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Question Number 45187 by MrW3 last updated on 10/Oct/18

Commented by MrW3 last updated on 12/Oct/18

Solution to Q45122.  Find the length of path from A to B  which has a constant slope.    A point P on the path can be described  by parameters (φ,ϕ) with   0≤φ≤2π, 0≤ϕ≤θ.    z=R sin ϕ  r=R cos ϕ    the slope k of the path is defined as   ratio from change in height to change  in horizontal distance between two  points:  k=tan α=((Δh)/(Δl))  let λ=(1/k)    Δh=dz=R cos ϕ dϕ  Δl=(√((Δl_φ )^2 +(Δl_r )^2 ))  Δl_φ =r dφ=R cos ϕ dφ=R cos ϕ (dφ/dϕ) dϕ  Δl_r =dr=−R sin ϕ dϕ  ⇒k=((R cos ϕ dϕ)/(R(√(cos^2  ϕ ((dφ/dϕ))^2 (dϕ)^2 +sin^2  ϕ (dϕ)^2 ))))  ⇒k=((cos ϕ)/(√(cos^2  ϕ ((dφ/dϕ))^2 +sin^2  ϕ)))=(1/(√(((dφ/dϕ))^2 +tan^2  ϕ)))  ⇒((dφ/dϕ))^2 +tan^2  ϕ=(1/k^2 )=λ^2   ⇒(dφ/dϕ)=(√(λ^2 −tan^2  ϕ))  ⇒φ=(√(1+λ^2 )) tan^(−1) ((((√(1+λ^2 )) tan ϕ)/(√(λ^2 −tan^2  ϕ))))−sin^(−1) (((tan ϕ)/λ))+C  at φ=0: ϕ=0 ⇒C=0  ⇒φ=(√(1+λ^2 )) tan^(−1) ((((√(1+λ^2 )) tan ϕ)/(√(λ^2 −tan^2  ϕ))))−sin^(−1) (((tan ϕ)/λ))    at ϕ=θ: φ=2π  ⇒2π=(√(1+λ^2 )) tan^(−1) ((((√(1+λ^2 )) tan θ)/(√(λ^2 −tan^2  θ))))−sin^(−1) (((tan θ)/λ))  ⇒we get λ=... in terms of θ  (this is only numerically possible)    the length of path from A to B:  s=((√(1+k^2 ))/k)h=(√(1+λ^2 ))h  h=R sin θ  ⇒s=R(√(1+λ^2 )) sin θ    example:  θ=60°  ⇒λ=6.0546 ⇒k=0.1652⇒α=9.38°  ⇒s=5.3145R    limit case:  from  2π=(√(1+λ^2 )) tan^(−1) ((((√(1+λ^2 )) tan ϕ)/(√(λ^2 −tan^2  ϕ))))−sin^(−1) (((tan ϕ)/λ))  we see the limit case is when λ→tan θ,  then we have  2π=(√(1+λ^2 )) (π/2)−(π/2)  (√(1+λ^2 ))=5  ⇒λ_(max) =tan θ_(max) =(√(24))  ⇒θ_(max) =tan^(−1) (√(24))=78.46°  ⇒k_(max) =(1/(√(24)))=0.204  ⇒α_(max) =tan^(−1) ((1/(√(24))))=11.54°

