Chapter 24: Problem 23
Write the homogeneous coordinates of the point at infinity on the line \(2 x-y=0\)
Short Answer
Step by step solution
Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Chapter 24: Problem 23
Write the homogeneous coordinates of the point at infinity on the line \(2 x-y=0\)
These are the key concepts you need to understand to accurately answer the question.
All the tools & learning materials you need for study success - in one app.
Get started for freeIf \(\omega\) is the differential one-form in three dimensions given by \(\omega=A d x+B d y+C d z\), show that $$ \begin{aligned} d \omega=&\left(\frac{\partial C}{\partial y}-\frac{\partial B}{\partial z}\right) d y d z+\left(\frac{\partial A}{\partial z}-\frac{\partial C}{\partial x}\right) d z d x \\ &+\left(\frac{\partial B}{\partial x}-\frac{\partial A}{\partial y}\right) d x d y \end{aligned} $$
If \(x=x(u, v), y=y(u, v), z=z(u, v)\) are the parametric equations of a surface and if \(E=x_{u}^{2}+y_{u}^{2}+z_{u}^{2}, F=x_{u} x_{v}+\) \(y_{u} y_{v}+z_{u} z_{v}\), and \(G=x_{v}^{2}+y_{v}^{2}+z_{v}^{2}\), show that \(d x^{2}+d y^{2}+d z^{2}=E d u^{2}+2 F d u d v+G d v^{2} .\)
Show that if a surface is given in the form \(z=z(x, y)\), then the measure of curvature \(k\) can be expressed as $$ k=\frac{z_{x x} z_{y y}-z_{x y}^{2}}{\left(1+z_{x}^{2}+z_{y}^{2}\right)^{2}} $$ Hint: Show first that if \(X, Y, Z\) are coordinates on the unit sphere corresponding to the point \((x, y, z(x, y))\) on the given surface, then $$ \begin{aligned} &X=\frac{-z_{x}}{\sqrt{1+z_{x}^{2}+z_{y}^{2}}}, \quad Y=\frac{-z_{y}}{\sqrt{1+z_{x}^{2}+z_{y}^{2}}} \\ &Z=\frac{1}{\sqrt{1+z_{x}^{2}+z_{y}^{2}}} \end{aligned} $$
Show that Beltrami's formulas for the lengths \(\rho, s, t\) of the sides of a right triangle on his pseudosphere transform into $$ \begin{aligned} &\frac{r}{a}=\tanh \frac{\rho}{k}, \quad \frac{r}{a} \cos \theta=\tanh \frac{s}{k}, \quad \text { and } \\ &\frac{v}{\sqrt{a^{2}-u^{2}}}=\tanh \frac{t}{k} \end{aligned} $$ and then show that $$ \cosh \frac{s}{k} \cosh \frac{t}{k}=\cosh \frac{\rho}{k} $$
Given that tan \(\frac{1}{2} \Pi(x)=e^{-x}\), where \(\Pi(x)\) is Lobachevsky's angle of parallelism, derive the formulas $$ \sin \Pi(x)=\frac{1}{\cosh x} \quad \text { and } \quad \cos \Pi(x)=\tanh x $$ and show that their power series expansions up to degree 2 are \(\sin \Pi(x)=1-\frac{1}{2} x^{2}\) and \(\cos \Pi(x)=x\), respectively.
What do you think about this solution?
We value your feedback to improve our textbook solutions.