water in cone-shaped watertank

N

noddy

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Oct 11, 2011
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My problem is
water flows out of a conical tank at a rate of 10cm^3/min. the cone stands vertically with the tip down. radius of the tank is 2m and the height is 6m.
a)how fast will the radius of the water surface change when the water is 4.5m deep?
b)how fast in cm / min decreases the water?

I got so faar
h/r = 6/2 =3 h=3r
du/dt = -10cm^3/min
v(r(t))= (pi * r^2 *h )/ 3 = pi*r^3
du/dt=du/dr*dr/dt = -10=3pi*(1.5)^2 dr/dt


anyone can help me?
 
noddy said:
My problem is
water flows out of a conical tank at a rate of 10cm^3/min. the cone stands vertically with the tip down. radius of the tank is 2m and the height is 6m.
a)how fast will the radius of the water surface change when the water is 4.5m deep?
b)how fast in cm / min decreases the water?

I got so faar
h/r = 6/2 =3 h=3r
du/dt = -10cm^3/min <<<< What is "u"?
v(r(t))= (pi * r^2 *h )/ 3 = pi*r^3
du/dt=du/dr*dr/dt = -10=3pi*(1.5)^2 dr/dt


anyone can help me?
Click to expand...

Exactly where are you Stuck?
 
at du/dt=du/dr*dr/dt = -10=3pi*(1.5)^2 dr/dt
don't know what to do next with
-10=3pi*(1,5)^2 dr /dt
 
this
a)how fast will the radius of the water surface change when the water is 4.5m deep?
b)how fast in cm / min decreases the water?
 
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