A catenoid is a surface of revolution. Points in a cateniod, $\displaystyle \tilde{p}\\\epsilon\\\\C$, can be written parametrically as

$\displaystyle \tilde{x}(\tilde{p})=(c\cosh(\frac{u}{c})\cos(v),c \cosh(\frac{u}{c})\sin(v),u)$

a. calculate $\displaystyle \frac{\partial\tilde{x}}{\partial\\u}\mid_{p}$ and $\displaystyle \frac{\partial\tilde{x}}{\partial\\v}\mid_{p}$

$\displaystyle \frac{\partial\tilde{x}}{\partial\\u}\mid_{p}=(\si nh(\frac{u}{c})\cos(v),\sinh(\frac{u}{c})\sin(v),1 )$

$\displaystyle \frac{\partial\tilde{x}}{\partial\\u}\mid_{p}=(-c\cosh(\frac{u}{c})\sin(v),c\\\cosh(\frac{u}{c})\c os(v),0)$

b. If a metric on C is defined by $\displaystyle g(\tilde{X},\tilde{Y}):=^{(3)}g(\tilde{X},\tilde{Y })$ where $\displaystyle \tilde{X},\tilde{Y}$ are vector fields tangent to C, show that the components $\displaystyle g'_{ij}$ of g relative to the basis $\displaystyle \{\frac{\partial\tilde{x}}{\partial\\u},\frac{\par tial\tilde{x}}{\partial\\v}\}$ for vector fields on C are

$\displaystyle [g'_{ij}]=\left[\begin{array}{cc}\cosh^{2}(\frac{u}{c})&0\\0&c^2\c osh^{2}(\frac{u}{c})\end{array}\right]$

$\displaystyle E=\|\frac{\partial\tilde{x}}{\partial\\u}\|^{2}=(\ sinh^{2}(\frac{u}{c})\cos^{2}(v))+(\sinh^{2}(\frac {u}{c})\sin^{2}(v))+1^{2}=\cosh^{2}(\frac{u}{c})$

$\displaystyle F=\frac{\partial\tilde{x}}{\partial\\u}.\frac{\par tial\tilde{x}}{\partial\\v}=0?$ ive tried to do the math, but cant get zero, but all

$\displaystyle G=\|\frac{\partial\tilde{x}}{\partial\\v}\|^{2}=(-c^{2}\cosh^{2}(\frac{u}{c})\sin(v))+(c^{2}\cosh^{2 }(\frac{u}{c})\cos(v))=c^{2}\cosh(\frac{u}{c})$

$\displaystyle e=\frac{1}{c}$

$\displaystyle f=0$

$\displaystyle g=-c$

I cant find out how to find these values properly, coz according to my maths wolfram and other are wrong, they a back to front from what i calculated?

c.Use the metric of part b. in Gauss's formula to calculate $\displaystyle K(\tilde{p})\\\epsilon\\H$

$\displaystyle K(\tilde{p})=\frac{eg-f}{EG-F}=\frac{-1}{c^{2}\cosh^{4}(\frac{u}{c})}=\frac{-\\sech^{4}(\frac{u}{c})}{c^{2}}$

d. Using the metric of part b. show that the only non-zero $\displaystyle \Gamma^i_{jk}$ are $\displaystyle \Gamma^1_{11}$=$\displaystyle \Gamma^2_{12}$=$\displaystyle \Gamma^2_{21}$=$\displaystyle \frac{\tanh(\frac{u}{c})}{c})$ and $\displaystyle \Gamma^1_{22}$$\displaystyle =-c\tanh(\frac{u}{c})$. Hence write the geodesic equations on the catenoid. Dont solve them

$\displaystyle \Gamma^1_{11}=\frac{1}{2}g^{11}\{\frac{\partial\\g _{11}}{\partial\\u}+\frac{\partial\\g_{11}}{\parti al\\u}-\frac{\partial\\g_{11}}{\partial\\u}\}=\frac{1}{2} \{\frac{2\cosh(\frac{u}{c})\sinh(\frac{u}{c}}{c}\} =\frac{\tanh(\frac{u}{c})}{c}$

then form there i have no idea... and i think the other questions are quite wrong...

e.are the v-coordinate curves geodesic curves? [hint use your geodesic equations from d. above]