A Glass Of Water Is Being Filled With Water At The Rate Of 12 Pi Cm^3/sec
And the radius of the top is 6 cm and the radius of the base is 3 cm and the height of the glass is 12 cm , find A) dh/dt and B) radius when the height of water is half?
well, this is a truncated cone. Volume of the full cone is 1/3 area base*height. So would be total height if it were not truncated:
By proportions: heighttotal/areatop=(heighttotal-12)/areabottom
height total=6(heighttotal-12)/3
h=2h-48
h=48,
but r is a function height.
r=3+h/4
dr=dh/4
now looking at incremental volumes
dVolume=PI*r^2*dh where r=3+h/4 or
dvolume=PI (3+h/4)^2 dh
change variables, let u= 3+h/4
du= dh/4 when h=0 u=3, when h=h, u=3+h/4
volume= INT(PI/4) u^2 du over limits
volume= PI/12 [(3+h/4)^3-3^3] check that.
a) dv/dt=PI/4 (3+h/4)^2 dh/dt
h=6, find dh/dt given dv/dt=12PI
b) r when h=6; too easy, we did that already. r=3+6/4