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 \(\displaystyle \frac{\partial\tilde{x}}{\partial\\v}\mid_{p}\)

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

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

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 \(\displaystyle \{\frac{\partial\tilde{x}}{\partial\\u},\frac{\partial\tilde{x}}{\partial\\v}\}\) for vector fields on C are

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

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

\(\displaystyle F=\frac{\partial\tilde{x}}{\partial\\u}.\frac{\partial\tilde{x}}{\partial\\v}=0?\) 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

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

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]

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

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

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

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

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 \(\displaystyle \{\frac{\partial\tilde{x}}{\partial\\u},\frac{\partial\tilde{x}}{\partial\\v}\}\) for vector fields on C are

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

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

\(\displaystyle F=\frac{\partial\tilde{x}}{\partial\\u}.\frac{\partial\tilde{x}}{\partial\\v}=0?\) 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

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

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]

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