Finding parametrized curve
Unit sphere parametrized as $x(θ, ϕ) = (sin (θ) cos (ϕ), sin (θ) sin (ϕ), cos (θ))$, how would one find a parametric curve which has $α(0)=(1,0,0)$ and $α'(0)=(0,3,4)$?
differential-geometry
edited Nov 27 '18 at 16:16
Tianlalu
3,09621038
asked Nov 27 '18 at 16:15
Jim
1
add a comment |
Unit sphere parametrized as $x(θ, ϕ) = (sin (θ) cos (ϕ), sin (θ) sin (ϕ), cos (θ))$, how would one find a parametric curve which has $α(0)=(1,0,0)$ and $α'(0)=(0,3,4)$?
differential-geometry
edited Nov 27 '18 at 16:16
Tianlalu
3,09621038
asked Nov 27 '18 at 16:15
Jim
1
Check out this post.
– MisterRiemann
Nov 27 '18 at 16:25
add a comment |
Unit sphere parametrized as $x(θ, ϕ) = (sin (θ) cos (ϕ), sin (θ) sin (ϕ), cos (θ))$, how would one find a parametric curve which has $α(0)=(1,0,0)$ and $α'(0)=(0,3,4)$?
differential-geometry
edited Nov 27 '18 at 16:16
Tianlalu
3,09621038
asked Nov 27 '18 at 16:15
Jim
1
Unit sphere parametrized as $x(θ, ϕ) = (sin (θ) cos (ϕ), sin (θ) sin (ϕ), cos (θ))$, how would one find a parametric curve which has $α(0)=(1,0,0)$ and $α'(0)=(0,3,4)$?
differential-geometry
differential-geometry
edited Nov 27 '18 at 16:16
Tianlalu
3,09621038
asked Nov 27 '18 at 16:15
Jim
1
edited Nov 27 '18 at 16:16
Tianlalu
3,09621038
asked Nov 27 '18 at 16:15
Jim
1
edited Nov 27 '18 at 16:16
Tianlalu
3,09621038
edited Nov 27 '18 at 16:16
Tianlalu
3,09621038
edited Nov 27 '18 at 16:16
Tianlalu
3,09621038
3,09621038
asked Nov 27 '18 at 16:15
Jim
1
asked Nov 27 '18 at 16:15
Jim
1
asked Nov 27 '18 at 16:15
Jim
1
1
Check out this post.
– MisterRiemann
Nov 27 '18 at 16:25
add a comment |
Check out this post.
– MisterRiemann
Nov 27 '18 at 16:25
Check out this post.
– MisterRiemann
Nov 27 '18 at 16:25
Check out this post.
– MisterRiemann
Nov 27 '18 at 16:25
add a comment |
1 Answer
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Define $v_0=(1,0,0)$ and $v_1=tfrac1{5}(0,3,4)$. Then
$$
alpha(t) = cos(5t)v_0+sin(5t)v_1
$$
is the curve you want.
answered Nov 27 '18 at 16:32
Federico
4,649514
add a comment |
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1 Answer
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1 Answer
1
active
oldest
votes
active
oldest
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Define $v_0=(1,0,0)$ and $v_1=tfrac1{5}(0,3,4)$. Then
$$
alpha(t) = cos(5t)v_0+sin(5t)v_1
$$
is the curve you want.
answered Nov 27 '18 at 16:32
Federico
4,649514
add a comment |
Define $v_0=(1,0,0)$ and $v_1=tfrac1{5}(0,3,4)$. Then
$$
alpha(t) = cos(5t)v_0+sin(5t)v_1
$$
is the curve you want.
answered Nov 27 '18 at 16:32
Federico
4,649514
add a comment |
Define $v_0=(1,0,0)$ and $v_1=tfrac1{5}(0,3,4)$. Then
$$
alpha(t) = cos(5t)v_0+sin(5t)v_1
$$
is the curve you want.
answered Nov 27 '18 at 16:32
Federico
4,649514
Define $v_0=(1,0,0)$ and $v_1=tfrac1{5}(0,3,4)$. Then
$$
alpha(t) = cos(5t)v_0+sin(5t)v_1
$$
is the curve you want.
answered Nov 27 '18 at 16:32
Federico
4,649514
answered Nov 27 '18 at 16:32
Federico
4,649514
answered Nov 27 '18 at 16:32
Federico
4,649514
answered Nov 27 '18 at 16:32
Federico
4,649514
4,649514
add a comment |
add a comment |
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Check out this post.
– MisterRiemann
Nov 27 '18 at 16:25