New Notes Blog: `y = 9 - |x| , y=0` Find b such that the line y = b divides the region bounded by the graphs of the equations into two regions of equal area.

Sunday, 18 October 2015

$\displaystyle{y}={9}-{\left|{x}\right|},{y}={0}$ Find b such that the line y = b divides the region bounded by the graphs of the equations into two regions of equal area.

Given ,

$\displaystyle{y}={9}-{\left|{x}\right|},{y}={0}$

first let us find the total area of the bounded by the curves.

so we shall proceed as follows

as given ,

$\displaystyle{y}={9}-{\left|{x}\right|},{y}={0}$

=> $\displaystyle{9}-{\left|{x}\right|}={0}$

=> $\displaystyle{\left|{x}\right|}-{9}={0}$

=> $\displaystyle{\left|{x}\right|}={9}$

so $\displaystyle{x}=\pm{9}$

the the area of the region is = $\displaystyle{\int_{{-{{9}}}}^{{9}}}{\left({9}-{\left|{x}\right|}-{0}\right)}{\left.{d}{x}\right.}$

=> $\displaystyle{\int_{{-{{9}}}}^{{0}}}{\left({9}+{x}-{0}\right)}{\left.{d}{x}\right.}$ + $\displaystyle{\int_{{0}}^{{9}}}{\left({9}-{x}-{0}\right)}{\left.{d}{x}\right.}$

=> $\displaystyle{{\left[{9}{x}+\frac{{{x}}^{{2}}}{{2}}\right]}_{{-{{9}}}}^{{0}}}+{{\left[{9}{x}-\frac{{{x}}^{{2}}}{{2}}\right]}_{{0}}^{{9}}}$

=> $\displaystyle{\left[{0}\right]}-{\left[-{81}+\frac{{81}}{{2}}\right]}+{\left[{81}-\frac{{81}}{{2}}\right]}-{\left[{0}\right]}$

=> $\displaystyle\frac{{81}}{{2}}+\frac{{81}}{{2}}={81}$

So now we have to find the horizonal line that splits the region into two regions with area 81/2

as when the line y=b intersects the curve $\displaystyle{y}={9}-{\left|{x}\right|}$ then the area bounded is 81,so

let us solve this as follows

first we shall find the intersecting points

as ,

$\displaystyle{9}-{\left|{x}\right|}={b}$

$\displaystyle{\left|{x}\right|}={9}-{b}$

$\displaystyle{x}=\pm{\left({9}-{b}\right)}$

so the area bound by these curves $\displaystyle{y}={b}$ and $\displaystyle{y}={9}-{\left|{x}\right|}$ is as follows

A= $\displaystyle{\int_{{-{{\left({9}-{b}\right)}}}}^{{{9}-{b}}}}{\left({9}-{\left|{x}\right|}-{b}\right)}{\left.{d}{x}\right.}=\frac{{81}}{{2}}$

=> $\displaystyle{\int_{{-{{\left({9}-{b}\right)}}}}^{{0}}}{\left({9}+{x}-{b}\right)}{\left.{d}{x}\right.}+{\int_{{0}}^{{{9}-{b}}}}{\left({9}-{x}-{b}\right)}{\left.{d}{x}\right.}=\frac{{81}}{{2}}$

=> $\displaystyle{{\left[{9}{x}+\frac{{{x}}^{{2}}}{{2}}-{b}{x}\right]}_{{-{{\left({9}-{b}\right)}}}}^{{0}}}+{{\left[{9}{x}-\frac{{{x}}^{{2}}}{{2}}-{b}{x}\right]}_{{0}}^{{{9}-{b}}}}=\frac{{81}}{{2}}$

=> $\displaystyle{\left[{0}\right]}-{\left[{9}{\left(-{\left({9}-{b}\right)}\right)}+\frac{{{\left(-{\left({9}-{b}\right)}\right)}}^{{2}}}{{2}}-{b}{\left(-{\left({9}-{b}\right)}\right)}\right]}+$

$\displaystyle{\left[{9}{\left({\left({9}-{b}\right)}\right)}-\frac{{{\left({\left({9}-{b}\right)}\right)}}^{{2}}}{{2}}-{b}{\left({\left({9}-{b}\right)}\right)}\right]}-{\left[{0}\right]}=\frac{{81}}{{2}}$

=> $\displaystyle{\left[{9}{\left({\left({9}-{b}\right)}\right)}-\frac{{{\left({\left({9}-{b}\right)}\right)}}^{{2}}}{{2}}-{b}{\left({\left({9}-{b}\right)}\right)}\right]}$

$\displaystyle-{\left[{9}{\left(-{\left({9}-{b}\right)}\right)}+\frac{{{\left(-{\left({9}-{b}\right)}\right)}}^{{2}}}{{2}}-{b}{\left(-{\left({9}-{b}\right)}\right)}\right]}=\frac{{81}}{{2}}$

let $\displaystyle{t}={9}-{b}$

=> $\displaystyle{\left[{9}{\left({t}\right)}-\frac{{{\left({t}\right)}}^{{2}}}{{2}}-{b}{\left({t}\right)}\right]}-{\left[{9}{\left(-{t}\right)}+\frac{{{\left(-{t}\right)}}^{{2}}}{{2}}-{b}{\left(-{t}\right)}\right]}=\frac{{81}}{{2}}$

