Answer to Question #148344 in Differential Geometry | Topology for anjali

Question #148344
Calculate the normal and the geodesic curvatures of the following curves on
the given surfaces:
(b) The right circular helix γ(θ) = (a cos θ, a sin θ, bθ) on the right circular cylinder
σ(u, v) = (a cos u, a sin u, v), where a, b > 0 are constants.
1
Expert's answer
2020-12-03T16:52:57-0500
"Solution"

The unit normal vector is given by (http://mathonline.wikidot.com/the-frenet-serret-formulas): "N(s)=\\frac{T'(s)}{||T'(s)||}" , where "T(s)=\\frac{\\vec{r}'(s)}{||\\vec{r}'(s)||},"

where "r(s)" is an arc-length parametrization of "\\vec{r}(t)"

"\\vec{r}(t)".


(b) We calculate the length of the curve between "[0,2\\pi)" and get:


For the the The normal curvature is related to the second fundamental form, and an expression for it is where "\\vec{N}"is related to the second fundamental form, and an expression for it is


"k_n=L \\dot{u}^2+2M\\dot{u}\\dot{v}+N\\dot{v}^2"


Where there is the surface on the paraboloid "\\sigma (u,v)=(u, v, u^2+v^2)" then

"L=\\sigma_{uu} \\cdot \\vec{N},\\ M=\\sigma_{uv} \\cdot \\vec{N}, L=\\sigma_{vv} \\cdot \\vec{N},"

where "\\vec{N}"is the normal vector to the surface.


But on the paraboloid, "z=0". So, "\\sigma_{uu}=\\sigma_{uv}=\\sigma_{vv}=0" and also "L=M=N=0" thus "k_n=0"



Geodesic Curvature;

"I=\\int_{0}^{2\\pi}\\sqrt{(x'(\\theta))^2+(y'(\\theta))^2+(z'(\\theta))^2}d\\theta=\\\\\\int_{0}^{2\\pi}\\sqrt{a^2sin^2(\\theta)+a^2cos^2(\\theta)+b^2}d\\theta=2\\pi\\sqrt{a^2+b^2}"


We make a change of variables: "s=\\theta\\sqrt{a^2+b^2}" . Then the curve takes the form:

"x(s)=a\\,cos(\\frac{s}{\\sqrt{a^2+b^2}}); y(s)=a\\,sin(\\frac{s}{\\sqrt{a^2+b^2}});\\,\\,z(s)=b\\frac{s}{\\sqrt{a^2+b^2}}"


In case we consider an interval "s\\in[0,2\\pi\\sqrt{a^2+b^2})""s\\in[0,2\\pi\\sqrt{a^2+b^2})" we receive an arc-length parametrization.

"\\vec{r}'(s)=(-\\frac{a}{\\sqrt{a^2+b^2}}\\,sin(\\frac{s}{\\sqrt{a^2+b^2}}),\\frac{a}{\\sqrt{a^2+b^2}}\\,cos(\\frac{s}{\\sqrt{a^2+b^2}}),\\frac{b}{\\sqrt{a^2+b^2}})";

"||\\vec{r}'(s)||=1""||\\vec{r}'(s)||=1".

"T(s)=\\vec{r}'(s)";

"T'(s)=(-\\frac{a}{{a^2+b^2}}\\,sin(\\frac{s}{\\sqrt{a^2+b^2}}),\\frac{a}{{a^2+b^2}}\\,cos(\\frac{s}{{a^2+b^2}}),0)" ;

"N'(s)=\\frac{T'(s)}{||T'(s)||}=\\frac{a^2+b^2}{|a|}(-\\frac{a}{{a^2+b^2}}\\,sin(\\frac{s}{\\sqrt{a^2+b^2}}),\\frac{a}{{a^2+b^2}}\\,cos(\\frac{s}{{a^2+b^2}}),0)"

The geodesic curvature is: "k_g=\\frac{(r',r'',n)}{|r'|^3}" .The normal vector for the parametric surface is: "n=r_u\\times r_v=\\begin{vmatrix}\n i & j& k \\\\\n -a\\,sin\\,u & a\\,cos\\,u&0\\\\\n 0 & 0&1\n\\end{vmatrix}=(a\\,cos\\,u,a\\,sin\\,u,0)" . We set "u=\\theta" and get the following curvature:

"k_g=\\frac{(r',r'',n)}{|r'|^3}=\\frac{\\begin{vmatrix}\n -a\\,sin\\,\\theta & a\\,cos\\,\\theta\\,&b \\\\\n -a\\,cos\\,\\theta & -a\\,sin\\,\\theta&0\\\\\na\\,cos\\,\\theta & a\\,sin\\,\\theta&0\n\\end{vmatrix}}{(a^2+b^2)^{3\/2}}=\\frac{1}{(a^2+b^2)^{3\/2}}b\\,\\,0=0"

Need a fast expert's response?

Submit order

and get a quick answer at the best price

for any assignment or question with DETAILED EXPLANATIONS!

Comments

No comments. Be the first!

Leave a comment

LATEST TUTORIALS
New on Blog
APPROVED BY CLIENTS