Question Number 46073 by ajfour last updated on 20/Oct/18
Commented by ajfour last updated on 20/Oct/18
$${Determine}\:{coordinates}\:{of}\:{A}; \\ $$$$\:\:\:{b}\:>\:{c}\:.\:{P}\:{is}\:{contact}\:{point}\:{of}\:{side} \\ $$$${AB}\:{of}\:\bigtriangleup{ABC}\:{with}\:{upper}\:{rim}\:{of} \\ $$$${cylinder}\:\left({height}\:{H},\:{radius}\:{R}\right). \\ $$
Answered by MrW3 last updated on 23/Oct/18
![let d=(√(R^2 −(a^2 /4))) x_B =(a/2) y_B =−d x_C =−(a/2) y_C =−d z_B =z_C =0 x_P =R cos φ y_P =R sin φ z_P =H x_A =(a/2)+(R cos φ−(a/2))λ ⇒x_A −x_B =(R cos φ−(a/2))λ ⇒x_A −x_C =a+(R cos φ−(a/2))λ y_A =−d+(R sin φ+d)λ ⇒y_A −y_B =(R sin φ+d)λ ⇒y_A −y_C =(R sin φ+d)λ z_A =Hλ ⇒z_A −z_B =Hλ ⇒z_A −z_C =Hλ AB=(√((x_A −x_B )^2 +(y_A −y_B )^2 +(z_A −z_B )^2 ))=c ⇒λ(√((R cos φ−(a/2))^2 +(R sin φ+d)^2 +H^2 ))=c ...(i) AC=(√((x_A −x_C )^2 +(y_A −y_C )^2 +(z_A −z_C )^2 ))=b ⇒λ(√(((a/λ)+R cos φ−(a/2))^2 +(R sin φ+d)^2 +H^2 ))=b ...(ii) ⇒λ(√(((a/λ))^2 +2((a/λ))(R cos φ−(a/2))+(R cos φ−(a/2))^2 +(R sin φ+d)^2 +H^2 ))=b ⇒λ(√(((a/λ))^2 +2((a/λ))(R cos φ−(a/2))+(c^2 /λ^2 )))=b ⇒(√(a^2 +c^2 +2aλ(R cos φ−(a/2))))=b ⇒2aλ(R cos φ−(a/2))=b^2 −(a^2 +c^2 ) ⇒cos φ=(1/(2R))[a−((a^2 +c^2 −b^2 )/(aλ))] ...(iii) (i): λ(√(R^2 −aR cos φ+(a^2 /4)+2dR sin φ+d^2 +H^2 ))=c λ(√(2R^2 +H^2 −R(a cos φ−2d sin φ)))=c H^2 +2R^2 [1−sin (α−φ)]=(c^2 /λ^2 ) sin (α−φ)=1−(1/(2R^2 ))((c^2 /λ^2 ) −H^2 ) ...(iv) with α=sin^(−1) (a/(2R)) put (iii) into (iv) we get an eqn. for λ. with λ and φ we can get the coordinates of point A. ⇒cos^(−1) [(1/(2aR))(a^2 −((a^2 +c^2 −b^2 )/λ))] +cos^(−1) [1−(1/(2R^2 ))((c^2 /λ^2 ) −H^2 )]=sin^(−1) (a/(2R))](https://www.tinkutara.com/question/Q46102.png)
$${let}\:{d}=\sqrt{{R}^{\mathrm{2}} −\frac{{a}^{\mathrm{2}} }{\mathrm{4}}} \\ $$$${x}_{{B}} =\frac{{a}}{\mathrm{2}} \\ $$$${y}_{{B}} =−{d} \\ $$$${x}_{{C}} =−\frac{{a}}{\mathrm{2}} \\ $$$${y}_{{C}} =−{d} \\ $$$${z}_{{B}} ={z}_{{C}} =\mathrm{0} \\ $$$${x}_{{P}} ={R}\:\mathrm{cos}\:\phi \\ $$$${y}_{{P}} ={R}\:\mathrm{sin}\:\phi \\ $$$${z}_{{P}} ={H} \\ $$$${x}_{{A}} =\frac{{a}}{\mathrm{2}}+\left({R}\:\mathrm{cos}\:\phi−\frac{{a}}{\mathrm{2}}\right)\lambda \\ $$$$\Rightarrow{x}_{{A}} −{x}_{{B}} =\left({R}\:\mathrm{cos}\:\phi−\frac{{a}}{\mathrm{2}}\right)\lambda \\ $$$$\Rightarrow{x}_{{A}} −{x}_{{C}} ={a}+\left({R}\:\mathrm{cos}\:\phi−\frac{{a}}{\mathrm{2}}\right)\lambda \\ $$$${y}_{{A}} =−{d}+\left({R}\:\mathrm{sin}\:\phi+{d}\right)\lambda \\ $$$$\Rightarrow{y}_{{A}} −{y}_{{B}} =\left({R}\:\mathrm{sin}\:\phi+{d}\right)\lambda \\ $$$$\Rightarrow{y}_{{A}} −{y}_{{C}} =\left({R}\:\mathrm{sin}\:\phi+{d}\right)\lambda \\ $$$${z}_{{A}} ={H}\lambda \\ $$$$\Rightarrow{z}_{{A}} −{z}_{{B}} ={H}\lambda \\ $$$$\Rightarrow{z}_{{A}} −{z}_{{C}} ={H}\lambda \\ $$$${AB}=\sqrt{\left({x}_{{A}} −{x}_{{B}} \right)^{\mathrm{2}} +\left({y}_{{A}} −{y}_{{B}} \right)^{\mathrm{2}} +\left({z}_{{A}} −{z}_{{B}} \right)^{\mathrm{2}} }={c} \\ $$$$\Rightarrow\lambda\sqrt{\left({R}\:\mathrm{cos}\:\phi−\frac{{a}}{\mathrm{2}}\right)^{\mathrm{2}} +\left({R}\:\mathrm{sin}\:\phi+{d}\right)^{\mathrm{2}} +{H}^{\mathrm{2}} }={c}\:\:\:…\left({i}\right) \\ $$$$ \\ $$$${AC}=\sqrt{\left({x}_{{A}} −{x}_{{C}} \right)^{\mathrm{2}} +\left({y}_{{A}} −{y}_{{C}} \right)^{\mathrm{2}} +\left({z}_{{A}} −{z}_{{C}} \right)^{\mathrm{2}} }={b} \\ $$$$\Rightarrow\lambda\sqrt{\left(\frac{{a}}{\lambda}+{R}\:\mathrm{cos}\:\phi−\frac{{a}}{\mathrm{2}}\right)^{\mathrm{2}} +\left({R}\:\mathrm{sin}\:\phi+{d}\right)^{\mathrm{2}} +{H}^{\mathrm{2}} }={b}\:\:\:\:…\left({ii}\right) \\ $$$$\Rightarrow\lambda\sqrt{\left(\frac{{a}}{\lambda}\right)^{\mathrm{2}} +\mathrm{2}\left(\frac{{a}}{\lambda}\right)\left({R}\:\mathrm{cos}\:\phi−\frac{{a}}{\mathrm{2}}\right)+\left({R}\:\mathrm{cos}\:\phi−\frac{{a}}{\mathrm{2}}\right)^{\mathrm{2}} +\left({R}\:\mathrm{sin}\:\phi+{d}\right)^{\mathrm{2}} +{H}^{\mathrm{2}} }={b} \\ $$$$\Rightarrow\lambda\sqrt{\left(\frac{{a}}{\lambda}\right)^{\mathrm{2}} +\mathrm{2}\left(\frac{{a}}{\lambda}\right)\left({R}\:\mathrm{cos}\:\phi−\frac{{a}}{\mathrm{2}}\right)+\frac{{c}^{\mathrm{2}} }{\lambda^{\mathrm{2}} }}={b} \\ $$$$\Rightarrow\sqrt{{a}^{\mathrm{2}} +{c}^{\mathrm{2}} +\mathrm{2}{a}\lambda\left({R}\:\mathrm{cos}\:\phi−\frac{{a}}{\mathrm{2}}\right)}={b} \\ $$$$\Rightarrow\mathrm{2}{a}\lambda\left({R}\:\mathrm{cos}\:\phi−\frac{{a}}{\mathrm{2}}\right)={b}^{\mathrm{2}} −\left({a}^{\mathrm{2}} +{c}^{\mathrm{2}} \right) \\ $$$$\Rightarrow\mathrm{cos}\:\phi=\frac{\mathrm{1}}{\mathrm{2}{R}}\left[{a}−\frac{{a}^{\mathrm{2}} +{c}^{\mathrm{2}} −{b}^{\mathrm{2}} }{{a}\lambda}\right]\:\:\:…\left({iii}\right) \\ $$$$\left({i}\right): \\ $$$$\lambda\sqrt{{R}^{\mathrm{2}} −{aR}\:\mathrm{cos}\:\phi+\frac{{a}^{\mathrm{2}} }{\mathrm{4}}+\mathrm{2}{dR}\:\mathrm{sin}\:\phi+{d}^{\mathrm{2}} +{H}^{\mathrm{2}} }={c} \\ $$$$\lambda\sqrt{\mathrm{2}{R}^{\mathrm{2}} +{H}^{\mathrm{2}} −{R}\left({a}\:\mathrm{cos}\:\phi−\mathrm{2}{d}\:\mathrm{sin}\:\phi\right)}={c} \\ $$$${H}^{\mathrm{2}} +\mathrm{2}{R}^{\mathrm{2}} \left[\mathrm{1}−\mathrm{sin}\:\left(\alpha−\phi\right)\right]=\frac{{c}^{\mathrm{2}} }{\lambda^{\mathrm{2}} } \\ $$$$\mathrm{sin}\:\left(\alpha−\phi\right)=\mathrm{1}−\frac{\mathrm{1}}{\mathrm{2}{R}^{\mathrm{2}} }\left(\frac{{c}^{\mathrm{2}} }{\lambda^{\mathrm{2}} }\:−{H}^{\mathrm{2}} \right)\:\:…\left({iv}\right) \\ $$$${with}\:\alpha=\mathrm{sin}^{−\mathrm{1}} \frac{{a}}{\mathrm{2}{R}} \\ $$$$ \\ $$$${put}\:\left({iii}\right)\:{into}\:\left({iv}\right)\:{we}\:{get}\:{an}\:{eqn}.\:{for}\:\lambda. \\ $$$${with}\:\lambda\:{and}\:\phi\:{we}\:{can}\:{get}\:{the}\:{coordinates} \\ $$$${of}\:{point}\:{A}. \\ $$$$ \\ $$$$\Rightarrow\mathrm{cos}^{−\mathrm{1}} \left[\frac{\mathrm{1}}{\mathrm{2}{aR}}\left({a}^{\mathrm{2}} −\frac{{a}^{\mathrm{2}} +{c}^{\mathrm{2}} −{b}^{\mathrm{2}} }{\lambda}\right)\right]\:+\mathrm{cos}^{−\mathrm{1}} \left[\mathrm{1}−\frac{\mathrm{1}}{\mathrm{2}{R}^{\mathrm{2}} }\left(\frac{{c}^{\mathrm{2}} }{\lambda^{\mathrm{2}} }\:−{H}^{\mathrm{2}} \right)\right]=\mathrm{sin}^{−\mathrm{1}} \frac{{a}}{\mathrm{2}{R}} \\ $$
Commented by ajfour last updated on 21/Oct/18
$${Thanks}\:{a}\:{lot}\:{Sir},\:{but}\:{can}\:{you} \\ $$$${please}\:{check}\:{my}\:{equation}\:{for}\:\boldsymbol{{z}}_{{A}} \:? \\ $$$${If}\:\:{c}={b},\:\:{and}\:{even}\:{a}=\mathrm{2}{R},\:{i}\:{think} \\ $$$${i}\:{get}\:{from}\:{my}\:{answer}, \\ $$$$\:\:\:\:\:{z}\:=\:\sqrt{\frac{{R}^{\:\mathrm{4}} }{{H}^{\mathrm{2}} }+{b}^{\mathrm{2}} }\:−\frac{{R}^{\:\mathrm{2}} }{{H}}\:. \\ $$$$\:{Does}\:{this}\:{seem}\:{true},\:{Sir}\:? \\ $$
Commented by MrW3 last updated on 21/Oct/18