$${Solution}\:{to}\:{Q}\mathrm{45122}. \\ $$$${Find}\:{the}\:{length}\:{of}\:{path}\:{from}\:{A}\:{to}\:{B} \\ $$$${which}\:{has}\:{a}\:{constant}\:{slope}. \\ $$$$ \\ $$$${A}\:{point}\:{P}\:{on}\:{the}\:{path}\:{can}\:{be}\:{described} \\ $$$${by}\:{parameters}\:\left(\phi,\varphi\right)\:{with}\: \\ $$$$\mathrm{0}\leqslant\phi\leqslant\mathrm{2}\pi,\:\mathrm{0}\leqslant\varphi\leqslant\theta. \\ $$$$ \\ $$$${z}={R}\:\mathrm{sin}\:\varphi \\ $$$${r}={R}\:\mathrm{cos}\:\varphi \\ $$$$ \\ $$$${the}\:{slope}\:{k}\:{of}\:{the}\:{path}\:{is}\:{defined}\:{as}\: \\ $$$${ratio}\:{from}\:{change}\:{in}\:{height}\:{to}\:{change} \\ $$$${in}\:{horizontal}\:{distance}\:{between}\:{two} \\ $$$${points}: \\ $$$${k}=\mathrm{tan}\:\alpha=\frac{\Delta{h}}{\Delta{l}} \\ $$$${let}\:\lambda=\frac{\mathrm{1}}{{k}} \\ $$$$ \\ $$$$\Delta{h}={dz}={R}\:\mathrm{cos}\:\varphi\:{d}\varphi \\ $$$$\Delta{l}=\sqrt{\left(\Delta{l}_{\phi} \right)^{\mathrm{2}} +\left(\Delta{l}_{{r}} \right)^{\mathrm{2}} } \\ $$$$\Delta{l}_{\phi} ={r}\:{d}\phi={R}\:\mathrm{cos}\:\varphi\:{d}\phi={R}\:\mathrm{cos}\:\varphi\:\frac{{d}\phi}{{d}\varphi}\:{d}\varphi \\ $$$$\Delta{l}_{{r}} ={dr}=−{R}\:\mathrm{sin}\:\varphi\:{d}\varphi \\ $$$$\Rightarrow{k}=\frac{{R}\:\mathrm{cos}\:\varphi\:{d}\varphi}{{R}\sqrt{\mathrm{cos}^{\mathrm{2}} \:\varphi\:\left(\frac{{d}\phi}{{d}\varphi}\right)^{\mathrm{2}} \left({d}\varphi\right)^{\mathrm{2}} +\mathrm{sin}^{\mathrm{2}} \:\varphi\:\left({d}\varphi\right)^{\mathrm{2}} }} \\ $$$$\Rightarrow{k}=\frac{\mathrm{cos}\:\varphi}{\sqrt{\mathrm{cos}^{\mathrm{2}} \:\varphi\:\left(\frac{{d}\phi}{{d}\varphi}\right)^{\mathrm{2}} +\mathrm{sin}^{\mathrm{2}} \:\varphi}}=\frac{\mathrm{1}}{\sqrt{\left(\frac{{d}\phi}{{d}\varphi}\right)^{\mathrm{2}} +\mathrm{tan}^{\mathrm{2}} \:\varphi}} \\ $$$$\Rightarrow\left(\frac{{d}\phi}{{d}\varphi}\right)^{\mathrm{2}} +\mathrm{tan}^{\mathrm{2}} \:\varphi=\frac{\mathrm{1}}{{k}^{\mathrm{2}} }=\lambda^{\mathrm{2}} \\ $$$$\Rightarrow\frac{{d}\phi}{{d}\varphi}=\sqrt{\lambda^{\mathrm{2}} −\mathrm{tan}^{\mathrm{2}} \:\varphi} \\ $$$$\Rightarrow\phi=\sqrt{\mathrm{1}+\lambda^{\mathrm{2}} }\:\mathrm{tan}^{−\mathrm{1}} \left(\frac{\sqrt{\mathrm{1}+\lambda^{\mathrm{2}} }\:\mathrm{tan}\:\varphi}{\sqrt{\lambda^{\mathrm{2}} −\mathrm{tan}^{\mathrm{2}} \:\varphi}}\right)−\mathrm{sin}^{−\mathrm{1}} \left(\frac{\mathrm{tan}\:\varphi}{\lambda}\right)+{C} \\ $$$${at}\:\phi=\mathrm{0}:\:\varphi=\mathrm{0}\:\Rightarrow{C}=\mathrm{0} \\ $$$$\Rightarrow\phi=\sqrt{\mathrm{1}+\lambda^{\mathrm{2}} }\:\mathrm{tan}^{−\mathrm{1}} \left(\frac{\sqrt{\mathrm{1}+\lambda^{\mathrm{2}} }\:\mathrm{tan}\:\varphi}{\sqrt{\lambda^{\mathrm{2}} −\mathrm{tan}^{\mathrm{2}} \:\varphi}}\right)−\mathrm{sin}^{−\mathrm{1}} \left(\frac{\mathrm{tan}\:\varphi}{\lambda}\right) \\ $$$$ \\ $$$${at}\:\varphi=\theta:\:\phi=\mathrm{2}\pi \\ $$$$\Rightarrow\mathrm{2}\pi=\sqrt{\mathrm{1}+\lambda^{\mathrm{2}} }\:\mathrm{tan}^{−\mathrm{1}} \left(\frac{\sqrt{\mathrm{1}+\lambda^{\mathrm{2}} }\:\mathrm{tan}\:\theta}{\sqrt{\lambda^{\mathrm{2}} −\mathrm{tan}^{\mathrm{2}} \:\theta}}\right)−\mathrm{sin}^{−\mathrm{1}} \left(\frac{\mathrm{tan}\:\theta}{\lambda}\right) \\ $$$$\Rightarrow{we}\:{get}\:\lambda=...\:{in}\:{terms}\:{of}\:\theta \\ $$$$\left({this}\:{is}\:{only}\:{numerically}\:{possible}\right) \\ $$$$ \\ $$$${the}\:{length}\:{of}\:{path}\:{from}\:{A}\:{to}\:{B}: \\ $$$${s}=\frac{\sqrt{\mathrm{1}+{k}^{\mathrm{2}} }}{{k}}{h}=\sqrt{\mathrm{1}+\lambda^{\mathrm{2}} }{h} \\ $$$${h}={R}\:\mathrm{sin}\:\theta \\ $$$$\Rightarrow{s}={R}\sqrt{\mathrm{1}+\lambda^{\mathrm{2}} }\:\mathrm{sin}\:\theta \\ $$$$ \\ $$$${example}: \\ $$$$\theta=\mathrm{60}° \\ $$$$\Rightarrow\lambda=\mathrm{6}.\mathrm{0546}\:\Rightarrow{k}=\mathrm{0}.\mathrm{1652}\Rightarrow\alpha=\mathrm{9}.\mathrm{38}° \\ $$$$\Rightarrow{s}=\mathrm{5}.\mathrm{3145}{R} \\ $$$$ \\ $$$${limit}\:{case}: \\ $$$${from} \\ $$$$\mathrm{2}\pi=\sqrt{\mathrm{1}+\lambda^{\mathrm{2}} }\:\mathrm{tan}^{−\mathrm{1}} \left(\frac{\sqrt{\mathrm{1}+\lambda^{\mathrm{2}} }\:\mathrm{tan}\:\varphi}{\sqrt{\lambda^{\mathrm{2}} −\mathrm{tan}^{\mathrm{2}} \:\varphi}}\right)−\mathrm{sin}^{−\mathrm{1}} \left(\frac{\mathrm{tan}\:\varphi}{\lambda}\right) \\ $$$${we}\:{see}\:{the}\:{limit}\:{case}\:{is}\:{when}\:\lambda\rightarrow\mathrm{tan}\:\theta, \\ $$$${then}\:{we}\:{have} \\ $$$$\mathrm{2}\pi=\sqrt{\mathrm{1}+\lambda^{\mathrm{2}} }\:\frac{\pi}{\mathrm{2}}−\frac{\pi}{\mathrm{2}} \\ $$$$\sqrt{\mathrm{1}+\lambda^{\mathrm{2}} }=\mathrm{5} \\ $$$$\Rightarrow\lambda_{{max}} =\mathrm{tan}\:\theta_{{max}} =\sqrt{\mathrm{24}} \\ $$$$\Rightarrow\theta_{{max}} =\mathrm{tan}^{−\mathrm{1}} \sqrt{\mathrm{24}}=\mathrm{78}.\mathrm{46}° \\ $$$$\Rightarrow{k}_{{max}} =\frac{\mathrm{1}}{\sqrt{\mathrm{24}}}=\mathrm{0}.\mathrm{204} \\ $$$$\Rightarrow\alpha_{{max}} =\mathrm{tan}^{−\mathrm{1}} \left(\frac{\mathrm{1}}{\sqrt{\mathrm{24}}}\right)=\mathrm{11}.\mathrm{54}° \\ $$

