anonymous
  • anonymous
A conical tank is leaking water at the rate of 75 cubic inc/min. At the same time water is being pumped into the tank at a constant rate.The tanks height is 60 in whiles its top diameter is 20 inches.If the water level is rising at a rate of 5in/min when the height of the water is 10 inches high,find the rate in which water is being pumped into the tank to the nearest cubic inhes/min.The volume of a cone is given by( v= 1/3pi*r^2*h)
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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amistre64
  • amistre64
A conical tank is leaking water at the rate of 75 cubic inc/min. At the same time water is being pumped into the tank at a constant rate. The tanks height is 60 in top diameter is 20 inches. If the water level is "rising" at a rate of 5in/min when the height of the water is 10 inches high find the rate in which water is being pumped into the tank to the nearest cubic inhes/min.The volume of a cone is given by( v= 1/3pi*r^2*h) thats easier to sort thru to me
amistre64
  • amistre64
Since the volume in the tank at any given time depends on how enters and exits it, we might be able to construct an equation to work with
amistre64
  • amistre64
we might as well start by deriving the given volume equation V= 1/3pi*r^2*h V' =2/3 pir h r' + 1/3 pir^2 h' h' is the rate of rising = 5; h is given to be 10; so the radius at that moment is: h/r = 6, r = h/6; r = 10/6, 5/3 V' =10(2)(5)/3(3) pi r' + 5/3 pi (5/3)^2

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amistre64
  • amistre64
since r = h/6 ; r' = h'/6; = 5/6 \[V' =\frac{10(2)(5)(5)}{3(3)(6)} pi + \frac{5\cdot 5^2}{3 \cdot 3^2} pi\] \[V' =\frac{250}{27} pi + \frac{125}{27} pi\] \[V' =\frac{375}{27} pi;\text{ at the given moment specfied}\]
amistre64
  • amistre64
we know the rate at which its leaving is: -75 so my hunch is to separate out the coming from the going to establish the rate of change that was determined from V' \[V' = in' - out'=\frac{3(125)}{3(9)}pi \] \[in' - 75=\frac{125}{9}pi \] \[in'=\frac{125}{9}pi+75 \]
amistre64
  • amistre64
dunno how right it is, but thats how i would go about trying to solve it
anonymous
  • anonymous
hey sorry im back i saw a few different ways but idk i first thought of doing some type of proportions?
anonymous
  • anonymous
amistre that was dead on just checked ur answer is one of the choices
amistre64
  • amistre64
cool :)

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