# Answer on Calculus Question for Muir

Question #2183

Water is leaking out of an inverted conical tank at a rate of 10000 cm

^{3}/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.(cm^{3}/min)Expert's answer

Tank volume -is the volume of the circular cone (1/3)*π*r

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*r

V(h) = (1/3)*π*((1/3)*h)

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)

dV = (1/9)*π*h

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)

"Water is leaking out of an inverted conical tank at a rate of 10 000 cm

dV = rate in - rate out = dIN - 10000 cm

dIN - 10000 cm

hence dIN=289 000 cm

^{2}*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*r

^{2}*hV(h) = (1/3)*π*((1/3)*h)

^{2}*h = (1/27)*π*h^{3}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*h

^{2}*dhNow 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 cm

^{3}/min at the same time that water is being pumped into the tank at a constant rate."dV = rate in - rate out = dIN - 10000 cm

^{3}/mindIN - 10000 cm

^{3}/min = (1/9)*π*(2 m)^{2}*(20 cm/min) =(1/9)*π*(200 cm)^{2}*(20 cm/min)=279 000 cm^{3}/minhence dIN=289 000 cm

^{3}/min - the rate at which water is being pumped into the tank .Need a fast expert's response?

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## Comments

Assignment Expert30.10.2012 09:22Great if it helped! You're welcome.

If you really liked our service please press like-button beside answer field. Thank you!

kathy29.10.2012 22:48Oh sorr, i got it now!

Assignment Expert26.10.2012 07:57If 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 Expert26.10.2012 07:52Dear visitor

Thanks for correcting us!

jojo26.10.2012 01:09where did dV = (1/9)*π*(2 m)2*(20 cm/min) come from? someone explain?

Jasmine23.10.2012 11:58Wouldn't you add 10,000 to both sides, making dIN 289,000?

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