Commented by MJS last updated on 10/Oct/18

great!  another aspect (sorry if it seems I′m always  looking for problems, borders or mistakes,  but I′m really interested)  you cannot reach some regions above a certain  θ within a turn of 2π because the slope is  limited by the slope of the sphere.  Can we find the maximum θ for this case?

$$\mathrm{great}! \\ $$$$\mathrm{another}\:\mathrm{aspect}\:\left(\mathrm{sorry}\:\mathrm{if}\:\mathrm{it}\:\mathrm{seems}\:\mathrm{I}'\mathrm{m}\:\mathrm{always}\right. \\ $$$$\mathrm{looking}\:\mathrm{for}\:\mathrm{problems},\:\mathrm{borders}\:\mathrm{or}\:\mathrm{mistakes}, \\ $$$$\left.\mathrm{but}\:\mathrm{I}'\mathrm{m}\:\mathrm{really}\:\mathrm{interested}\right) \\ $$$$\mathrm{you}\:\mathrm{cannot}\:\mathrm{reach}\:\mathrm{some}\:\mathrm{regions}\:\mathrm{above}\:\mathrm{a}\:\mathrm{certain} \\ $$$$\theta\:\mathrm{within}\:\mathrm{a}\:\mathrm{turn}\:\mathrm{of}\:\mathrm{2}\pi\:\mathrm{because}\:\mathrm{the}\:\mathrm{slope}\:\mathrm{is} \\ $$$$\mathrm{limited}\:\mathrm{by}\:\mathrm{the}\:\mathrm{slope}\:\mathrm{of}\:\mathrm{the}\:\mathrm{sphere}. \\ $$$$\mathrm{Can}\:\mathrm{we}\:\mathrm{find}\:\mathrm{the}\:\mathrm{maximum}\:\theta\:\mathrm{for}\:\mathrm{this}\:\mathrm{case}? \\ $$

Commented by ajfour last updated on 10/Oct/18

Wonderful analysis Sir,   but how was the integration done  Sir ?    ∫_0 ^(  θ) (√(λ^2 −tan^2 ϕ)) dϕ

$${Wonderful}\:{analysis}\:{Sir},\: \\ $$$${but}\:{how}\:{was}\:{the}\:{integration}\:{done} \\ $$$${Sir}\:? \\ $$$$\:\:\int_{\mathrm{0}} ^{\:\:\theta} \sqrt{\lambda^{\mathrm{2}} −\mathrm{tan}\:^{\mathrm{2}} \varphi}\:{d}\varphi\:\: \\ $$

Commented by MJS last updated on 10/Oct/18

...you cannot reach the peak with a constant  slope...

$$...\mathrm{you}\:\mathrm{cannot}\:\mathrm{reach}\:\mathrm{the}\:\mathrm{peak}\:\mathrm{with}\:\mathrm{a}\:\mathrm{constant} \\ $$$$\mathrm{slope}... \\ $$

Commented by MrW3 last updated on 10/Oct/18

MJS sir, you are absolutely right.  with a constant slope one can reach  a point at max. θ≈78.5°  I have added this into my working  above. please review, thank you sir.

$${MJS}\:{sir},\:{you}\:{are}\:{absolutely}\:{right}. \\ $$$${with}\:{a}\:{constant}\:{slope}\:{one}\:{can}\:{reach} \\ $$$${a}\:{point}\:{at}\:{max}.\:\theta\approx\mathrm{78}.\mathrm{5}° \\ $$$${I}\:{have}\:{added}\:{this}\:{into}\:{my}\:{working} \\ $$$${above}.\:{please}\:{review},\:{thank}\:{you}\:{sir}. \\ $$

Commented by MrW3 last updated on 10/Oct/18

To ajfour sir:  for integral please see Q45213.

$${To}\:{ajfour}\:{sir}: \\ $$$${for}\:{integral}\:{please}\:{see}\:{Q}\mathrm{45213}. \\ $$

Answered by ajitbh last updated on 10/Oct/18

  let me take θ in question as 𝛂.  (dz/(rdφ)) = ((Rcos θdθ)/(Rcos θdφ)) = k  ⇒  ∫_0 ^(  α) dθ =∫_0 ^(  2π) kdφ  ⇒  k = 𝛂/2𝛑  ∫ds = ∫(√((Rcos θdφ)^2 +(Rcos θdθ)^2 ))        = ∫_0 ^(  α) (√(1+(1/k^2 ))) (Rcos θ)dθ     s = R(√(1+((4𝛑^2 )/𝛂^2 ))) sin 𝛂 .

