Question and Answers Forum

All Questions      Topic List

Geometry Questions

Previous in All Question      Next in All Question      

Previous in Geometry      Next in Geometry      

Question Number 46073 by ajfour last updated on 20/Oct/18

Commented by ajfour last updated on 20/Oct/18

$$DeterminecoordinatesofA;$$$$b>c.Piscontactpointofside$$$$ABof△ABCwithupperrimof$$$$cylinder(heightH,radiusR).$$

Answered by MrW3 last updated on 23/Oct/18

$$letd=\sqrt{R^2-\frac{a^2}{4}}$$$$x_B=\frac{a}{2}$$$$y_B=-d$$$$x_C=-\frac{a}{2}$$$$y_C=-d$$$$z_B=z_C=0$$$$x_P=R\cos \varphi$$$$y_P=R\sin \varphi$$$$z_P=H$$$$x_A=\frac{a}{2}+(R\cos \varphi -\frac{a}{2})\lambda$$$$\Rightarrow x_A-x_B=(R\cos \varphi -\frac{a}{2})\lambda$$$$\Rightarrow x_A-x_C=a+(R\cos \varphi -\frac{a}{2})\lambda$$$$y_A=-d+(R\sin \varphi +d)\lambda$$$$\Rightarrow y_A-y_B=(R\sin \varphi +d)\lambda$$$$\Rightarrow y_A-y_C=(R\sin \varphi +d)\lambda$$$$z_A=H\lambda$$$$\Rightarrow z_A-z_B=H\lambda$$$$\Rightarrow z_A-z_C=H\lambda$$$$AB=\sqrt{{(x_A-x_B)}^2+{(y_A-y_B)}^2+{(z_A-z_B)}^2}=c$$$$\Rightarrow \lambda \sqrt{{(R\cos \varphi -\frac{a}{2})}^2+{(R\sin \varphi +d)}^2+H^2}=c...\left(i\right)$$$$$$$$AC=\sqrt{{(x_A-x_C)}^2+{(y_A-y_C)}^2+{(z_A-z_C)}^2}=b$$$$\Rightarrow \lambda \sqrt{{(\frac{a}{\lambda }+R\cos \varphi -\frac{a}{2})}^2+{(R\sin \varphi +d)}^2+H^2}=b...\left(ii\right)$$$$\Rightarrow \lambda \sqrt{{\left(\frac{a}{\lambda }\right)}^2+2\left(\frac{a}{\lambda }\right)(R\cos \varphi -\frac{a}{2})+{(R\cos \varphi -\frac{a}{2})}^2+{(R\sin \varphi +d)}^2+H^2}=b$$$$\Rightarrow \lambda \sqrt{{\left(\frac{a}{\lambda }\right)}^2+2\left(\frac{a}{\lambda }\right)(R\cos \varphi -\frac{a}{2})+\frac{c^2}{{\lambda }^2}}=b$$$$\Rightarrow \sqrt{a^2+c^2+2a\lambda (R\cos \varphi -\frac{a}{2})}=b$$$$\Rightarrow 2a\lambda (R\cos \varphi -\frac{a}{2})=b^2-(a^2+c^2)$$$$\Rightarrow \cos \varphi =\frac{1}{2R}[a-\frac{a^2+c^2-b^2}{a\lambda }]...\left(iii\right)$$$$\left(i\right):$$$$\lambda \sqrt{R^2-aR\cos \varphi +\frac{a^2}{4}+2dR\sin \varphi +d^2+H^2}=c$$$$\lambda \sqrt{2R^2+H^2-R(a\cos \varphi -2d\sin \varphi )}=c$$$$H^2+2R^2[1-\sin (\alpha -\varphi )]=\frac{c^2}{{\lambda }^2}$$$$\sin (\alpha -\varphi )=1-\frac{1}{2R^2}(\frac{c^2}{{\lambda }^2}-H^2)...\left(iv\right)$$$$with\alpha ={\sin }^{-1}\frac{a}{2R}$$$$$$$$put\left(iii\right)into\left(iv\right)wegetaneqn.for\lambda .$$$$with\lambda and\varphi wecangetthecoordinates$$$$ofpointA.$$$$$$$$\Rightarrow {\cos }^{-1}\left[\frac{1}{2aR}(a^2-\frac{a^2+c^2-b^2}{\lambda })\right]+{\cos }^{-1}[1-\frac{1}{2R^2}(\frac{c^2}{{\lambda }^2}-H^2)]={\sin }^{-1}\frac{a}{2R}$$

