Problem of the Week #6

■  No GDC  ■

The diagram below shows two circles each with a radius of one unit.  One circle is fixed with its centre at the origin \(\left( {0,\,0} \right)\).  The other circle has its centre at the point B\(\left( {b,\,0} \right)\) such that \(0 \le b \le 2,\;b \in \mathbb{R}\).  Changing the value of \(b\) causes the circle with centre B to move along the x-axis and for the area of the region where the circles overlap to increase or decrease accordingly.  Let \(A\left( b \right)\) be a function that expresses the area of the overlapping region in terms of \(b\).  [*part (c) is for HL only]

(a)   Show that \(A\left( b \right)\) can be written as \(A\left( b \right) = 2{\cos ^{ - 1}}\left( {\frac{b}{2}} \right) - \frac{b}{2}\sqrt {4 - {b^2}} \)
(b)   Evaluate \(A\left( 0 \right)\) and \(A\left( 2 \right)\).

(c)   Show that \(\frac{{{\textrm{d}}A}}{{{\textrm{d}}b}} = - \sqrt {4 - {b^2}} \).   [HL only]

(d)   As the value of \(b\) increases from 0 to 2 the area of the overlapping region continually decreases.  Find the exact value of \(b\) such that \(\frac{{{\textrm{d}}A}}{{{\textrm{d}}b}} = - 1\).