Chapter 6, Problem 8P

To determine

To find : The height $h$ at which a horizontal line is placed so that the shaded area is minimized.

Answer to Problem 8P

The required height is $\frac{\sqrt{3}}{2}$ at which shadow area is minimum.

Explanation of Solution

Given information :

Semicircle with radius $1$ , horizontal diameter $PQ$ , and the tangent lines at $P$ , $Q$ .

  

Formula used :

  $\begin{array}{l}A={\displaystyle \underset{0}{\overset{b}{\int }}f\left(t\right)}dt\\ \Rightarrow \frac{dA}{dt}=f\left(b\right)\end{array}$

To find extreme values we put first order derivative as $0$

Calculation :

We shift the origin at centre of the semicircle considering $PQ$ AS X-axis and a vertical line through centre of the semicircle as Y-axis.

  

Now, the figure shows that the shadow region has been divided into four parts $A_1,A_2,A_3,A_4$

Note that, $A_1=A_4$ and $A_2=A_3$

Therefore, the area of the shadow region $A=2\left(A_3+A_4\right)$

  $A_3={\displaystyle \underset{0}{\overset{h}{\int }}\left[1-\sqrt{1-y^2}\right]}dy$ and $A_4={\displaystyle \underset{1}{\overset{h}{\int }}\sqrt{1-y^2}}dy$

So, $A\left(h\right)=2\left[{\displaystyle \underset{0}{\overset{h}{\int }}\left[1-\sqrt{1-y^2}\right]}dy-{\displaystyle \underset{1}{\overset{h}{\int }}\sqrt{1-y^2}}dy\right]$

To minimize the shadow area, differentiate with respect to $h$

  $\frac{dA\left(h\right)}{dh}=2\left[\left(1-\sqrt{1-h^2}\right)-\sqrt{1-h^2}\right]\text{ [using fundamental theorem of calculus]}$

Now, $\frac{dA\left(h\right)}{dh}=0$

  $\begin{array}{l}\Rightarrow \left(1-\sqrt{1-h^2}\right)-\sqrt{1-h^2}=0\\ \Rightarrow \sqrt{1-h^2}=\frac{1}{2}\\ \Rightarrow 1-h^2=\frac{1}{4}\\ \Rightarrow h^2=\frac{3}{4}\\ \Rightarrow h=\frac{\sqrt{3}}{2}\text{ [}h\text{ cannot be negative for first quadrant]}\end{array}$

So, the required height is $\frac{\sqrt{3}}{2}$ at which shadow area is minimum.