$$\:\:{let}\:{me}\:{take}\:\theta\:{in}\:{question}\:{as}\:\boldsymbol{\alpha}. \\ $$$$\frac{{dz}}{{rd}\phi}\:=\:\frac{{R}\mathrm{cos}\:\theta{d}\theta}{{R}\mathrm{cos}\:\theta{d}\phi}\:=\:\boldsymbol{{k}} \\ $$$$\Rightarrow\:\:\int_{\mathrm{0}} ^{\:\:\alpha} {d}\theta\:=\int_{\mathrm{0}} ^{\:\:\mathrm{2}\pi} \boldsymbol{{k}}{d}\phi \\ $$$$\Rightarrow\:\:\boldsymbol{{k}}\:=\:\boldsymbol{\alpha}/\mathrm{2}\boldsymbol{\pi} \\ $$$$\int\boldsymbol{{ds}}\:=\:\int\sqrt{\left({R}\mathrm{cos}\:\theta{d}\phi\right)^{\mathrm{2}} +\left({R}\mathrm{cos}\:\theta{d}\theta\right)^{\mathrm{2}} } \\ $$$$\:\:\:\:\:\:=\:\int_{\mathrm{0}} ^{\:\:\alpha} \sqrt{\mathrm{1}+\frac{\mathrm{1}}{{k}^{\mathrm{2}} }}\:\left({R}\mathrm{cos}\:\theta\right){d}\theta \\ $$$$\:\:\:\boldsymbol{{s}}\:=\:\boldsymbol{{R}}\sqrt{\mathrm{1}+\frac{\mathrm{4}\boldsymbol{\pi}^{\mathrm{2}} }{\boldsymbol{\alpha}^{\mathrm{2}} }}\:\mathrm{sin}\:\boldsymbol{\alpha}\:. \\ $$

Commented by ajitbh last updated on 10/Oct/18

Sir, my tinkutara app is crashing  i changed id, yet the problem  persists..

$${Sir},\:{my}\:{tinkutara}\:{app}\:{is}\:{crashing} \\ $$$${i}\:{changed}\:{id},\:{yet}\:{the}\:{problem} \\ $$$${persists}.. \\ $$

Commented by MrW3 last updated on 10/Oct/18

sir, I see, hope the app problem will be  solved soon.

$${sir},\:{I}\:{see},\:{hope}\:{the}\:{app}\:{problem}\:{will}\:{be} \\ $$$${solved}\:{soon}. \\ $$

Commented by MrW3 last updated on 10/Oct/18

one question to your solution:  in (dz/(rdφ)) = ((Rcos θdθ)/(Rcos θdφ)) = k  you assumed that r is constant. but  if r is not constant, as in this case,  dl≠rdφ, but dl=(√((rdφ)^2 +(dr)^2 )), as  we know from polar coordinate system  when calculating length of curve.  can you please check sir. thank you.

$${one}\:{question}\:{to}\:{your}\:{solution}: \\ $$$${in}\:\frac{{dz}}{{rd}\phi}\:=\:\frac{{R}\mathrm{cos}\:\theta{d}\theta}{{R}\mathrm{cos}\:\theta{d}\phi}\:=\:\boldsymbol{{k}} \\ $$$${you}\:{assumed}\:{that}\:{r}\:{is}\:{constant}.\:{but} \\ $$$${if}\:{r}\:{is}\:{not}\:{constant},\:{as}\:{in}\:{this}\:{case}, \\ $$$${dl}\neq{rd}\phi,\:{but}\:{dl}=\sqrt{\left({rd}\phi\right)^{\mathrm{2}} +\left({dr}\right)^{\mathrm{2}} },\:{as} \\ $$$${we}\:{know}\:{from}\:{polar}\:{coordinate}\:{system} \\ $$$${when}\:{calculating}\:{length}\:{of}\:{curve}. \\ $$$${can}\:{you}\:{please}\:{check}\:{sir}.\:{thank}\:{you}. \\ $$

Commented by ajfour last updated on 10/Oct/18

checked sir, but the solution cant  be deleted very soon..!

$${checked}\:{sir},\:{but}\:{the}\:{solution}\:{cant} \\ $$$${be}\:{deleted}\:{very}\:{soon}..! \\ $$

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