- #1
brotherbobby
- 699
- 163
- 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.
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##.
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?