Mixing
Easier way to parametrize an ellipse.
$begingroup$
I want to calculate the parametrization of the curve $$E={(x,y,z):z=x^2+y^2,x+y+z=1}$$
To do so I did a change of variables taking as basis the vectors $frac{1}{sqrt{2}}(-1,1,0),frac{1}{sqrt{6}}(-1,-1,2),frac{1}{sqrt{3}}(1,1,1)$ which are the normal vector of the plane, and two other perpendicular vecors of the plane. With this basis the plane $x+y+z=1$ becomes $z'=sqrt{3}/3$, and $$begin{cases}x=-x'/sqrt{2}-y'/sqrt{6}+z'/sqrt{3}\y=x'/sqrt{2}-y'/sqrt{6}+z'/sqrt{3} \z=2y'/sqrt{6}+z'/sqrt{3}end{cases}$$
Now using this change of variables into $E$ I have that $$frac{(x')^2}{frac{1}{(3^{(1/4)})^2}}+frac{(y'-sqrt{2/3})^2}{(3^{(1/4)})^2}=1$$ and therefore the parametrization of the ellipse is $$x'=3^{1/4}costheta,;y'=sqrt{2/3}+3^{1/4}sintheta,; z'=sqrt{3}/3,; 0leq theta leq2pi$$
As I checked with wolfram alpha, this is correct. However it requires a lot of calculations to just parametrize this curve, is there any faster way to do it?
multivariable-calculus parametrization
edited Jan 16 at 13:46
user376343
3,8133828
asked Jan 16 at 13:37
John KeeperJohn Keeper
532315
$endgroup$
add a comment |
$begingroup$
I want to calculate the parametrization of the curve $$E={(x,y,z):z=x^2+y^2,x+y+z=1}$$
To do so I did a change of variables taking as basis the vectors $frac{1}{sqrt{2}}(-1,1,0),frac{1}{sqrt{6}}(-1,-1,2),frac{1}{sqrt{3}}(1,1,1)$ which are the normal vector of the plane, and two other perpendicular vecors of the plane. With this basis the plane $x+y+z=1$ becomes $z'=sqrt{3}/3$, and $$begin{cases}x=-x'/sqrt{2}-y'/sqrt{6}+z'/sqrt{3}\y=x'/sqrt{2}-y'/sqrt{6}+z'/sqrt{3} \z=2y'/sqrt{6}+z'/sqrt{3}end{cases}$$
Now using this change of variables into $E$ I have that $$frac{(x')^2}{frac{1}{(3^{(1/4)})^2}}+frac{(y'-sqrt{2/3})^2}{(3^{(1/4)})^2}=1$$ and therefore the parametrization of the ellipse is $$x'=3^{1/4}costheta,;y'=sqrt{2/3}+3^{1/4}sintheta,; z'=sqrt{3}/3,; 0leq theta leq2pi$$
As I checked with wolfram alpha, this is correct. However it requires a lot of calculations to just parametrize this curve, is there any faster way to do it?
multivariable-calculus parametrization
edited Jan 16 at 13:46
user376343
3,8133828
asked Jan 16 at 13:37
John KeeperJohn Keeper
532315
$endgroup$
$begingroup$
There are other parametrization options. Do you want this particular one?
$endgroup$
– Andrei
Jan 16 at 14:01
add a comment |
$begingroup$
I want to calculate the parametrization of the curve $$E={(x,y,z):z=x^2+y^2,x+y+z=1}$$
To do so I did a change of variables taking as basis the vectors $frac{1}{sqrt{2}}(-1,1,0),frac{1}{sqrt{6}}(-1,-1,2),frac{1}{sqrt{3}}(1,1,1)$ which are the normal vector of the plane, and two other perpendicular vecors of the plane. With this basis the plane $x+y+z=1$ becomes $z'=sqrt{3}/3$, and $$begin{cases}x=-x'/sqrt{2}-y'/sqrt{6}+z'/sqrt{3}\y=x'/sqrt{2}-y'/sqrt{6}+z'/sqrt{3} \z=2y'/sqrt{6}+z'/sqrt{3}end{cases}$$
Now using this change of variables into $E$ I have that $$frac{(x')^2}{frac{1}{(3^{(1/4)})^2}}+frac{(y'-sqrt{2/3})^2}{(3^{(1/4)})^2}=1$$ and therefore the parametrization of the ellipse is $$x'=3^{1/4}costheta,;y'=sqrt{2/3}+3^{1/4}sintheta,; z'=sqrt{3}/3,; 0leq theta leq2pi$$
As I checked with wolfram alpha, this is correct. However it requires a lot of calculations to just parametrize this curve, is there any faster way to do it?
