# P.o.t.W. #9 Solution

SOLUTIONgradient of line, $$m = \frac{{a{c^2} - a{b^2}}}{{c - b}} = \frac{{a\left( {c + b} \right)\left( {c - b} \right)}}{{c - b}} = ac + ab$$equation of line: $$y - a{b^2} = \left( {ac + ab} \right)\left( {x - b} \right)$$$$y = \left( {ac + ab} \right)x - abc - a{b^2} + a{b^2}$$$$y = \left( {ac + ab} \right)x - abc$$area of bounded region $$= \int_b^c {\left\{ {\left[ {\left( {ac + ab} \right)x - abc} \right] - a{x^2}} \right\}} \,dx$$$$= \int_b^c {\left\{ { - a{x^2} + \left( {ac + ab} \right)x - abc} \right\}dx}$$$$= \left. { - \frac{1}{3}a{x^3} + \frac{1}{2}\left( {ac + ab} \right){x^2} - abcx} \right]_b^c$$$$= \left[ { - \frac{1}{3}a{c^3} + \frac{1}{2}\left( {ac + ab} \right){c^2} - ab{c^2}} \right] - \left[ { - \frac{1}{3}a{b^3} + \frac{1}{2}\left( {ac + ab} \right){b^2} - a{b^2}c} \right]$$$$= - \frac{1}{3}a{c^3} + \frac{1}{2}a{c^3} + \frac{1}{2}ab{c^2} - ab{c^2} + \frac{1}{3}a{b^3} - \frac{1}{2}a{b^3} - \frac{1}{2}a{b^2}c + a{b^2}c$$$$= \frac{1}{6}a{c^3} - \frac{1}{6}a{b^3} - \frac{1}{2}ab{c^2} + \frac{1}{2}a{b^2}c$$$$= \frac{a}{6}\left( {{c^3} - {b^3}} \right) - \frac{{abc}}{2}\left( {c - b} \right)$$$$= \frac{a}{6}\left( {c - b} \right)\left( {{c^2} + bc + {b^2}} \right) - \frac{{abc}}{2}\left( {c - b} \right)$$$$= \frac{a}{6}\left( {c - b} \right)\left[ {\left( {{c^2} + bc + {b^2}} \right) - 3bc} \right]$$$$= \frac{a}{6}\left( {c - b} \right)\left( {{c^2} - 2bc + {b^2}} \right)$$$$= \frac{a}{6}\left( {c - b} \right){\left( {c - b} \right)^2}\;\;\; \Rightarrow \;\;\;$$thus, area of bounded region $$= \frac{a}{6}{\left( {c - b} \right)^3}$$ Q.E.D.Comment: For a given value of a, the area of the parabolic segment is determined by the value of $$c - b$$. If the horizontal distance between the points of intersection is constant...

To access the entire contents of this site, you need to log in or subscribe to it.

You can also request a Free trial or check the blog (which is also free)

All materials on this website are for the exclusive use of teachers and students at subscribing schools for the period of their subscription. Any unauthorised copying or posting of materials on other websites is an infringement of our copyright and could result in your account being blocked and legal action being taken against you.