=> $\displaystyle{\left[{9}{t}-\frac{{{t}}^{{2}}}{{2}}-{b}{t}\right]}+{\left[{9}{t}-\frac{{{t}}^{{2}}}{{2}}-{b}{t}\right]}=\frac{{81}}{{2}}$

=> $\displaystyle{18}{t}-{{t}}^{{2}}-{2}{b}{t}=\frac{{81}}{{2}}$

but we know half the Area of the region between $\displaystyle{y}={9}-{\left|{x}\right|},{y}={0}$ curves = $\displaystyle\frac{{81}}{{2}}$

so now ,

$\displaystyle{18}{t}-{{t}}^{{2}}-{2}{b}{t}=\frac{{81}}{{2}}$

$\displaystyle{18}{t}-{2}{b}{t}-{{t}}^{{2}}=\frac{{81}}{{2}}$

=> $\displaystyle{t}{\left({18}-{2}{b}\right)}-{{t}}^{{2}}=\frac{{81}}{{2}}$

=> $\displaystyle{{t}}^{{2}}={t}{\left({18}-{2}{b}\right)}-\frac{{81}}{{2}}$

=> $\displaystyle{{t}}^{{2}}-{t}{\left({18}-{2}{b}\right)}+\frac{{81}}{{2}}={0}$

this is like quadratic equation

$\displaystyle{a}{{x}}^{{2}}+{b}{x}+{c}={0}$

so $\displaystyle{t}=\frac{{-{b}\pm\sqrt{{{{b}}^{{2}}-{4}{a}{c}}}}}{{2}}{a}$

$\displaystyle=\frac{{{18}-{2}{b}\pm\sqrt{{{{\left(-{\left({18}-{2}{b}\right)}\right)}}^{{2}}-{4}\cdot{\left(\frac{{81}}{{2}}\right)}}}}}{{2}}$

but

$\displaystyle{t}={9}-{b}$

so,

$\displaystyle{9}-{b}=\frac{{{18}-{2}{b}\pm\sqrt{{{{\left(-{\left({18}-{2}{b}\right)}\right)}}^{{2}}-{4}\cdot{\left(\frac{{81}}{{2}}\right)}}}}}{{2}}$

=> $\displaystyle{18}-{2}{b}={\left({18}-{2}{b}\pm\sqrt{{{{\left(-{\left({18}-{2}{b}\right)}\right)}}^{{2}}-{4}\cdot{\left(\frac{{81}}{{2}}\right)}}}\right)}$

=> $\displaystyle\pm\sqrt{{{{\left(-{\left({18}-{2}{b}\right)}\right)}}^{{2}}-{4}\cdot{\left(\frac{{81}}{{2}}\right)}}}={0}$

=> $\displaystyle\sqrt{{{{\left(-{\left({18}-{2}{b}\right)}\right)}}^{{2}}-{4}\cdot{\left(\frac{{81}}{{2}}\right)}}}={0}$

=> $\displaystyle{{\left(-{\left({18}-{2}{b}\right)}\right)}}^{{2}}-{4}\cdot{\left(\frac{{81}}{{2}}\right)}={0}$

=> $\displaystyle{{\left(-{\left({18}-{2}{b}\right)}\right)}}^{{2}}={4}\cdot{\left(\frac{{81}}{{2}}\right)}$

=> $\displaystyle{{\left(-{\left({18}-{2}{b}\right)}\right)}}^{{2}}={2}\cdot{\left({81}\right)}$

=> $\displaystyle{\left(-{\left({18}-{2}{b}\right)}\right)}=\pm\sqrt{{{2}}}\cdot{9}$

=> $\displaystyle-{18}+{2}{b}=\pm{9}\sqrt{{{2}}}$

=> $\displaystyle{2}{b}=\pm{9}\sqrt{{{2}}}+{18}$

=> $\displaystyle{b}=\frac{{{18}\pm{9}\sqrt{{{2}}}}}{{2}}$

so $\displaystyle{b}=\frac{{{18}\pm{9}\sqrt{{{2}}}}}{{2}}$

New Notes Blog

Sunday, 18 October 2015

$\displaystyle{y}={9}-{\left|{x}\right|},{y}={0}$ Find b such that the line y = b divides the region bounded by the graphs of the equations into two regions of equal area.

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In "By the Waters of Babylon," under the leadership of John, what do you think the Hill People will do with their society?

Sunday, 18 October 2015

Find b such that the line y = b divides the region bounded by the graphs of the equations into two regions of equal area.

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Post a Comment

In &quot;By the Waters of Babylon,&quot; under the leadership of John, what do you think the Hill People will do with their society?

$\displaystyle{y}={9}-{\left|{x}\right|},{y}={0}$ Find b such that the line y = b divides the region bounded by the graphs of the equations into two regions of equal area.

In "By the Waters of Babylon," under the leadership of John, what do you think the Hill People will do with their society?