multivariable-calculus parametrization
edited Jan 16 at 13:46
user376343
3,8133828
asked Jan 16 at 13:37
John KeeperJohn Keeper
532315
$endgroup$
I want to calculate the parametrization of the curve $$E={(x,y,z):z=x^2+y^2,x+y+z=1}$$
To do so I did a change of variables taking as basis the vectors $frac{1}{sqrt{2}}(-1,1,0),frac{1}{sqrt{6}}(-1,-1,2),frac{1}{sqrt{3}}(1,1,1)$ which are the normal vector of the plane, and two other perpendicular vecors of the plane. With this basis the plane $x+y+z=1$ becomes $z'=sqrt{3}/3$, and $$begin{cases}x=-x'/sqrt{2}-y'/sqrt{6}+z'/sqrt{3}\y=x'/sqrt{2}-y'/sqrt{6}+z'/sqrt{3} \z=2y'/sqrt{6}+z'/sqrt{3}end{cases}$$
Now using this change of variables into $E$ I have that $$frac{(x')^2}{frac{1}{(3^{(1/4)})^2}}+frac{(y'-sqrt{2/3})^2}{(3^{(1/4)})^2}=1$$ and therefore the parametrization of the ellipse is $$x'=3^{1/4}costheta,;y'=sqrt{2/3}+3^{1/4}sintheta,; z'=sqrt{3}/3,; 0leq theta leq2pi$$
As I checked with wolfram alpha, this is correct. However it requires a lot of calculations to just parametrize this curve, is there any faster way to do it?
multivariable-calculus parametrization
multivariable-calculus parametrization
edited Jan 16 at 13:46
user376343
3,8133828
asked Jan 16 at 13:37
John KeeperJohn Keeper
532315
edited Jan 16 at 13:46
user376343
3,8133828
asked Jan 16 at 13:37
John KeeperJohn Keeper
532315
edited Jan 16 at 13:46
user376343
3,8133828
edited Jan 16 at 13:46
user376343
3,8133828
edited Jan 16 at 13:46
user376343
3,8133828
3,8133828
asked Jan 16 at 13:37
John KeeperJohn Keeper
532315
asked Jan 16 at 13:37
John KeeperJohn Keeper
532315
asked Jan 16 at 13:37
John KeeperJohn Keeper
532315
532315
$begingroup$
There are other parametrization options. Do you want this particular one?
$endgroup$
– Andrei
Jan 16 at 14:01
add a comment |
$begingroup$
There are other parametrization options. Do you want this particular one?
$endgroup$
– Andrei
Jan 16 at 14:01
$begingroup$
There are other parametrization options. Do you want this particular one?
$endgroup$
– Andrei
Jan 16 at 14:01
$begingroup$
There are other parametrization options. Do you want this particular one?
$endgroup$
– Andrei
Jan 16 at 14:01
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
You can parametrize in the original coordinates. Solving the plane for $z$ you have:
$$x+y+z=1 iff z = 1-x-y tag{$star$}$$
so substituting into the paraboloid and rewriting gives:
$$z=x^2+y^2 implies 1-x-y=x^2+y^2 iff left(x+tfrac{1}{2}right)^2+left(y+tfrac{1}{2}right)^2=tfrac{3}{2}$$
This is easily parametrized as $x+tfrac{1}{2}=sqrt{tfrac{3}{2}}cos t$ and $y+tfrac{1}{2}=sqrt{tfrac{3}{2}}sin t$ and $z$ follows from $(star)$:
$$z = 1-x-y=1-left(sqrt{tfrac{3}{2}}cos t-tfrac{1}{2}right)-left(sqrt{tfrac{3}{2}}sin t-tfrac{1}{2}right)=2-sqrt{tfrac{3}{2}}cos t-sqrt{tfrac{3}{2}}sin t$$
to get:
$$gamma : [0,2pi]tomathbb{R}^3: t mapsto
left( sqrt{tfrac{3}{2}}cos t-tfrac{1}{2} ,
sqrt{tfrac{3}{2}}sin t-tfrac{1}{2} ,
2-sqrt{tfrac{3}{2}}cos t-sqrt{tfrac{3}{2}}sin t right)$$
answered Jan 16 at 14:17
StackTDStackTD
22.9k2152
$endgroup$
add a comment |
$begingroup$
$$z=x^2+y^2$$ hints the use of polar coordinates and the equation of the plane becomes $$rho^2+rho(costheta+sintheta)-1=0.$$
Solve for $rho$ in terms of $theta$ and you are done.
