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water flows into an inverted right circular conical tank at the rate of 2 cubic feet per minute. If the altitude of the tank is 20 ft and the radius of its base is 10ft., at what rate is the water level rising when the tank is 1/8 full?

So I was able to obtain V when it is 1/8 full. Which is 261.799. dv/dt is 2ft^3/min as in V for volume.

Seeking dh/dt when V is 261.799. Where h is the height. I also notice the relationship between h and r. h=2r.

I'm having trouble find h and r at 1/8 full.

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  • $\begingroup$ thank you all of those who helped me $\endgroup$ – EagleTamer Dec 18 '13 at 1:54
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The volume is given by:

$V=\frac{1}{3}\pi r^2h$

You are given $\frac{dV}{dt}$, so you will have to take the derivative of the volume function with respect to time. Keep in mind that the radius and the height are NOT constants, they are variables. However, they are proportional:

$\frac{h_1}{r_1}=\frac{h_2}{r_2}$

This is because, if you draw a triangle straight down in the cone, you'll get similar triangles, as I, a true artist, have demonstrated below:

similar triangles

This is true for RIGHT cones.

We are given the cone's actual height and radius, and so we know that:

$\frac{h}{r}=\frac{20}{10}=2$

Hence:

$r=\frac{h}{2}$

I found what r is because I want to use it in the equation because we are seeking for the change in the height over time. Let's plug it into the original equation, then:

$V=\frac{1}{3}\pi (\frac{h}{2})^2h=\frac{1}{12}\pi h^3$

Take the derivative of this with respect to time:

$\frac{dV}{dt}=\frac{1}{4}\pi h^2 * \frac{dh}{dt}$

This by the way always happens, we just never see it. Take for instance:

$y= 5x^2+x$

When you differentiate, the answer is really:

$\frac{dy}{dx}=10x*\frac{dx}{dx}+\frac{dx}{dx}$

But $\frac{dx}{dx} = 1$ so we never show it.

Back to what I was saying, the next step is to determine $\frac{dh}{dt}$, which is what we are trying to solve. So we need the instantaneous height of the cone when it's 1/8 filled. 1/8 filled means the cone's volume is 1/8 of its original volume. Its original volume is (and you can use $V=\frac{1}{3}\pi r^2h$, but I find the function of volume against height only easier since there's only 1 variable):

$V=\frac{1}{12}\pi h^3$

$V_{full}=\frac{1}{12}\pi (20)^3 =$ some value I stored on my calculator.

Multiply that by 1/8 and you get the volume of the tank at that time. Use the equation to find the instantaneous height:

$V_{instantaneous}=\frac{1}{12}\pi h^3$

The height is 10. You can now plug it into the derivative equation we had:

$\frac{dV}{dt}=\frac{1}{4}\pi h^2 * \frac{dh}{dt}$

We are given that $\frac{dV}{dt}=2$, and we now know the instantaneous height. The answer is then:

$\frac{dh}{dt}=\frac{2}{25 \pi} \approx 0.025 \frac{ft}{min}$

Let me know if I made any mistakes before you down vote me into oblivion.

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  • $\begingroup$ I almost did the same stuff as you, Ross was helping me as I did it. I differentiated the function when there was still r in it I also had dr/dt, which made things more complicated. Well I have like 3 more related problems left to do :( $\endgroup$ – EagleTamer Dec 18 '13 at 1:54
  • $\begingroup$ They're easy once you get used to them! And this is super important for physics if you're planning to take higher-leveled physics. $\endgroup$ – Shahar Dec 18 '13 at 2:07
  • $\begingroup$ i'm in Ap physics C mechanics atm and let me tell you it is not easy $\endgroup$ – EagleTamer Dec 18 '13 at 2:15
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You have $V=\frac 13\pi r^2 h$. Now use the relation you found between $r$ and $h$ to find that $V$ is proportional to the cube of one of them. You can then solve for it.

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  • $\begingroup$ So since i'm looking for h I substituted r for h. would it be V= 1/3(1/4h^3). (I used 1/3pier^2h) $\endgroup$ – EagleTamer Dec 17 '13 at 22:44
  • $\begingroup$ That is correct except you lost the $\pi$. Now solve for $h$-multiply by $\frac {12}\pi$ and take the cube root. $\endgroup$ – Ross Millikan Dec 17 '13 at 22:56
  • $\begingroup$ so I found that h=10 and r=5. dV/Dt= 2/3πrh(dr/dt) + 1/3πr^2(dh/dt) I have two unknown variables which are dr/dt and dh/dt. I'm trying to find dh/dt, so how can I get dr/dt? $\endgroup$ – EagleTamer Dec 17 '13 at 23:20
  • $\begingroup$ sorry, I meant to write more, but I didn't know you couldn't go to the next line by pressing enter. I added my full statement $\endgroup$ – EagleTamer Dec 17 '13 at 23:25
  • $\begingroup$ Because$ h=2r$, by the chain rule $\frac {dh}{dt}=2\frac {dr}{dt}$ $\endgroup$ – Ross Millikan Dec 17 '13 at 23:30

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