Question Number 206645 by Red1ight last updated on 21/Apr/24
$$\mathrm{Let}\:{f}\left({x}\right)={x}\left({x}−\mathrm{10}\right) \\ $$$$\mathrm{and}\:\mathrm{let}\:\mathrm{A}\:\mathrm{be}\:\mathrm{the}\:\mathrm{region}\:\mathrm{enclosed}\:\mathrm{within} \\ $$$$\mathrm{the}\:\mathrm{following}\:\mathrm{points} \\ $$$$\left(\mathrm{2},\mathrm{7}\right),\left(\mathrm{8},\mathrm{7}\right),\left(\mathrm{2},\mathrm{4}\right),\left(\mathrm{8},\mathrm{4}\right) \\ $$$$\mathrm{what}\:\mathrm{is}\:\mathrm{the}\:\mathrm{average}\:\mathrm{arc}\:\mathrm{length}\:\mathrm{of}\:\mathrm{a}\centerdot{f}\left({x}\right) \\ $$$$\mathrm{inside}\:\mathrm{A},{a}\in\mathbb{R}^{−} \\ $$
Commented by mr W last updated on 21/Apr/24
$${what}'{s}\:{your}\:{definition}\:{for}\:“{average} \\ $$$${arc}\:{length}''? \\ $$
Commented by Red1ight last updated on 22/Apr/24
$$\mathrm{Sum}\:\mathrm{of}\:\mathrm{lengths}\:\mathrm{for}\:\mathrm{all}\:\mathrm{possible}\:\mathrm{arcs} \\ $$$$\mathrm{that}\:\mathrm{can}\:\mathrm{be}\:\mathrm{drawn} \\ $$$$\mathrm{within}\:\mathrm{that}\:\mathrm{region} \\ $$$$\mathrm{devided}\:\mathrm{by}\:\mathrm{the}\:\mathrm{range}\:\mathrm{of}\:\mathrm{a} \\ $$$$\mathrm{avg}=\frac{\Sigma{L}_{{i}} }{{a}_{{max}} −{a}_{{min}} } \\ $$$$\mathrm{where}\:\mathrm{L}_{{i}} \:\mathrm{is}\:\mathrm{a}\:\mathrm{length}\:\mathrm{of}\:\mathrm{an}\:\mathrm{arc}\:\mathrm{when}\:\mathrm{a}=\mathrm{a}_{\mathrm{i}} \\ $$$$\mathrm{where}\:\mathrm{it}\:\mathrm{is}\:\mathrm{also}\:\mathrm{within}\:\mathrm{the}\:\mathrm{region} \\ $$$$\mathrm{where}\:{a}_{{min}} \leqslant\mathrm{a}_{\mathrm{i}} \leqslant{a}_{{max}} \\ $$
Commented by Red1ight last updated on 22/Apr/24
$$ \\ $$for example the green parabola represents a=-0.1 which from yhe plot is it not within the region
while the red and blue (a=-0.2,a=-0.3) parabola are within. so their arc length is included in the summation
Commented by Red1ight last updated on 22/Apr/24
Commented by mr W last updated on 22/Apr/24
$${i}\:{asked}\:{this}\:{because}\:{you}\:{gave}\:{a}\in{R}^{−} , \\ $$$${which}\:{means}\:{a}_{{min}} =−\infty,\:{a}_{{max}} =\mathrm{0}. \\ $$
Answered by Frix last updated on 22/Apr/24
$${y}={ax}\left({x}−\mathrm{10}\right) \\ $$$$\mathrm{Arc}\:\mathrm{length}\:\alpha\leqslant{x}\leqslant\beta:\:\underset{\alpha} {\overset{\beta} {\int}}\sqrt{\mathrm{1}+\left({y}'\right)^{\mathrm{2}} }{dx} \\ $$$$\mathrm{1}.\:−\frac{\mathrm{7}}{\mathrm{16}}\leqslant{a}\leqslant−\frac{\mathrm{7}}{\mathrm{25}} \\ $$$$\:\:\:\:\:\mathrm{2}\:\mathrm{interections}\:\mathrm{with}\:{y}=\mathrm{7}:\:{x}=\mathrm{5}\pm\sqrt{\frac{\mathrm{25}{a}+\mathrm{7}}{{a}}} \\ $$$$\:\:\:\:\:\frac{\mathrm{400}}{\mathrm{63}}\underset{−\frac{\mathrm{7}}{\mathrm{16}}} {\overset{−\frac{\mathrm{7}}{\mathrm{25}}} {\int}}\left(\mathrm{2}\underset{\mathrm{2}} {\overset{\mathrm{5}−\sqrt{\frac{\mathrm{25}{a}+\mathrm{7}}{{a}}}} {\int}}\sqrt{\mathrm{1}+\left({y}'\right)^{\mathrm{2}} }{dx}\right){da}\approx\mathrm{3}.\mathrm{05122} \\ $$$$\mathrm{2}.\:−\frac{\mathrm{7}}{\mathrm{25}}\leqslant{a}\leqslant−\frac{\mathrm{1}}{\mathrm{4}} \\ $$$$\:\:\:\:\:\frac{\mathrm{100}}{\mathrm{3}}\underset{−\frac{\mathrm{7}}{\mathrm{25}}} {\overset{−\frac{\mathrm{1}}{\mathrm{4}}} {\int}}\left(\mathrm{2}\underset{\mathrm{2}} {\overset{\mathrm{5}} {\int}}\sqrt{\mathrm{1}+\left({y}'\right)^{\mathrm{2}} }{dx}\right){da}\approx\mathrm{7}.\mathrm{98232} \\ $$$$\mathrm{3}.\:−\frac{\mathrm{1}}{\mathrm{4}}\leqslant{a}\leqslant−\frac{\mathrm{4}}{\mathrm{25}} \\ $$$$\:\:\:\:\:\mathrm{2}\:\mathrm{intersections}\:\mathrm{with}\:{y}=\mathrm{4}:\:{x}=\mathrm{5}\pm\sqrt{\frac{\mathrm{25}{a}+\mathrm{4}}{{a}}} \\ $$$$\:\:\:\:\:\frac{\mathrm{100}}{\mathrm{9}}\underset{−\frac{\mathrm{1}}{\mathrm{4}}} {\overset{−\frac{\mathrm{4}}{\mathrm{25}}} {\int}}\left(\mathrm{2}\underset{\mathrm{5}−\sqrt{\frac{\mathrm{25}{a}+\mathrm{4}}{{a}}}} {\overset{\mathrm{5}} {\int}}\sqrt{\mathrm{1}+\left({y}'\right)^{\mathrm{2}} }{dx}\right){da}\approx\mathrm{5}.\mathrm{06862} \\ $$$$\mathrm{Sum}\:\mathrm{of}\:\mathrm{these}\:\approx\:\mathrm{16}.\mathrm{1022} \\ $$