Question
A reservoir is in the form og the frustum of a cone with upper base of radius 8ft and lower base radius of 4 ft and altitude of 10ft. The water in the reservoir is xft deep. If the level of the water is increasing at 4ft/min., how fast is the volume of water in the reservoir is increasing when its depth is 2ft.?
Answer: 100pi ft^3/min..
Help please, i need to see the solution...
Answer: 100pi ft^3/min..
Help please, i need to see the solution...
Answers
when the water is x feet deep, the surface has radius 4+(8-4)x/10 = 4+x/5
The volume of the cut-off part of the cone is 1/3 pi *16*10 = 160/3 pi
So, the volume of water is
v = 1/3 pi (4+x/5)^2 (x+10) - 160pi/3
= pi/75 x^3 + 2pi/3 x^2 + 32pi/3 x
dv/dt = (pi/25 x^2 + 4pi/3 x + 32pi/3) dx/dt
when x=2,
dv/dt = (4pi/25 + 8pi/3 + 32pi/3)(4)
= 4048/75 pi
The volume of the cut-off part of the cone is 1/3 pi *16*10 = 160/3 pi
So, the volume of water is
v = 1/3 pi (4+x/5)^2 (x+10) - 160pi/3
= pi/75 x^3 + 2pi/3 x^2 + 32pi/3 x
dv/dt = (pi/25 x^2 + 4pi/3 x + 32pi/3) dx/dt
when x=2,
dv/dt = (4pi/25 + 8pi/3 + 32pi/3)(4)
= 4048/75 pi
100pi ft^3/min
Solution pls
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