Commented by ajfour last updated on 21/Oct/18

$$ThanksalotSir,butcanyou$$$$pleasecheckmyequationfor{\mathit{z}}_A?$$$$Ifc=b,andevena=2R,ithink$$$$igetfrommyanswer,$$$$z=\sqrt{\frac{R^4}{H^2}+b^2}-\frac{R^2}{H}.$$$$Doesthisseemtrue,Sir?$$

Commented by MrW3 last updated on 21/Oct/18

$$a=2R\Rightarrow d=0$$$$c=b$$$$\Rightarrow \cos \varphi =1-\frac{1}{\lambda }$$$$\lambda \sqrt{H^2+\frac{2R^2}{\lambda }}=b$$$${\lambda }^2(H^2+\frac{2R^2}{\lambda })=b^2$$$$H^2{\lambda }^2+2R^2\lambda -b^2=0$$$$\lambda =\frac{-R^2+\sqrt{R^4+H^2{b}^2}}{H^2}=\sqrt{{\left(\frac{R}{H}\right)}^4+{\left(\frac{b}{H}\right)}^2}-{\left(\frac{R}{H}\right)}^2$$$$z_A=\lambda H=\sqrt{\frac{R^4}{H^2}+b^2}-\frac{R^2}{H}$$$$\Rightarrow youransweriscorrect.$$

Commented by ajfour last updated on 23/Oct/18

$$QuitecompactSir!$$

Answered by ajfour last updated on 21/Oct/18

$$[4R^2+4(c^2-z^2)\left(\frac{z-H}{z}\right)-{\left(\frac{b^2-c^2}{a}\right)}^2]$$$$\times (R^2-\frac{a^2}{4})$$$$={[\left(\frac{b^2+c^2}{2}\right)-2R^2-zH-c^2\left(\frac{z-H}{z}\right)]}^2$$$$$$

Answered by ajfour last updated on 21/Oct/18

$$A(r\sin \theta ,r\cos \theta ,z)$$$$\theta beingangleofradiusfrom$$$$y-axes.$$$$R\sin \alpha =\frac{a}{2},R\cos \alpha =\sqrt{R^2-\frac{a^2}{4}}$$$$B(\frac{a}{2},-R\cos \alpha ,0)$$$$C(-\frac{a}{2},-R\cos \alpha ,0)$$$${(r\sin \theta -\frac{a}{2})}^2+{(r\cos \theta +R\cos \alpha )}^2$$$$+z^2=c^2...\left(i\right)$$$${(r\sin \theta +\frac{a}{2})}^2+{(r\cos \theta +R\cos \alpha )}^2$$$$+z^2=b^2....\left(ii\right)$$$$\left(i\right)-\left(ii\right)gives$$$$2\mathit{a}\mathit{r}\sin \mathit{\theta }={\mathit{b}}^2-{\mathit{c}}^2....\left(I\right)$$$$2r\cos \theta =\sqrt{4r^2-{\left(\frac{b^2-c^2}{a}\right)}^2}..\left(II\right)$$$$APBisstraight,hence$$$$\frac{r\sin \theta -R\sin \varphi }{r\sin \theta -\frac{a}{2}}=\frac{r\cos \theta -R\cos \varphi }{r\cos \theta +R\cos \alpha }$$$$=\frac{z-H}{z}....\left(iii\right)$$$${\left(R\sin \varphi \right)}^2+{\left(R\cos \varphi \right)}^2=R^2,$$$$so$$$${[r\sin \theta -\left(\frac{z-H}{z}\right)(r\sin \theta -\frac{a}{2})]}^2$$$$+{[r\cos \theta -\left(\frac{z-H}{z}\right)(r\cos \theta +R\cos \alpha )]}^2=R^2$$$$.....$$$$.....$$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com