2 -32h = 12/pi  right?

anonymous · Answer

ang gulo wah

amistre64 · Answer

maybe i mess that up alittle

dv/dh = that up there

dv/dt = 1 and  dv/dh dh/dt = dv/dt

amistre64 · Answer

we wanna find dh/dt sooo

dh/dt = dh/dv dv/dt ; gotta take the inverse if thats gonna work

amistre64 · Answer

dh/dt is the inverse of dh/dv

right :)

amistre64 · Answer

on paper i get dh/dt = -1/(9pi)