answered Jan 16 at 14:50
Yves DaoustYves Daoust
128k675227
$endgroup$
add a comment |
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2 Answers
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2 Answers
2
active
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votes
$begingroup$
You can parametrize in the original coordinates. Solving the plane for $z$ you have:
$$x+y+z=1 iff z = 1-x-y tag{$star$}$$
so substituting into the paraboloid and rewriting gives:
$$z=x^2+y^2 implies 1-x-y=x^2+y^2 iff left(x+tfrac{1}{2}right)^2+left(y+tfrac{1}{2}right)^2=tfrac{3}{2}$$
This is easily parametrized as $x+tfrac{1}{2}=sqrt{tfrac{3}{2}}cos t$ and $y+tfrac{1}{2}=sqrt{tfrac{3}{2}}sin t$ and $z$ follows from $(star)$:
$$z = 1-x-y=1-left(sqrt{tfrac{3}{2}}cos t-tfrac{1}{2}right)-left(sqrt{tfrac{3}{2}}sin t-tfrac{1}{2}right)=2-sqrt{tfrac{3}{2}}cos t-sqrt{tfrac{3}{2}}sin t$$
to get:
$$gamma : [0,2pi]tomathbb{R}^3: t mapsto
left( sqrt{tfrac{3}{2}}cos t-tfrac{1}{2} ,
sqrt{tfrac{3}{2}}sin t-tfrac{1}{2} ,
2-sqrt{tfrac{3}{2}}cos t-sqrt{tfrac{3}{2}}sin t right)$$
answered Jan 16 at 14:17
StackTDStackTD
22.9k2152
$endgroup$
add a comment |
$begingroup$
You can parametrize in the original coordinates. Solving the plane for $z$ you have:
$$x+y+z=1 iff z = 1-x-y tag{$star$}$$
so substituting into the paraboloid and rewriting gives:
$$z=x^2+y^2 implies 1-x-y=x^2+y^2 iff left(x+tfrac{1}{2}right)^2+left(y+tfrac{1}{2}right)^2=tfrac{3}{2}$$
This is easily parametrized as $x+tfrac{1}{2}=sqrt{tfrac{3}{2}}cos t$ and $y+tfrac{1}{2}=sqrt{tfrac{3}{2}}sin t$ and $z$ follows from $(star)$:
$$z = 1-x-y=1-left(sqrt{tfrac{3}{2}}cos t-tfrac{1}{2}right)-left(sqrt{tfrac{3}{2}}sin t-tfrac{1}{2}right)=2-sqrt{tfrac{3}{2}}cos t-sqrt{tfrac{3}{2}}sin t$$
to get:
$$gamma : [0,2pi]tomathbb{R}^3: t mapsto
left( sqrt{tfrac{3}{2}}cos t-tfrac{1}{2} ,
sqrt{tfrac{3}{2}}sin t-tfrac{1}{2} ,
2-sqrt{tfrac{3}{2}}cos t-sqrt{tfrac{3}{2}}sin t right)$$
answered Jan 16 at 14:17
StackTDStackTD
22.9k2152
$endgroup$
add a comment |
$begingroup$
You can parametrize in the original coordinates. Solving the plane for $z$ you have:
$$x+y+z=1 iff z = 1-x-y tag{$star$}$$
so substituting into the paraboloid and rewriting gives:
$$z=x^2+y^2 implies 1-x-y=x^2+y^2 iff left(x+tfrac{1}{2}right)^2+left(y+tfrac{1}{2}right)^2=tfrac{3}{2}$$
This is easily parametrized as $x+tfrac{1}{2}=sqrt{tfrac{3}{2}}cos t$ and $y+tfrac{1}{2}=sqrt{tfrac{3}{2}}sin t$ and $z$ follows from $(star)$:
$$z = 1-x-y=1-left(sqrt{tfrac{3}{2}}cos t-tfrac{1}{2}right)-left(sqrt{tfrac{3}{2}}sin t-tfrac{1}{2}right)=2-sqrt{tfrac{3}{2}}cos t-sqrt{tfrac{3}{2}}sin t$$
to get:
$$gamma : [0,2pi]tomathbb{R}^3: t mapsto
