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Answer on Calculus Question for Muir

Question #2183
Water is leaking out of an inverted conical tank at a rate of 10000 cm3/min at the same time that water is being pumped into the tank at a constant rate. The tank has height 6 m and the diameter at the top is 4 m. If the water level is rising at a rate of 20 cm/min when the height of the water is 2 m, find the rate at which water is being pumped into the tank.(cm3/min)
Expert's answer
Tank volume -is the volume of the circular cone (1/3)*π*r2*h.

There's a secret in most such problems. It's the constant ratio of the height of the water to the radius of the top of the water, no matter the depth of the water. That's why the cone is inverted. It doesn't work the other way.

"The tank has height 6 m and the diameter at the top is 4 m."

Height/Radius = 6/2, making Radius = (1/3)*Height.

"If the water level is rising at a rate of 20 cm/min when the height of the water is 2 m"

V(r,h) = (1/3)*π*r2*h

V(h) = (1/3)*π*((1/3)*h)2*h = (1/27)*π*h3

We're talking about change, so we'll need the derivative. Change is over time, so we'll assume that the volume and the height are functions of time.

V(t) = (1/27)*π*h(t)3
dV = (1/9)*π*h2*dh

Now we need dV or dh

"If the water level is rising at a rate of 20 cm/min when the height of the water is 2 m"

then h = 2 m and dh = 20 cm/min

dV = (1/9)*π*(2 m)2*(20 cm/min)

"Water is leaking out of an inverted conical tank at a rate of 10 000 cm3/min at the same time that water is being pumped into the tank at a constant rate."

dV = rate in - rate out = dIN - 10000 cm3/min

dIN - 10000 cm3/min = (1/9)*π*(2 m)2*(20 cm/min) =(1/9)*π*(200 cm)2*(20 cm/min)=279 000 cm3/min
hence dIN=289 000 cm3/min - the rate at which water is being pumped into the tank .

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Comments

Assignment Expert
30.10.2012 09:22

Great if it helped! You're welcome.
If you really liked our service please press like-button beside answer field. Thank you!

kathy
29.10.2012 22:48

Oh sorr, i got it now!

Assignment Expert
26.10.2012 07:57

If we have that V(t) = (1/27)*π*h<sup>3</sup>
then to obtain dV we differentiate V with respect to h:
dV =((1/27)*π*h(t)<sup>3</sup> )'= (1/27)*π*3h(t)<sup>2</sup>*dh= (1/9)*π*h(t)<sup>2</sup>*dh
Also we've mentioned that
"If the water level is rising at a rate of 20 cm/min when the height of the water is 2 m"
then h = 2 m and dh = 20 cm/min

Assignment Expert
26.10.2012 07:52

Dear visitor
Thanks for correcting us!

jojo
26.10.2012 01:09

where did dV = (1/9)*π*(2 m)2*(20 cm/min) come from? someone explain?

Jasmine
23.10.2012 11:58

Wouldn't you add 10,000 to both sides, making dIN 289,000?

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