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Water is leaking out of an inverted conical tank at a rate of 9,500 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. (Round your answer to the nearest integer)

My answer was 288,752 cm3/min

However, the answer is wrong. Can someone help me?

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  • $\begingroup$ my answer is 269753, let me know if it is right please $\endgroup$ – Mirko Mar 16 '15 at 20:47
  • $\begingroup$ No, it is wromg $\endgroup$ – user137452 Mar 16 '15 at 20:51
  • $\begingroup$ actually I got your answer (when I added the 9500 which I should have done, instead of subtracting it at a certain step), except you may not have correctly rounded. Writing it down, will post in a minute. $\endgroup$ – Mirko Mar 16 '15 at 20:59
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Let $h$ denote the height, and $r$ the radius of the part of the cone that is filled with water. Since the total height is 6 and the radius at the top is 2, using proportions we have $r=\frac h3$. The volume of cone is $V=\pi r^2 h/3 = \pi h^3/27$. Here both $V$ and $h$ are functions of the time $t$, so differentiating both sides with respect to $t$ we get $V'=3h^2\pi h'/27=h^2\pi h'/9$. It is given that $h'=20$ when $h=2 m = 200 cm$, so we have $V'= 200^2\cdot 20\pi /9 \approx 279252.68$. Adding the 9500 we get $288752.68\approx 288753$. (It appears to me you got the answer mostly right, except the "Round your answer to the nearest integer" part.)

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  • $\begingroup$ If we set up (but do not solve) for the volume of the cone using the disc method, one of the Fundamental Theorems of Calculus tells us that $dV/dh$ is equal to the area of the surface of the water when it reaches height $h$. Calculating the radius at $h=2$ to be $200/3$, and given $h'=20$, we get $V'= 200^2\cdot 20\pi/9$ by simple application of the chain rule. $\endgroup$ – David K Mar 16 '15 at 21:12
  • $\begingroup$ @DavidK yes, but the title said "Related Rate Question" so I felt obliged to use related rates. Just kidding :) It's a bit difficult for me to formally follow the use of FTC, though it is easier to convince myself of what you are saying taking $\Delta V \approx S\cdot \Delta h$ (just thinking of the picture geometrically, using infinitesimals), where $S$ denotes the surface area of water. $\endgroup$ – Mirko Mar 16 '15 at 21:26
  • $\begingroup$ The FTC in this case just formalizes what your intuition says correctly about $\Delta V\approx S$. My comment was merely a (hopefully interesting) footnote to your answer, not a criticism (in fact I had already given a +1 to the answer). $\endgroup$ – David K Mar 16 '15 at 21:27

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