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- 1A water tank is shaped like a cone with vertex downward. The tank has a radius of 3m & is 5m high. The tank is initially full of water but a valve at the bottom of the tank is opened & water begins to drain. When the height of the water is dropped to 3m, water is flowing out of the tank at a rate of 1.02 m^3/2. How fast is the water level dropping at the moment? Note: Volume of a cone of radius r & height is v=(1/3)(pi)r^2h. A diagram will help solve this."^1" simply means prime
know: V^1(To) = -.02
h(To)= 3
Need to know: h^1(To)
Find an equation relating - Apply volume formula to the cone of water V(t)=(1/3)(pi)*(r(t))^2(h(t))
Law of similar triangles:
3/5 = r(t)/h(t) => r(t) = 3/5h(t)
V(t) = (1/3)(pi)(3/5h(t))^2*h(t)
V(t)=(3pi/25)h(t)^3
V^1(t)=(3pi/25)(3(h(t))^3-1*h^1(t)
Now plug in the numbers
-.02 =(3pi/25)*(3(3)^2)*h^1(To)
h^1(To) = (-.02)(25/81pi)
h(To)= -.002m/s
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- Gosh this stuff is hard
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- Charles Duffy
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