New Notes Blog: `(0,0) , (8,15)` Find the distance between the two points using integration.

Monday, 12 August 2013

$\displaystyle{\left({0},{0}\right)},{\left({8},{15}\right)}$ Find the distance between the two points using integration.

Given the equation of a line $\displaystyle{y}={m}{x}+{b},$

=> slope = $\displaystyle\frac{{\left.{d}{y}}}{{\left.{d}{x}}}={m}$ . Thus, the distance is:

$\displaystyle{L}={\int_{{a}}^{{b}}}\sqrt{{{1}+{{\left(\frac{{\left.{d}{y}}}{{\left.{d}{x}}}\right)}}^{{2}}}}{\left.{d}{x}\right.},{a}\le{x}\le{b}$

we know the two points $\displaystyle{\left({x}_{{1}},{y}_{{1}}\right)}={\left({0},{0}\right)}$

$\displaystyle{\left({x}_{{2}},{y}_{{2}}\right)}={\left({8},{15}\right)}$

$\displaystyle{m}=\frac{{{y}_{{2}}-{y}_{{1}}}}{{{x}_{{2}}-{x}_{{1}}}}=\frac{{{15}-{0}}}{{{8}-{0}}}=\frac{{15}}{{8}}$

so now the length is $\displaystyle{L}={\int_{{0}}^{{8}}}\sqrt{{{1}+{{\left(\frac{{15}}{{8}}\right)}}^{{2}}}}{\left.{d}{x}\right.}$

$\displaystyle={\int_{{0}}^{{8}}}\sqrt{{{1}+{\left(\frac{{225}}{{64}}\right)}}}{\left.{d}{x}\right.}$

= $\displaystyle{\int_{{0}}^{{8}}}\sqrt{{\frac{{{64}+{225}}}{{64}}}}{\)}{\left.{d}{x}\right.}$

= $\displaystyle{\int_{{0}}^{{8}}}\sqrt{{\frac{{{289}}}{{64}}}}{\)}{\left.{d}{x}\right.}$

= $\displaystyle{\int_{{0}}^{{8}}}{\left(\frac{{17}}{{8}}\right)}{\left.{d}{x}\right.}$

= $\displaystyle{\left(\frac{{17}}{{8}}\right)}{\int_{{0}}^{{8}}}{1}\ldots$

Given the equation of a line $\displaystyle{y}={m}{x}+{b},$

=> slope = $\displaystyle\frac{{\left.{d}{y}}}{{\left.{d}{x}}}={m}$ . Thus, the distance is:

$\displaystyle{L}={\int_{{a}}^{{b}}}\sqrt{{{1}+{{\left(\frac{{\left.{d}{y}}}{{\left.{d}{x}}}\right)}}^{{2}}}}{\left.{d}{x}\right.},{a}\le{x}\le{b}$

we know the two points $\displaystyle{\left({x}_{{1}},{y}_{{1}}\right)}={\left({0},{0}\right)}$

$\displaystyle{\left({x}_{{2}},{y}_{{2}}\right)}={\left({8},{15}\right)}$

$\displaystyle{m}=\frac{{{y}_{{2}}-{y}_{{1}}}}{{{x}_{{2}}-{x}_{{1}}}}=\frac{{{15}-{0}}}{{{8}-{0}}}=\frac{{15}}{{8}}$

so now the length is $\displaystyle{L}={\int_{{0}}^{{8}}}\sqrt{{{1}+{{\left(\frac{{15}}{{8}}\right)}}^{{2}}}}{\left.{d}{x}\right.}$

$\displaystyle={\int_{{0}}^{{8}}}\sqrt{{{1}+{\left(\frac{{225}}{{64}}\right)}}}{\left.{d}{x}\right.}$

= $\displaystyle{\int_{{0}}^{{8}}}\sqrt{{\frac{{{64}+{225}}}{{64}}}}{\)}{\left.{d}{x}\right.}$

= $\displaystyle{\int_{{0}}^{{8}}}\sqrt{{\frac{{{289}}}{{64}}}}{\)}{\left.{d}{x}\right.}$

= $\displaystyle{\int_{{0}}^{{8}}}{\left(\frac{{17}}{{8}}\right)}{\left.{d}{x}\right.}$

= $\displaystyle{\left(\frac{{17}}{{8}}\right)}{\int_{{0}}^{{8}}}{1}{\left.{d}{x}\right.}$

= $\displaystyle{\left(\frac{{17}}{{8}}\right)}{{\mid}_{{0}}^{{8}}}{x}$

= $\displaystyle{\left(\frac{{17}}{{8}}\right)}{\left[{8}-{0}\right]}$

= $\displaystyle{17}$

so the distance between the two points = 17

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