$${a}=\mathrm{2}{R}\Rightarrow{d}=\mathrm{0} \\ $$$${c}={b} \\ $$$$\Rightarrow\mathrm{cos}\:\phi=\mathrm{1}−\frac{\mathrm{1}}{\lambda} \\ $$$$\lambda\sqrt{{H}^{\mathrm{2}} +\frac{\mathrm{2}{R}^{\mathrm{2}} }{\lambda}}={b} \\ $$$$\lambda^{\mathrm{2}} \left({H}^{\mathrm{2}} +\frac{\mathrm{2}{R}^{\mathrm{2}} }{\lambda}\right)={b}^{\mathrm{2}} \\ $$$${H}^{\mathrm{2}} \lambda^{\mathrm{2}} +\mathrm{2}{R}^{\mathrm{2}} \lambda−{b}^{\mathrm{2}} =\mathrm{0} \\ $$$$\lambda=\frac{−{R}^{\mathrm{2}} +\sqrt{{R}^{\mathrm{4}} +{H}^{\mathrm{2}} {b}^{\mathrm{2}} }}{{H}^{\mathrm{2}} }=\sqrt{\left(\frac{{R}}{{H}}\right)^{\mathrm{4}} +\left(\frac{{b}}{{H}}\right)^{\mathrm{2}} }−\left(\frac{{R}}{{H}}\right)^{\mathrm{2}} \\ $$$${z}_{{A}} =\lambda{H}=\sqrt{\frac{{R}^{\mathrm{4}} }{{H}^{\mathrm{2}} }+{b}^{\mathrm{2}} }−\frac{{R}^{\mathrm{2}} }{{H}} \\ $$$$\Rightarrow{your}\:{answer}\:{is}\:{correct}. \\ $$
Commented by ajfour last updated on 23/Oct/18
$${Quite}\:{compact}\:{Sir}! \\ $$
Answered by ajfour last updated on 21/Oct/18
![[4R^2 +4(c^2 −z^2 )(((z−H)/z))−(((b^2 −c^2 )/a))^2 ] ×(R^2 −(a^2 /4)) =[(((b^2 +c^2 )/2))−2R^2 −zH−c^2 (((z−H)/z))]^2](https://www.tinkutara.com/question/Q46108.png)
$$\left[\mathrm{4}{R}^{\mathrm{2}} +\mathrm{4}\left({c}^{\mathrm{2}} −{z}^{\mathrm{2}} \right)\left(\frac{{z}−{H}}{{z}}\right)−\left(\frac{{b}^{\mathrm{2}} −{c}^{\mathrm{2}} }{{a}}\right)^{\mathrm{2}} \right] \\ $$$$×\left({R}^{\mathrm{2}} −\frac{{a}^{\mathrm{2}} }{\mathrm{4}}\right) \\ $$$$=\left[\left(\frac{{b}^{\mathrm{2}} +{c}^{\mathrm{2}} }{\mathrm{2}}\right)−\mathrm{2}{R}^{\mathrm{2}} −{zH}−{c}^{\mathrm{2}} \left(\frac{{z}−{H}}{{z}}\right)\right]^{\mathrm{2}} \\ $$$$ \\ $$
Answered by ajfour last updated on 21/Oct/18
![A(rsin θ, rcos θ, z) θ being angle of radius from y-axes. R sin α = (a/2) , Rcos α = (√(R^2 −(a^2 /4))) B((a/2), −Rcos α, 0) C(−(a/2), −Rcos α, 0) (rsin θ−(a/2))^2 +(rcos θ+Rcos α)^2 +z^2 = c^2 ...(i) (rsin θ+(a/2))^2 +(rcos θ+Rcos α)^2 +z^2 = b^2 ....(ii) (i)−(ii) gives 2arsin 𝛉 = b^2 −c^2 ....(I) 2rcos θ = (√(4r^2 −(((b^2 −c^2 )/a))^2 )) ..(II) APB is straight, hence ((rsin θ−Rsin φ)/(rsin θ−(a/2))) = ((rcos θ−Rcos φ)/(rcos θ+Rcos α)) = ((z−H)/z) ....(iii) (Rsin φ)^2 +(Rcos φ)^2 = R^2 , so [rsin θ−(((z−H)/z))(rsin θ−(a/2))]^2 +[rcos θ−(((z−H)/z))(rcos θ+Rcos α)]^2 = R^2 ..... .....](https://www.tinkutara.com/question/Q46149.png)
