A conical party hat made out of cardboard has a radius of 4cm and a height of 12cm. When filled with beer, it leaks at the rate of 4cm^3/min. (a)At what rate is the level of the beer failing when the beer is 6cm deep? (b) When is the hat half empty?
voluemcone=1/3 h*PI*r^2
but r=h/4
volume=1/3 PI h^3/16 check that
dv/dt= you do it, solve for dh/dt
To find the rate at which the level of the beer is falling, we need to differentiate the volume of the beer with respect to time.
The volume of a cone can be given by the formula V = (1/3)πr^2h, where V is the volume, r is the radius, and h is the height.
(a) To find the rate at which the level of the beer is falling when the beer is 6cm deep, we need to differentiate the volume formula with respect to time and then substitute the given values.
First, differentiate the volume formula:
dV/dt = (1/3)π(2rh(dr/dt) + r^2(dh/dt))
Given values:
r = 4cm
h = 6cm (since the beer is 6cm deep)
dr/dt = 0cm/min (since there is no change in the radius)
dh/dt = ??? (rate at which the height is changing)
Now substitute the values into the formula and solve for dh/dt:
dV/dt = (1/3)π(2(4)(0) + 4^2(dh/dt))
-4 = 16(dh/dt)
dh/dt = -4/16 = -1/4 cm/min
Therefore, the level of the beer is falling at a rate of 1/4 cm/min when the beer is 6cm deep.
(b) To find when the hat is half empty, we need to determine the height at which the volume of the beer is half of the total volume of the cone.
Total volume of the cone:
V = (1/3)π(4^2)(12) = 64π cm^3
Half of the volume:
64π/2 = 32π cm^3
Using the volume formula, solve for the height when the volume is 32π cm^3:
32π = (1/3)π(4^2)(h)
32 = 8h
h = 4 cm
Therefore, the hat is half empty when the height of the beer is 4 cm.