Doubt about the procedure of parametrization
0
$begingroup$
I can't understand how parametrics equations are found. For example, I realize that the parametrization of the curve given by the intersection of the plane $ 2x+2y+z=2$ and $z=x^2+y^2$ is:
- $x=-1+cos(t)$
- $y=-1+sin(t)$
- $z=6-2cos(t)-2sin(t)$
- $0leqslant tleqslant2pi$
Or that the surface of $x^2+y^2=2$ delimited by $x^2+y^2+z^2=4$ is:
- $x=sqrt2cos(u)$
- $y=sqrt2sin(u)$
- $z=v$
$0leqslant uleqslantpi/2$ and $0leqslant vleqslantsqrt2$
But what is the step by step to find those equations?
parametrization
edited Dec 6 '18 at 6:29
Tianlalu
3,08621038
asked Dec 6 '18 at 6:09
J.DoeJ.Doe
1
$endgroup$
add a comment |
0
$begingroup$
I can't understand how parametrics equations are found. For example, I realize that the parametrization of the curve given by the intersection of the plane $ 2x+2y+z=2$ and $z=x^2+y^2$ is:
- $x=-1+cos(t)$
- $y=-1+sin(t)$
- $z=6-2cos(t)-2sin(t)$
- $0leqslant tleqslant2pi$
Or that the surface of $x^2+y^2=2$ delimited by $x^2+y^2+z^2=4$ is:
- $x=sqrt2cos(u)$
- $y=sqrt2sin(u)$
- $z=v$
$0leqslant uleqslantpi/2$ and $0leqslant vleqslantsqrt2$
But what is the step by step to find those equations?
parametrization
edited Dec 6 '18 at 6:29
Tianlalu
3,08621038
asked Dec 6 '18 at 6:09
J.DoeJ.Doe
1
$endgroup$
$begingroup$
Are you sure you got these right? For the first problem $x^2+y^2=2-2x-2yimplies (x+1)^2+(y+1)^2=4$, which is inconsistent with your first two bullet points.
$endgroup$
– J.G.
Dec 6 '18 at 6:36
$begingroup$
Now I see it.I got them from a list and didn't check. What would be a correct parametrization then?
$endgroup$
– J.Doe
Dec 6 '18 at 21:36
$begingroup$
You need to double the trigonometric functions' coefficients.
$endgroup$
– J.G.
Dec 6 '18 at 23:30
add a comment |
0
0
0
$begingroup$
I can't understand how parametrics equations are found. For example, I realize that the parametrization of the curve given by the intersection of the plane $ 2x+2y+z=2$ and $z=x^2+y^2$ is:
- $x=-1+cos(t)$
- $y=-1+sin(t)$
- $z=6-2cos(t)-2sin(t)$
- $0leqslant tleqslant2pi$
Or that the surface of $x^2+y^2=2$ delimited by $x^2+y^2+z^2=4$ is:
- $x=sqrt2cos(u)$
- $y=sqrt2sin(u)$
- $z=v$
$0leqslant uleqslantpi/2$ and $0leqslant vleqslantsqrt2$
But what is the step by step to find those equations?
parametrization
edited Dec 6 '18 at 6:29
Tianlalu
3,08621038
asked Dec 6 '18 at 6:09
J.DoeJ.Doe
1
$endgroup$
I can't understand how parametrics equations are found. For example, I realize that the parametrization of the curve given by the intersection of the plane $ 2x+2y+z=2$ and $z=x^2+y^2$ is:
- $x=-1+cos(t)$
- $y=-1+sin(t)$
- $z=6-2cos(t)-2sin(t)$
- $0leqslant tleqslant2pi$
Or that the surface of $x^2+y^2=2$ delimited by $x^2+y^2+z^2=4$ is:
- $x=sqrt2cos(u)$
- $y=sqrt2sin(u)$
- $z=v$
$0leqslant uleqslantpi/2$ and $0leqslant vleqslantsqrt2$
But what is the step by step to find those equations?
parametrization
parametrization
edited Dec 6 '18 at 6:29
Tianlalu
3,08621038
asked Dec 6 '18 at 6:09
J.DoeJ.Doe
1
edited Dec 6 '18 at 6:29
Tianlalu
3,08621038
asked Dec 6 '18 at 6:09
J.DoeJ.Doe
1
edited Dec 6 '18 at 6:29
Tianlalu
3,08621038
edited Dec 6 '18 at 6:29
Tianlalu
3,08621038
edited Dec 6 '18 at 6:29
Tianlalu
3,08621038
3,08621038
asked Dec 6 '18 at 6:09
J.DoeJ.Doe
1
asked Dec 6 '18 at 6:09
J.DoeJ.Doe
1
asked Dec 6 '18 at 6:09
J.DoeJ.Doe
1
1
$begingroup$
Are you sure you got these right? For the first problem $x^2+y^2=2-2x-2yimplies (x+1)^2+(y+1)^2=4$, which is inconsistent with your first two bullet points.
$endgroup$
– J.G.
Dec 6 '18 at 6:36
$begingroup$
Now I see it.I got them from a list and didn't check. What would be a correct parametrization then?
$endgroup$
– J.Doe
Dec 6 '18 at 21:36
$begingroup$
You need to double the trigonometric functions' coefficients.
$endgroup$
– J.G.
Dec 6 '18 at 23:30
add a comment |
$begingroup$
Are you sure you got these right? For the first problem $x^2+y^2=2-2x-2yimplies (x+1)^2+(y+1)^2=4$, which is inconsistent with your first two bullet points.
$endgroup$
– J.G.
Dec 6 '18 at 6:36
$begingroup$
Now I see it.I got them from a list and didn't check. What would be a correct parametrization then?
$endgroup$
– J.Doe
Dec 6 '18 at 21:36
$begingroup$
You need to double the trigonometric functions' coefficients.
$endgroup$
– J.G.
Dec 6 '18 at 23:30
$begingroup$
Are you sure you got these right? For the first problem $x^2+y^2=2-2x-2yimplies (x+1)^2+(y+1)^2=4$, which is inconsistent with your first two bullet points.
$endgroup$
– J.G.
Dec 6 '18 at 6:36
$begingroup$
Are you sure you got these right? For the first problem $x^2+y^2=2-2x-2yimplies (x+1)^2+(y+1)^2=4$, which is inconsistent with your first two bullet points.
$endgroup$
– J.G.
Dec 6 '18 at 6:36
$begingroup$
Now I see it.I got them from a list and didn't check. What would be a correct parametrization then?
$endgroup$
– J.Doe
Dec 6 '18 at 21:36
$begingroup$
Now I see it.I got them from a list and didn't check. What would be a correct parametrization then?
$endgroup$
– J.Doe
Dec 6 '18 at 21:36
$begingroup$
You need to double the trigonometric functions' coefficients.
$endgroup$
– J.G.
Dec 6 '18 at 23:30
$begingroup$
You need to double the trigonometric functions' coefficients.
$endgroup$
– J.G.
Dec 6 '18 at 23:30
add a comment |
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$begingroup$
Are you sure you got these right? For the first problem $x^2+y^2=2-2x-2yimplies (x+1)^2+(y+1)^2=4$, which is inconsistent with your first two bullet points.
$endgroup$
– J.G.
Dec 6 '18 at 6:36
$begingroup$
Now I see it.I got them from a list and didn't check. What would be a correct parametrization then?
$endgroup$
– J.Doe
Dec 6 '18 at 21:36
$begingroup$
You need to double the trigonometric functions' coefficients.
$endgroup$
– J.G.
Dec 6 '18 at 23:30