$${A}\left({r}\mathrm{sin}\:\theta,\:{r}\mathrm{cos}\:\theta,\:{z}\right) \\ $$$$\:\theta\:{being}\:{angle}\:{of}\:{radius}\:{from} \\ $$$${y}-{axes}. \\ $$$${R}\:\mathrm{sin}\:\alpha\:=\:\frac{{a}}{\mathrm{2}}\:\:,\:{R}\mathrm{cos}\:\alpha\:=\:\sqrt{{R}^{\mathrm{2}} −\frac{{a}^{\mathrm{2}} }{\mathrm{4}}} \\ $$$${B}\left(\frac{{a}}{\mathrm{2}},\:−{R}\mathrm{cos}\:\alpha,\:\mathrm{0}\right) \\ $$$${C}\left(−\frac{{a}}{\mathrm{2}},\:−{R}\mathrm{cos}\:\alpha,\:\mathrm{0}\right) \\ $$$$\left({r}\mathrm{sin}\:\theta−\frac{{a}}{\mathrm{2}}\right)^{\mathrm{2}} +\left({r}\mathrm{cos}\:\theta+{R}\mathrm{cos}\:\alpha\right)^{\mathrm{2}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:+{z}^{\mathrm{2}} \:=\:{c}^{\mathrm{2}} \:\:\:\:\:\:…\left({i}\right) \\ $$$$\left({r}\mathrm{sin}\:\theta+\frac{{a}}{\mathrm{2}}\right)^{\mathrm{2}} +\left({r}\mathrm{cos}\:\theta+{R}\mathrm{cos}\:\alpha\right)^{\mathrm{2}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:+{z}^{\mathrm{2}} \:=\:{b}^{\mathrm{2}} \:\:\:\:\:\:….\left({ii}\right) \\ $$$$\left({i}\right)−\left({ii}\right)\:{gives} \\ $$$$\:\:\:\:\:\:\mathrm{2}\boldsymbol{{ar}}\mathrm{sin}\:\boldsymbol{\theta}\:=\:\boldsymbol{{b}}^{\mathrm{2}} −\boldsymbol{{c}}^{\mathrm{2}} \:\:\:\:\:….\left({I}\right) \\ $$$$\:\mathrm{2}{r}\mathrm{cos}\:\theta\:=\:\sqrt{\mathrm{4}{r}^{\mathrm{2}} −\left(\frac{{b}^{\mathrm{2}} −{c}^{\mathrm{2}} }{{a}}\right)^{\mathrm{2}} }\:\:..\left({II}\right) \\ $$$${APB}\:{is}\:{straight},\:{hence} \\ $$$$\frac{{r}\mathrm{sin}\:\theta−{R}\mathrm{sin}\:\phi}{{r}\mathrm{sin}\:\theta−\frac{{a}}{\mathrm{2}}}\:=\:\frac{{r}\mathrm{cos}\:\theta−{R}\mathrm{cos}\:\phi}{{r}\mathrm{cos}\:\theta+{R}\mathrm{cos}\:\alpha} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\:\frac{{z}−{H}}{{z}}\:\:\:\:\:\:\:….\left({iii}\right) \\ $$$$\:\:\:\:\left({R}\mathrm{sin}\:\phi\right)^{\mathrm{2}} +\left({R}\mathrm{cos}\:\phi\right)^{\mathrm{2}} =\:{R}^{\mathrm{2}} \:\:\:, \\ $$$${so} \\ $$$$\left[{r}\mathrm{sin}\:\theta−\left(\frac{{z}−{H}}{{z}}\right)\left({r}\mathrm{sin}\:\theta−\frac{{a}}{\mathrm{2}}\right)\right]^{\mathrm{2}} \\ $$$$\:+\left[{r}\mathrm{cos}\:\theta−\left(\frac{{z}−{H}}{{z}}\right)\left({r}\mathrm{cos}\:\theta+{R}\mathrm{cos}\:\alpha\right)\right]^{\mathrm{2}} =\:{R}^{\mathrm{2}} \\ $$$$….. \\ $$$$….. \\ $$