# Answer to Question #2183 in Calculus for Muir

Question #2183
Water is leaking out of an inverted conical tank at a rate of 10000 cm[sup]3[/sup]/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[sup]3[/sup]/min)
1
2012-10-26T07:50:58-0400
Tank volume -is the volume of the circular cone (1/3)*&pi;*r2*h.

There&#039;s a secret in most such problems. It&#039;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&#039;s why the cone is inverted. It doesn&#039;t work the other way.

&quot;The tank has height 6 m and the diameter at the top is 4 m.&quot;

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

V(r,h) = (1/3)*&pi;*r2*h

V(h) = (1/3)*&pi;*((1/3)*h)2*h = (1/27)*&pi;*h3

We&#039;re talking about change, so we&#039;ll need the derivative. Change is over time, so we&#039;ll assume that the volume and the height are functions of time.

V(t) = (1/27)*&pi;*h(t)3
dV = (1/9)*&pi;*h2*dh

Now we need dV or dh

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

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

dV = (1/9)*&pi;*(2 m)2*(20 cm/min)

&quot;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.&quot;

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

dIN - 10000 cm3/min = (1/9)*&pi;*(2 m)2*(20 cm/min) =(1/9)*&pi;*(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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Assignment Expert
30.10.12, 16:22

Great if it helped! You're welcome.
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kathy
30.10.12, 05:48

Oh sorr, i got it now!

Assignment Expert
26.10.12, 15:57

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
26.10.12, 15:52

Dear visitor
Thanks for correcting us!

jojo
26.10.12, 09:09

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

Jasmine
23.10.12, 19:58

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