curves and multivariable function
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let $g:Rrightarrow R^3$ be a curve such that for every $textbf{t}$ the vector $g(textbf{t})$ is orthogonal to the tanget line to the curve at the point
prove that the curve lies on a sphere centered at $(0,0)$.
help please heres what i did :
$$
g(t) = (x(t),y(t),z(t)) \
g'(t) = (x' , y' , z') \
<g(t),g'(t)> = 0 \
xx' + yy' + zz' = 0
$$
now how to continue ? through integrals ?
functions curves
edited Nov 12 at 17:48
Joey Kilpatrick
977121
asked Nov 12 at 17:44
Razi Awad
236
add a comment |
up vote
0
down vote
favorite
let $g:Rrightarrow R^3$ be a curve such that for every $textbf{t}$ the vector $g(textbf{t})$ is orthogonal to the tanget line to the curve at the point
prove that the curve lies on a sphere centered at $(0,0)$.
help please heres what i did :
$$
g(t) = (x(t),y(t),z(t)) \
g'(t) = (x' , y' , z') \
<g(t),g'(t)> = 0 \
xx' + yy' + zz' = 0
$$
now how to continue ? through integrals ?
functions curves
edited Nov 12 at 17:48
Joey Kilpatrick
977121
asked Nov 12 at 17:44
Razi Awad
236
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
let $g:Rrightarrow R^3$ be a curve such that for every $textbf{t}$ the vector $g(textbf{t})$ is orthogonal to the tanget line to the curve at the point
prove that the curve lies on a sphere centered at $(0,0)$.
help please heres what i did :
$$
g(t) = (x(t),y(t),z(t)) \
g'(t) = (x' , y' , z') \
<g(t),g'(t)> = 0 \
xx' + yy' + zz' = 0
$$
now how to continue ? through integrals ?
functions curves
edited Nov 12 at 17:48
Joey Kilpatrick
977121
asked Nov 12 at 17:44
Razi Awad
236
let $g:Rrightarrow R^3$ be a curve such that for every $textbf{t}$ the vector $g(textbf{t})$ is orthogonal to the tanget line to the curve at the point
prove that the curve lies on a sphere centered at $(0,0)$.
help please heres what i did :
$$
g(t) = (x(t),y(t),z(t)) \
g'(t) = (x' , y' , z') \
<g(t),g'(t)> = 0 \
xx' + yy' + zz' = 0
$$
now how to continue ? through integrals ?
functions curves
functions curves
edited Nov 12 at 17:48
Joey Kilpatrick
977121
asked Nov 12 at 17:44
Razi Awad
236
edited Nov 12 at 17:48
Joey Kilpatrick
977121
asked Nov 12 at 17:44
Razi Awad
236
edited Nov 12 at 17:48
Joey Kilpatrick
977121
edited Nov 12 at 17:48
Joey Kilpatrick
977121
edited Nov 12 at 17:48
Joey Kilpatrick
977121
977121
asked Nov 12 at 17:44
Razi Awad
236
asked Nov 12 at 17:44
Razi Awad
236
asked Nov 12 at 17:44
Razi Awad
236
236
add a comment |
add a comment |
1 Answer
1
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votes
up vote
1
down vote
Once you have $x(t)x'(t)+y(t)y'(t)+z(t)z'(t)=0$, youn only have to look for a Primitive.
$$0=int_0^t 0ds=int_0^t(x(s)x'(s)+y(s)y'(s)+z(s)z'(s))dt=frac{1}{2}left(x(t)^2+y(t)^2+z(t)^2right)-C$$ where $C=frac{1}{2}left(x(0)^2+y(0)^2+z(0)^2right)$ is some real constant.
If $C$ where negative, your curve wiil lie in the circle $x^2+y^2+z^2=2C$.
answered Nov 12 at 17:57
Tito Eliatron
810418
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
Once you have $x(t)x'(t)+y(t)y'(t)+z(t)z'(t)=0$, youn only have to look for a Primitive.
$$0=int_0^t 0ds=int_0^t(x(s)x'(s)+y(s)y'(s)+z(s)z'(s))dt=frac{1}{2}left(x(t)^2+y(t)^2+z(t)^2right)-C$$ where $C=frac{1}{2}left(x(0)^2+y(0)^2+z(0)^2right)$ is some real constant.
If $C$ where negative, your curve wiil lie in the circle $x^2+y^2+z^2=2C$.
answered Nov 12 at 17:57
Tito Eliatron
810418
add a comment |
up vote
1
down vote
Once you have $x(t)x'(t)+y(t)y'(t)+z(t)z'(t)=0$, youn only have to look for a Primitive.
$$0=int_0^t 0ds=int_0^t(x(s)x'(s)+y(s)y'(s)+z(s)z'(s))dt=frac{1}{2}left(x(t)^2+y(t)^2+z(t)^2right)-C$$ where $C=frac{1}{2}left(x(0)^2+y(0)^2+z(0)^2right)$ is some real constant.
If $C$ where negative, your curve wiil lie in the circle $x^2+y^2+z^2=2C$.
answered Nov 12 at 17:57
Tito Eliatron
810418
add a comment |
up vote
1
down vote
up vote
1
down vote
Once you have $x(t)x'(t)+y(t)y'(t)+z(t)z'(t)=0$, youn only have to look for a Primitive.
$$0=int_0^t 0ds=int_0^t(x(s)x'(s)+y(s)y'(s)+z(s)z'(s))dt=frac{1}{2}left(x(t)^2+y(t)^2+z(t)^2right)-C$$ where $C=frac{1}{2}left(x(0)^2+y(0)^2+z(0)^2right)$ is some real constant.
If $C$ where negative, your curve wiil lie in the circle $x^2+y^2+z^2=2C$.
answered Nov 12 at 17:57
Tito Eliatron
810418
Once you have $x(t)x'(t)+y(t)y'(t)+z(t)z'(t)=0$, youn only have to look for a Primitive.
$$0=int_0^t 0ds=int_0^t(x(s)x'(s)+y(s)y'(s)+z(s)z'(s))dt=frac{1}{2}left(x(t)^2+y(t)^2+z(t)^2right)-C$$ where $C=frac{1}{2}left(x(0)^2+y(0)^2+z(0)^2right)$ is some real constant.
If $C$ where negative, your curve wiil lie in the circle $x^2+y^2+z^2=2C$.
answered Nov 12 at 17:57
Tito Eliatron
810418
answered Nov 12 at 17:57
Tito Eliatron
810418
answered Nov 12 at 17:57
Tito Eliatron
810418
answered Nov 12 at 17:57
Tito Eliatron
810418
810418
add a comment |
add a comment |
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