A-telephone-wire-hangs-from-two-points-P-Q-60m-apart-P-Q-are-on-the-same-level-the-mid-point-of-the-telephone-wire-is-3m-below-the-level-of-P-Q-Assuming-that-it-hangs-in-form-of-a-curve-

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Question Number 6169 by sanusihammed last updated on 17/Jun/16

$$AtelephonewirehangsfromtwopointsP,Q60mapart$$$$P,Qareonthesamelevel.themidpointofthetelephone$$$$wireis3mbelowthelevelofP,Q.Assumingthatithangsin$$$$formofacurve,finditequation.$$$$$$$$pleasehelp.thanksforyourtime.$$

Answered by Yozzii last updated on 17/Jun/16

$$Letcurvebequadraticwhere0⩽x⩽60$$$$f\left(x\right)={ax}^2+bx,a>0,P(0,0),Q(60,0)(P\mathrm{\&}Qlieonx-axis)$$$$\Rightarrow f\left(60\right)=0or3600a+60b=0$$$$b=-60a\dots \dots ..\left(i\right)$$$$Atx=0.5\mid PQ\mid =30,f^{\prime }\left(30\right)=0if$$$$weassumeauniformwireliesbetween$$$$P\mathrm{\&}Q.$$$$\therefore 2a\left(30\right)+b=0\Rightarrow b=-60a(nonewinformationbasedonquadraticassumption.)$$$$\therefore Atx=30,f\left(30\right)=-3.$$$$\therefore 900a+30b=-3$$$$300a+10b=-1\dots \dots \dots .\left(ii\right)$$$$From\left(i\right),in\left(ii\right)300a+10(-60a)=-1$$$$-300a=-1\Rightarrow a=\frac{1}{300}\Rightarrow b=\frac{-60}{300}=\frac{-1}{5}.$$$$\therefore f\left(x\right)=\frac{1}{300}{x}^2-\frac{1}{5}x(0⩽x⩽60)$$$$------------------------$$$$Ingeneral,inx-yCartesianco-ordinates,$$$$wecanletP(a,b)andQ(a+60,b)where$$$$a,b\in \mathbb{R}.$$$$Thismeansthatonecanchooseanyarbitrary$$$$placeinthex-yplanetoputthiscurve$$$$forx\in [a,a+60].$$$$Assumingthewireisuniform,the$$$$exampleaboveindicatesthatthe$$$$quadraticcurveisasuitablemodel.$$$$Letf\left(x\right)={cx}^2+dx(c>0).$$$$Atx=a,f\left(x\right)=b\therefore b={ca}^2+da$$$$ord=\frac{b-{ca}^2}{a}(a\ne 0)\dots \dots .\left(i\right)$$$$Atx=a+60,f(a+60)=b$$$$\therefore c{(a+60)}^2+d(a+60)=b$$$${ca}^2+c(120a+3600)+da+60d=b$$$$120c(a+30)+60d=0$$$$2c(a+30)+d=0$$$$2ca+60c+\frac{b}{a}-ca=0\left(from\left(i\right)\right)$$$$ca+60c=\frac{-b}{a}$$$$c=\frac{-b}{a(a+60)}$$$$\Rightarrow d=\frac{b}{a}-a\times \frac{-b}{a(a+60)}$$$$d=\frac{b}{a}(1+\frac{a}{a+60})=\frac{b(a+60+a)}{a(a+60)}$$$$d=\frac{2b(a+30)}{a(a+60)}$$$$Sincec>0\Rightarrow -60<a<0\mathrm{\&}b>0orb<0\mathrm{\&}a>0orb<0\mathrm{\&}a<-60.$$$$\therefore f\left(x\right)=\frac{bx}{a(a+60)}(-x+2a+60)$$$$Atx=a+30,$$$$f(a+30)=\frac{b(a+30)(-a-30+2a+60)}{a(a+60)}$$$$f(a+30)=\frac{b{(a+30)}^2}{a(a+60)}$$$$f(a+30)-b=\frac{900b}{a(a+60)}$$$$Butf(a+30)-f\left(a\right)=-3.$$$$\Rightarrow 300b=-a(a+60)$$$$b=\frac{-a(a+60)}{300}.$$$$\therefore f\left(x\right)=\frac{-a(a+60)}{300}\times \frac{x}{a(a+60)}(-x+2a+60)$$$$f\left(x\right)=\frac{x(x-2a-60)}{300}=\frac{1}{300}{x}^2-\frac{x(a+30)}{150}$$$$f\left(a\right)=\frac{a(a-2a-60)}{300}=\frac{-a(a+60)}{300}$$$$f(a+60)=\frac{(a+60)(-a)}{300}=b$$$$f^{\prime }\left(x\right)=\frac{x-a-30}{150}\Rightarrow atmidpoint,f^{\prime }\left(x\right)=0$$$$orx=a+30.$$$$Wemaychoosetoshiftthiscurve$$$$byanamounthverticallysothatin$$$$general$$$$f\left(x\right)=\frac{1}{300}{x}^2-\frac{x(a+30)}{150}+h(a⩽x⩽a+60,a,h\in \mathbb{R}).$$$$a=0andh=0\Rightarrow f\left(x\right)=\frac{1}{300}{x}^2-\frac{1}{5}xaswasfound.$$

Commented by sanusihammed last updated on 17/Jun/16

$$Thankssomuch$$

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