Question Number 60439 by ajfour last updated on 21/May/19
Commented by ajfour last updated on 21/May/19
$$\mathrm{Find}\:\mathrm{length}\:\mathrm{AB}.\:\mathrm{The}\:\mathrm{curve}\:\mathrm{is}\:\mathrm{cubic}. \\ $$
Answered by tanmay last updated on 21/May/19
Commented by ajfour last updated on 21/May/19
$$\mathrm{thanks}\:\mathrm{Tanmay}\:\mathrm{Sir}. \\ $$
Commented by tanmay last updated on 21/May/19
$${most}\:{welcome}\:{sir}… \\ $$
Answered by mr W last updated on 21/May/19
$${f}\left({x}\right)={a}\left({x}−\mathrm{4}\right)\left({x}−\mathrm{2}\right)\left({x}+\mathrm{1}\right) \\ $$$${f}\left(\mathrm{0}\right)={a}\left(−\mathrm{4}\right)\left(−\mathrm{2}\right)\left(\mathrm{1}\right)=\mathrm{1} \\ $$$$\Rightarrow{a}=\frac{\mathrm{1}}{\mathrm{8}} \\ $$$$\Rightarrow{f}\left({x}\right)=\frac{\mathrm{1}}{\mathrm{8}}\left({x}−\mathrm{4}\right)\left({x}−\mathrm{2}\right)\left({x}+\mathrm{1}\right) \\ $$$$\Rightarrow{f}\left({x}\right)=\frac{\mathrm{1}}{\mathrm{8}}\left({x}^{\mathrm{3}} −\mathrm{5}{x}^{\mathrm{2}} +\mathrm{2}{x}+\mathrm{8}\right) \\ $$$${f}'\left({x}\right)=\frac{\mathrm{1}}{\mathrm{8}}\left(\mathrm{3}{x}^{\mathrm{2}} −\mathrm{10}{x}+\mathrm{2}\right) \\ $$$${A}\left({h},{k}\right) \\ $$$${k}=\frac{\mathrm{1}}{\mathrm{8}}\left({h}^{\mathrm{3}} −\mathrm{5}{h}^{\mathrm{2}} +\mathrm{2}{h}+\mathrm{8}\right) \\ $$$${m}_{{A}} =\frac{\mathrm{1}}{\mathrm{8}}\left(\mathrm{3}{h}^{\mathrm{2}} −\mathrm{10}{h}+\mathrm{2}\right) \\ $$$${eqn}.\:{of}\:{AB}: \\ $$$${y}−{k}={m}_{{A}} \left({x}−{h}\right) \\ $$$${B}\left(\mathrm{4},\mathrm{0}\right) \\ $$$$\mathrm{0}−{k}={m}_{{A}} \left(\mathrm{4}−{h}\right) \\ $$$$\mathrm{0}−\frac{\mathrm{1}}{\mathrm{8}}\left({h}^{\mathrm{3}} −\mathrm{5}{h}^{\mathrm{2}} +\mathrm{2}{h}+\mathrm{8}\right)=\frac{\mathrm{1}}{\mathrm{8}}\left(\mathrm{3}{h}^{\mathrm{2}} −\mathrm{10}{h}+\mathrm{2}\right)\left(\mathrm{4}−{h}\right) \\ $$$$−{h}^{\mathrm{3}} +\mathrm{5}{h}^{\mathrm{2}} −\mathrm{2}{h}−\mathrm{8}=−\mathrm{3}{h}^{\mathrm{3}} +\mathrm{22}{h}^{\mathrm{2}} −\mathrm{42}{h}+\mathrm{8} \\ $$$$\Rightarrow\mathrm{2}{h}^{\mathrm{3}} −\mathrm{17}{h}^{\mathrm{2}} +\mathrm{40}{h}−\mathrm{16}=\mathrm{0} \\ $$$$\Rightarrow{h}=\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$\Rightarrow{k}=\frac{\mathrm{1}}{\mathrm{8}}\left(\frac{\mathrm{1}}{\mathrm{8}}−\frac{\mathrm{5}}{\mathrm{4}}+\mathrm{1}+\mathrm{8}\right)=\frac{\mathrm{63}}{\mathrm{64}} \\ $$$$\Rightarrow{AB}=\sqrt{\left(\mathrm{4}−\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{2}} +\left(\frac{\mathrm{63}}{\mathrm{64}}\right)^{\mathrm{2}} }=\frac{\mathrm{7}\sqrt{\mathrm{1105}}}{\mathrm{64}}=\mathrm{3}.\mathrm{64} \\ $$
Commented by ajfour last updated on 21/May/19
$$\mathrm{thanks}\:\mathrm{sir}. \\ $$