Rate at which a bucket gets filled

[brotherbobby]

Homework Statement

Water from a tap drops into a bucket at the rate of $1\;cm^3/s$. The bucket is in the shape of an overturned right circular "cone", with a base diameter of 16 cm and its top diameter of 40 cm, as the picture below shows. It is 30 cm tall. At a height of $10\; cm$, calculate the $\mathbf{\text{rate}}$ at which water climbs up the sides of the bucket in $cm/s$.

Relevant Equations

1. $y(x)$ is a function, given or found. The rate at which variable $y$ changes with $x$ at a given point $x_0$ is given by $\left.\frac{dy}{dx} \right|_{x=x_0}$.
2. If triangle $ABC$ is similar to triangle $DEF$, the ratio of their corresponding sides are proportional. Thus, for example, $\frac{AB}{DE} = \frac{BC}{EF}$.
3. The volume of a (right circular) cone is given by : $V_{\text{cone}} = \dfrac{1}{3} \pi r^2 h$, the symbols having their usual meanings.

Problem statement : The problem statement is given above. I draw the image to the right. Note I have drawn the radii which is half the given diameters.

Attempt : I wonder right at the start if the height of the bucket, 30 cm, is relevant. Though I am tentative, it is also clear to me right away that, because the volume flow rate is a constant ($1\;cm^3/s$), the rate of increase of water is highest at the base (smallest area) and lowest at the top (largest area).

What is the volume $V(t)$ of the liquid at any given instant of time? That seems to be the problem. How is this volume related to the height of the liquid at that time, $y(t)$?
We know the rate at which volume increases : $\frac{dV}{dt} = 1\;cm^3/s$.

To progress further, for my own convenience, I extend the the bucket "beyond" its base where its slant height and "midrib" meet. I paste the figure to the left. The bucket is given by ABCD and the water surface by UV. Both DB and CA meet "below" the bucket at O. The bucket has radii $\mathbf{R_1,R_2}$ and height $\mathbf{H}$; the water surface a momentary radius of $\mathbf{r}$ and a height $\mathbf{y}$. The point O is at a depth $\mathbf{x}$ from the base AB.

From similar triangles AOB and COD, we have $\frac{R_2}{x+H} = \frac{R_1}{x}\Rightarrow x = \frac{R_1}{R_2-R_1}H= \frac{8}{20-8}\times 30= 20\;\text{cm}$.

What is the (instantaneous) volume of water in the bucket? Calling that volume as $V$ and realising that it is a function of the time $t$, we see that this volume is given by the "difference" in volumes of the "cones" formed by rotating UVO and ABO about the vertial axis. Remembering the formula for the volume of a cone, we find that the instantaneous volume of water $$V= \dfrac{\pi}{3}r^2(x+y)-\dfrac{\pi}{3}R_1^2x= \dfrac{\pi}{3}\left[ r^2(x+y) - R_1^2x \right].$$ From similar $\triangle's\; \mathbf{UVO}\; \text{and}\; \mathbf{ABO}$, we have $\frac{r}{x+y} = \frac{R_1}{x}\Rightarrow r^2 = \left(1+\frac{y}{x}\right)^2 R_1^2$. Substituting this value for $r$ in the expression for the volume of water above, $$V = \dfrac{\pi}{3} R_1^2 \left[ (x+y) (1+y/x)^2-x\right] = \dfrac{\pi}{3} R_1^2\left[ (x+y) \left(1+\frac{2y}{x}+\frac{y^2}{x^2}\right) - x\right]= \dfrac{\pi}{3} R_1^2 \left[3y+\frac{3y^2}{x}+\frac{y^3}{x^2}\right]$$ after expanding the brackets and some algebra.
Differentiating both sides relative to time, and using $\dot x=\frac{dx}{dt}$, we have $$\dot V = \dfrac{\pi}{3} R_1^2\left[3\dot y+\frac{6y\dot y}{x}+\frac{3y^2\dot y}{x^2}\right]=\pi R_1^2 \dot y\left(1+\frac{y}{x}\right)^2\Rightarrow \dot y = \frac{\dot V}{\pi R_1^2} \frac{1}{\left(1+\frac{y}{x} \right)^2}$$

Putting values, we get $\dot y = \frac{1}{\pi \times 8^2}\times \frac{1}{\left(1+\frac{10}{20} \right)^2} = \boxed{2.21\times 10^{-3}\; \text{cm/s}}$.

Ignoring the answer, is my line of reasoning correct?

 

[.Scott]

You have the right answer (although not precisely).

Calculate the area of the water surface.
PS: You have already seriously over-thought this thing.

 

[pbuk]

Calculate the area of the water surface.
PS: You have already seriously over-thought this thing.

Yes, @brotherbobby your line of reasoning is over-complicated: you don't need calculus for this. You should be able to write down the value "12cm" pretty much straight away: can you see how? And can you see why it is important?

 

[brotherbobby]

You have the right answer (although not precisely).

Calculate the area of the water surface.
PS: You have already seriously over-thought this thing.

I suppose what you are trying to make me realize is that we could use the fact that the rate of volume flow (which is a constant) at any point $\dot V = A(t) v(t)$, where $A(t)$ is the instantaneous cross sectional area of water and $v(t)$ the instantaneous velocity of water.

Attempt : I put the image again. To find the area of cross section of the water ($r$), we need to use the similar triangles UVO and ABO. I have done that above to get $r = \left(1+\frac{y}{x}\right)R_1$.

The area of cross section of the water front : $A(t) = \pi \left(1+\frac{y}{x}\right)^2R^2_1= \pi \times \left(1+\frac{10}{20}\right)^2\times 8^2 = 452.4\; \text{cm}^2$.

Given $\dot V = A(t)v(t)\Rightarrow v(t) = \frac{\dot V}{A(t)} = \frac{1\;\text{cm}^3\text{/s}}{452.4\;\text{cm}^2} = \boxed{2.210\times 10^{-3}\;\text{cm/s}}$.

It is the same to the answer I got but with much less effort. Thank you for the help.

 

[.Scott]

I took note of the fact that the radius was 1/3 of the way from 8cm to 20cm, thus 8+(1/3)(20-8) = 8+4 = 12cm
So the area is 144pi cm².

Also: My comment about imprecise was inaccurate. I must have mistyped pi.