Chapter 6, Problem 10P

To determine

To find : The radius of the ball that causes the greatest volume of water to spill out of the cup.

Answer to Problem 10P

The required $R$ is $R=\frac{h\sin \theta }{1+\sin \theta -2{\sin }^2\theta }$

Explanation of Solution

Given information :

A paper drinking cup filled with water has the shape of a cone with height $h$ and semi-vertical angle $\theta$ .

A ball is placed carefully in the cup, thereby displacing some of the water and making it overflow.

  

Formula used :

  $\sin \theta =\frac{perpendicular}{hypotenuse}$

  

Volume of the displaced water $={\displaystyle \int \pi y^{2}dyy=f\left(x\right)}$

For critical point of volume $\frac{dv}{dx}=0$

Calculation :

As in the diagram we place the cone horizontally with its tip at the origin. Let, the radius of the ball be $R$ and the x-coordinate of its centre be $L$ . The volume of displaced water is the volume of revolution that is obtained by rotating the upper part of the circle, we have

  $\frac{R}{L}=\sin \theta \Rightarrow R=L\sin \theta$

The equation of the circle is ${\left(x-L\right)}^2=R^2$ and the upper half is the graph of the function $y=\sqrt{R^2-{\left(x-L\right)}^2}$

Using the washer method, we find the volume of displaced water

  $\begin{array}{c}v={\displaystyle \underset{L-R}{\overset{h}{\int }}\pi \left(\sqrt{R^2-{\left(x-L\right)}^2}\right)}dx\\ =\pi {\displaystyle \underset{L-R}{\overset{h}{\int }}\left(\sqrt{R^2-{\left(x-L\right)}^2}\right)}dx\\ =\pi {{\displaystyle \left[R^{2}x-\frac{{\left(x-L\right)}^2}{3}\right]}}_{L-R}^h\\ =\pi \left[R^{2}h-\frac{{\left(h-L\right)}^2}{3}-R^2\left(L-R\right)-\frac{R^3}{3}\right]\end{array}$

  $\begin{array}{l}=\pi \left[{\left(L\sin \theta \right)}^{2}h-\frac{{\left(h-L\right)}^3}{3}-\left(L\sin \theta \right)\left(L-L\sin \theta \right)-{\left(\frac{L\sin \theta }{3}\right)}^3\right]\text{ Putting }R=L\sin \theta \\ =\pi \left[L^2\sin \theta \cdot h-\frac{{\left(h-L\right)}^3}{3}-L^3{\sin }^2\theta \left(1-\sin \theta \right)-\frac{L^3{\sin }^3\theta }{3}\right]\mathrm{............}\left(i\right)\end{array}$

Volume of water spill is greatest or least when $\frac{dv}{dL}=0$

  $\begin{array}{l}\frac{dv}{dL}=0\\ \Rightarrow \pi \left[2Lh{\sin }^2\theta +{\left(h-L\right)}^2-3L^2{\sin }^2\theta \left(1-\sin \theta \right)-L^2{\sin }^3\theta \right]=0\\ \Rightarrow {\left(h-L\right)}^2-3L^2{\sin }^2\theta +2Lh{\sin }^2\theta +2L^2{\sin }^3\theta =0\\ \Rightarrow {\left(h-L\right)}^2-L^2{\sin }^2\theta -2L^2{\sin }^2\theta +2Lh{\sin }^2\theta +2L^2{\sin }^3\theta =0\\ \Rightarrow \left[{\left(h-L\right)}^2-L^2{\sin }^2\theta \right]+\left[-2L^2{\sin }^2\theta +2Lh{\sin }^2\theta +2L^2{\sin }^3\theta \right]=0\\ \Rightarrow \left(h-L-L\sin \theta \right)\left(h-L-L\sin \theta \right)+2L{\sin }^2\theta \left(h-L-L\sin \theta \right)=0\\ \Rightarrow \left(h-L-L\sin \theta \right)\left[h-L-L\sin \theta +2L{\sin }^2\theta \right]=0\end{array}$

  $\begin{array}{l}\Rightarrow \text{Either }L\left(1-\sin \theta \right)=h\\ \text{ L=}\frac{h}{1-\sin \theta }\end{array}$

  $\begin{array}{l}\Rightarrow Or,L-h+L\sin \theta -2L{\sin }^2\theta =0\\ \Rightarrow L=\frac{h}{1+\sin \theta -2{\sin }^2\theta }\end{array}$

When, $\text{L=}\frac{h}{1-\sin \theta }$ ,

  $\begin{array}{l}v=\pi \left[{\left(\frac{h}{1-\sin \theta }\right)}^2\cdot {\sin }^2\theta h-\frac{{\left(h-\frac{h}{1-\sin \theta }\right)}^3}{3}-{\left(\frac{h}{1-\sin \theta }\right)}^3{\sin }^2\theta \left(1-\sin \theta \right)-\frac{{\left(1-\frac{h}{\sin \theta }\right)}^3{\sin }^3\theta }{3}\right]\\ =\pi \left[\frac{h^3{\sin }^2\theta }{{\left(1-\sin \theta \right)}^2}+\frac{h^3{\sin }^3\theta }{3{\left(1-\sin \theta \right)}^2}-\frac{h^3{\sin }^2\theta }{{\left(1-\sin \theta \right)}^2}-\frac{h^3{\sin }^3\theta }{3{\left(1-\sin \theta \right)}^2}\right]=0\end{array}$

So, at $L=\frac{h}{1+\sin \theta -2{\sin }^2\theta }$ , $v$ reaches its maximum value,

i.e. when $R=L\sin \theta$

  $\Rightarrow R=\frac{h\sin \theta }{1+\sin \theta -2{\sin }^2\theta }$ , the maximum water will be displaced.