Question

2. Prove that the vector field obtained on the torus by parametrizing all its meridians by arc length and taking their tangent vectors (Example 1) is differentiable.

Answer 1

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ANSWER:

Given that,

the vector field obtained on the torus by parametrizing all its meridians by arc length.

Here we need to prove that the vector field obtained on the torus by parametrizing all its meridians by arc length and their tangent vectors are differentiable.

Explicitly,a differentiable vector field W in the torus.

Meridians of $T^2$ parameterized by circular segment length, for all
pET
characterize W(p) as the speed vector of the meridian going through p.

x = (R+r tanϕ)cosθ

y = (R+r tanϕ)sinθ

z = r sinϕ

Where θ,ϕ∈[0,2π]

$\frac{\mathrm{d} }{\mathrm{d} \Theta }(x)=\frac{\mathrm{d} }{\mathrm{d} \Theta }(R+r tan \phi )cos \Theta$

(y) = de(R+ rtano)sine de

lo(~) _ (r.sino)

Thus, the vector field obtained on the torus by parametrizing all its meridians by arc length and their tangent vectors are differentiable.

Hence proved.

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