# What is the rate of change of the height of water in a conical frustum bucket after one minute if it is being filled at a constant rate?

I have an embarrassing question to ask. Embarrassing in the sense that I should be able to confirm or refute the solutions I got but I just can't seem to rationalise the results I worked out.

I am working on a tutorial for some of my students and I want to know if I have the Mathematics right because the numbers I end up with seem a bit small.

And before people start telling me, I know there are better ways to do this problem but it is more about showing the students how similar triangles can be used in these related rates type problems that they will get in their exam. Don't shoot the messenger - I don't write the curriculum.

The problem is set out in this PowerPoint (or see below because apparently PPT is the tool of the devil) as a number of tutors will deliver this to the different classes.

I greatly appreciate any help I may get.

Pete

[edited here - lucky I do LateX]

The volume of a conical frustum bucket is given by $$V=\frac{\pi h}{3}(R^2+Rr+r^2)$$ given $r=15$ cm, $R=22$ cm & $h=56$ cm what is the rate of change of the height of water in the bucket after one minute if it is being filled at a constant rate $\frac{dV}{dt}=0.3$ L/s, from empty.

Eliminating $R$ from the equation above using similar triangles and letting $R=15+\frac{h}{8}$ we get $$V=225\pi h + \tfrac{15\pi h^2}{8}+\tfrac{\,\pi h^3}{192}$$ $$\Rightarrow \tfrac{dV}{dt}=(225\pi+\tfrac{15\pi h}{4}+\tfrac{\pi h^2}{64})\tfrac{dh}{dt}$$

After 1 minute $V=18$ litres $=18000$ cm$^3 \Rightarrow h\approx 21.4$ cm.

Solving for $\frac{dh}{dt}$ gives $$\tfrac{dh}{dt} =\frac{0.3}{225\pi+\tfrac{15\pi h}{4}+\tfrac{\pi h^2}{64}}$$ and hence $\frac{dh}{dt}\approx0.0031$cm/s

This is what seems a bit 'low' to me - I am looking at the horse bucket in my mind as I fill it ans it goes faster than that - doesn't it?

• Please can you write the question in your question .... please don't refer us to PowerPoints ... that is an evil medium $\ddot\frown$. – Lord Shark the Unknown Sep 25 '17 at 6:45
• Reminds me of joffrey shouting "I am the king!" – samjoe Sep 25 '17 at 6:53
• If you can't be bothered to write out the problem here, why should we be bothered to follow a link? – Qudit Sep 25 '17 at 6:53
• no need to get snippy people ... I have edited my original post. – Pete Langshaw Sep 25 '17 at 7:00
• Welcome to MSE, below there is an answer, here there are some advices for you: 1) always edit a title summarizing your problem, for example Calculating a volume... and add those tags that you need; 2) you can write an introductory paragraph (saying why you need to solve the problem) or provide your explanation in comments if some user asks you about if it is an assignment, 3) also it is good if emphasize your question inside of a section Question. I hope don't disturb you with this comment. Good luck. – user243301 Sep 25 '17 at 7:29

## 1 Answer

With $R = 15 + h/8$ and $r = 15$ you should get $$V = {\frac {3\,\pi\,{h}^{3}}{64}}+{\frac {25\,\pi\,{h}^{2}}{4}}+ 150 \pi h$$ Note also that this volume is measured in cubic centimetres, not litres, so $dV/dt = 300\; cm^3/s$, not $0.3$.

• The question asks about rate of change in the height of water. How can the answer be volume/second? – Stephen Friedrich Sep 25 '17 at 11:38
• @StephenFriedrich the answer points out that dV/dt is in cubic cm per second not that h is in those units. It says nothing about the units of the answer except by implication. – MD-Tech Sep 25 '17 at 12:35
• Wolfram Alpha agrees with the volume formula given in the question: wolframalpha.com/input/… – David K Sep 25 '17 at 13:07
• I think the units mentioned here are the real answer to the question. OP introduced a mistaken factor of 1000, and noticed that the result wasn't right. – David K Sep 25 '17 at 13:08