left( sqrt{tfrac{3}{2}}cos t-tfrac{1}{2} ,
sqrt{tfrac{3}{2}}sin t-tfrac{1}{2} ,
2-sqrt{tfrac{3}{2}}cos t-sqrt{tfrac{3}{2}}sin t right)$$
answered Jan 16 at 14:17
StackTDStackTD
22.9k2152
$endgroup$
You can parametrize in the original coordinates. Solving the plane for $z$ you have:
$$x+y+z=1 iff z = 1-x-y tag{$star$}$$
so substituting into the paraboloid and rewriting gives:
$$z=x^2+y^2 implies 1-x-y=x^2+y^2 iff left(x+tfrac{1}{2}right)^2+left(y+tfrac{1}{2}right)^2=tfrac{3}{2}$$
This is easily parametrized as $x+tfrac{1}{2}=sqrt{tfrac{3}{2}}cos t$ and $y+tfrac{1}{2}=sqrt{tfrac{3}{2}}sin t$ and $z$ follows from $(star)$:
$$z = 1-x-y=1-left(sqrt{tfrac{3}{2}}cos t-tfrac{1}{2}right)-left(sqrt{tfrac{3}{2}}sin t-tfrac{1}{2}right)=2-sqrt{tfrac{3}{2}}cos t-sqrt{tfrac{3}{2}}sin t$$
to get:
$$gamma : [0,2pi]tomathbb{R}^3: t mapsto
left( sqrt{tfrac{3}{2}}cos t-tfrac{1}{2} ,
sqrt{tfrac{3}{2}}sin t-tfrac{1}{2} ,
2-sqrt{tfrac{3}{2}}cos t-sqrt{tfrac{3}{2}}sin t right)$$
answered Jan 16 at 14:17
StackTDStackTD
22.9k2152
edited Jan 16 at 14:29
answered Jan 16 at 14:17
StackTDStackTD
22.9k2152
answered Jan 16 at 14:17
StackTDStackTD
22.9k2152
answered Jan 16 at 14:17
StackTDStackTD
22.9k2152
22.9k2152
add a comment |
add a comment |
$begingroup$
$$z=x^2+y^2$$ hints the use of polar coordinates and the equation of the plane becomes $$rho^2+rho(costheta+sintheta)-1=0.$$
Solve for $rho$ in terms of $theta$ and you are done.
answered Jan 16 at 14:50
Yves DaoustYves Daoust
128k675227
$endgroup$
add a comment |
$begingroup$
$$z=x^2+y^2$$ hints the use of polar coordinates and the equation of the plane becomes $$rho^2+rho(costheta+sintheta)-1=0.$$
Solve for $rho$ in terms of $theta$ and you are done.
answered Jan 16 at 14:50
Yves DaoustYves Daoust
128k675227
$endgroup$
add a comment |
$begingroup$
$$z=x^2+y^2$$ hints the use of polar coordinates and the equation of the plane becomes $$rho^2+rho(costheta+sintheta)-1=0.$$
Solve for $rho$ in terms of $theta$ and you are done.
answered Jan 16 at 14:50
Yves DaoustYves Daoust
128k675227
$endgroup$
$$z=x^2+y^2$$ hints the use of polar coordinates and the equation of the plane becomes $$rho^2+rho(costheta+sintheta)-1=0.$$
Solve for $rho$ in terms of $theta$ and you are done.
answered Jan 16 at 14:50
Yves DaoustYves Daoust
128k675227
answered Jan 16 at 14:50
Yves DaoustYves Daoust
128k675227
answered Jan 16 at 14:50
Yves DaoustYves Daoust
128k675227
answered Jan 16 at 14:50
Yves DaoustYves Daoust
128k675227
128k675227
add a comment |
add a comment |
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$begingroup$
There are other parametrization options. Do you want this particular one?
$endgroup$
– Andrei
Jan 16 at 14:01