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![]() 750 gallon cone bottom tank only NO STAND US $704.00
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rates of change?
a conical tank (upside down circular prism, cone) has radius 160cm and height 800cm. water is running out of a small of a small hole in the bottom of the tank. when the height 'h' of the water in the tank is 600cm , and the radius is 120 cm, what is the rate of change in is volume 'V' with respect to h?
neat setting out is great appreciate, it does justice to your great minds!
r=radius
h=height
v=volume
Triangle1 (with r=160,h=800) is similar to Triangle2 (with radius r & height h).
So,
r/h=160/800=120/600
r/h=1/5
r=h/5
The volume of a cone =(1/3)(pi)(r^2)h
Substituting r=h/5,
Volume =(1/3)(pi)[(h/5)^2]h
=(1/3)(pi)(h^3/25)
=(pi/75)(h^3)
Differntiating with respect to time,t
dv/dt=(pi/75)[3(h^2)]
dv/dt =(pi/25)(h^2)
At h=600 cm,
dv/dt=(pi/25)(600^2)
dv/dt=45238.93421 cubic cm/s.(I hope the unit of time used here, is seconds)
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