Queen's College, Hong Kong



An example in calculating Arc LengthYue Kwok Choy(1)FormulasLet y=f(x) be differentiable function, arc length L=ab1+dydx2dxIf x=h(y), L=cd1+dxdy2dyFor parametric equation: x=ut, y=vt, α≤t≤β, L=αβdxdt2+dydt2dt.For polar equation: r=fθ, α≤t≤β,L=αβr2+drdθ2dθ(2)ExampleFind the arc length of the parabola y2=x from 0,0 to 1,1.This problem looks easy, but it needs quite a lot of integration techniques.You need to pay more attention on how to handle the problem rather than the tedious calculations in the followings. Method 1y2=x?dxdy=2yArc length , L = 011+2y2dyLet 2y=tanθ, dy=12sec2θdθ.When y=0,θ=0. When y=1,θ=tan-12.L = 0tan-121+tanθ212sec2θdθ=120tan-12sec3θdθ=120tan-12secθdtanθ=12secθtanθ0tan-12-120tan-12tanθdsecθ=1252-120tan-12tan2θsecθdθ=5-120tan-12sec2θ-1secθdθ=5-120tan-12sec3θdθ+120tan-12secθdθ=5-L+120tan-12secθsecθ+tanθsecθ+tanθdθ=5-L+120tan-12dsecθ+tanθsecθ+tanθ=5-L+12lnsecθ+tanθ0tan-122L=5+12ln5+2L=125+12ln5+2≈1.4789428575446Method 2For those who understand hyperbolic functions, the calculations are easier.Use the substitution 2y=sinhθ. dy=12coshθdθ.Arc length , L = 011+2y2dy=0sinh-121+sinhθ212coshθdθ=120sinh-12cosh2θdθ=120sinh-121+cosh2θ2dθ=14θ+sinh2θ20sinh-12=14θ+2sinhθcoshθ20sinh-12=14θ+sinhθ1+sinhθ20sinh-12=14sinh-12+25=1425+sinh-12≈1.4789428575446I love this method, short and nice!Method 3y2=x?y=x?dydx=12x (Note that we only need the positive branch of the parabola.)L = 011+12x2dx=011+14xdx=014x+14xdxLet t=4x+14x?x=-14 1-t2, t≠-1 and t≠1dx=- t2 t2-12dtWhen x = 0, t =∞. When x = 1, t=52L =∞52t- t2 t2-12dt=-18∞524 t2 t2-12dtWe have to express the integrand in partial fractions, observe that:4 t2 t2-12=2tt2-12=1t+1+1t-12=1t+12+1t-12+2t+1t-1=1t+12+1t-12+1t-1-1t+1L =-18∞521t+12+1t-12+1t-1-1t+1dt=181t+1+1t-1-lnt-1t+1∞52=18152+1+152-1-ln52-152+1≈1.4789428575446Method 4y2=x?y=x?dydx=12x L = 011+12x2dx=011+14xdx=014x+14xdxConsider I = 124x+1xdx. Put u=4x+1x, du=-1 x2dxI = 124x+1xdx=12x4x+1x-x d4x+1x=12x4x+1x-x 124x+1x-1 x2dx=12x4x+1x+12 1x4x+1dx=12x4x+1x+14 1x2+x4dx=12x4x+1x+14 1x+182-182dxThe last integral is a standard integral E=dxx2-a2 which can be solved easily as:Put u2=x2-a2?2udu=2xdx?dxu=dux=dx+duu+x=dx+ux+uE=dxx2-a2=dxu=dx+ux+u=lnx+u+c=lnx+x2-a2+cTherefore,I =12x4x+1x+14lnx+18+x+182-18201=125+14ln1+18+1+182-182-18ln18≈1.4789428575446 ................
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