Parametrization for the curve on cylinder $y = 7 - x^4$ that passes through the point $(0, 7, -3) $when t = 0...
$begingroup$
Can you help me?
So far I have turned $y = 7-x^4$ into $langle1, 1, 0rangle$ and used it to make the equation $L = (0, 7, -3) + t(1, 1, 0)$. I know this is wrong, but I just don't know what, and I know it has to do with $y = 7-x^4$.
Thank you in advance.
parametric
edited Mar 9 '16 at 15:59
Narasimham
20.6k52158
asked Sep 29 '13 at 1:59
anon12345anon12345
42
$endgroup$
add a comment |
$begingroup$
Can you help me?
So far I have turned $y = 7-x^4$ into $langle1, 1, 0rangle$ and used it to make the equation $L = (0, 7, -3) + t(1, 1, 0)$. I know this is wrong, but I just don't know what, and I know it has to do with $y = 7-x^4$.
Thank you in advance.
parametric
edited Mar 9 '16 at 15:59
Narasimham
20.6k52158
asked Sep 29 '13 at 1:59
anon12345anon12345
42
$endgroup$
$begingroup$
$y=7-x^4$ defines a surface, not a curve, given the implication that this is all in $mathbb{R}^3$.
$endgroup$
– alex.jordan
Mar 14 '14 at 3:30
$begingroup$
@alex.jordan yes, yet he is clearly asking for the curve defined by intersecting said surface with the plane $z=-3$
$endgroup$
– John C
Oct 3 '14 at 13:22
add a comment |
$begingroup$
Can you help me?
So far I have turned $y = 7-x^4$ into $langle1, 1, 0rangle$ and used it to make the equation $L = (0, 7, -3) + t(1, 1, 0)$. I know this is wrong, but I just don't know what, and I know it has to do with $y = 7-x^4$.
Thank you in advance.
parametric
edited Mar 9 '16 at 15:59
Narasimham
20.6k52158
asked Sep 29 '13 at 1:59
anon12345anon12345
42
$endgroup$
Can you help me?
So far I have turned $y = 7-x^4$ into $langle1, 1, 0rangle$ and used it to make the equation $L = (0, 7, -3) + t(1, 1, 0)$. I know this is wrong, but I just don't know what, and I know it has to do with $y = 7-x^4$.
Thank you in advance.
parametric
parametric
edited Mar 9 '16 at 15:59
Narasimham
20.6k52158
asked Sep 29 '13 at 1:59
anon12345anon12345
42
edited Mar 9 '16 at 15:59
Narasimham
20.6k52158
asked Sep 29 '13 at 1:59
anon12345anon12345
42
edited Mar 9 '16 at 15:59
Narasimham
20.6k52158
edited Mar 9 '16 at 15:59
Narasimham
20.6k52158
edited Mar 9 '16 at 15:59
Narasimham
20.6k52158
20.6k52158
asked Sep 29 '13 at 1:59
anon12345anon12345
42
asked Sep 29 '13 at 1:59
anon12345anon12345
42
asked Sep 29 '13 at 1:59
anon12345anon12345
42
42
$begingroup$
$y=7-x^4$ defines a surface, not a curve, given the implication that this is all in $mathbb{R}^3$.
$endgroup$
– alex.jordan
Mar 14 '14 at 3:30
$begingroup$
@alex.jordan yes, yet he is clearly asking for the curve defined by intersecting said surface with the plane $z=-3$
$endgroup$
– John C
Oct 3 '14 at 13:22
add a comment |
$begingroup$
$y=7-x^4$ defines a surface, not a curve, given the implication that this is all in $mathbb{R}^3$.
$endgroup$
– alex.jordan
Mar 14 '14 at 3:30
$begingroup$
@alex.jordan yes, yet he is clearly asking for the curve defined by intersecting said surface with the plane $z=-3$
$endgroup$
– John C
Oct 3 '14 at 13:22
$begingroup$
$y=7-x^4$ defines a surface, not a curve, given the implication that this is all in $mathbb{R}^3$.
$endgroup$
– alex.jordan
Mar 14 '14 at 3:30
$begingroup$
$y=7-x^4$ defines a surface, not a curve, given the implication that this is all in $mathbb{R}^3$.
$endgroup$
– alex.jordan
Mar 14 '14 at 3:30
$begingroup$
@alex.jordan yes, yet he is clearly asking for the curve defined by intersecting said surface with the plane $z=-3$
$endgroup$
– John C
Oct 3 '14 at 13:22
$begingroup$
@alex.jordan yes, yet he is clearly asking for the curve defined by intersecting said surface with the plane $z=-3$
$endgroup$
– John C
Oct 3 '14 at 13:22
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
When you choose your curve, $z$ can take any value, so just choose $f(t)=(t, 7-t^4,-3)$. Clearly, the curve is parallel to the xy-plane (a curve is parallel to a plane if it lies in a plane parallel to the plane)...no reparametrization is necessary and certainly we don't need to think about lines through the point $(0,7,-3)$.
edited Oct 3 '14 at 13:40
John C
796311
answered Sep 29 '13 at 2:32
Matt RMatt R
233116
$endgroup$
add a comment |
$begingroup$
Required parametrization of curve parallel to base curve defined by first two coordinates is
$$ x= t , y = 7 -t^4 , z = -3 $$
$x,y $ relation shows that it is a fourth order parabola. Also no $x,y$ is involved in $z$. So it is a surface obtained by extruding the higher order parabola parallelly drawn along $z.$ Such independent two coordinates satisfy not a curve but the full surface, which are called cylinders.
The base parabola curve is shown in green, and the surface generated upto $z=-3$ is yellow.The upper edge $z=-3$ is the curve your question.
answered Mar 9 '16 at 16:06
NarasimhamNarasimham
20.6k52158
$endgroup$
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
When you choose your curve, $z$ can take any value, so just choose $f(t)=(t, 7-t^4,-3)$. Clearly, the curve is parallel to the xy-plane (a curve is parallel to a plane if it lies in a plane parallel to the plane)...no reparametrization is necessary and certainly we don't need to think about lines through the point $(0,7,-3)$.
edited Oct 3 '14 at 13:40
John C
796311
answered Sep 29 '13 at 2:32
Matt RMatt R
233116
$endgroup$
add a comment |
$begingroup$
When you choose your curve, $z$ can take any value, so just choose $f(t)=(t, 7-t^4,-3)$. Clearly, the curve is parallel to the xy-plane (a curve is parallel to a plane if it lies in a plane parallel to the plane)...no reparametrization is necessary and certainly we don't need to think about lines through the point $(0,7,-3)$.
edited Oct 3 '14 at 13:40
John C
796311
answered Sep 29 '13 at 2:32
Matt RMatt R
233116
$endgroup$
add a comment |
$begingroup$
When you choose your curve, $z$ can take any value, so just choose $f(t)=(t, 7-t^4,-3)$. Clearly, the curve is parallel to the xy-plane (a curve is parallel to a plane if it lies in a plane parallel to the plane)...no reparametrization is necessary and certainly we don't need to think about lines through the point $(0,7,-3)$.
edited Oct 3 '14 at 13:40
John C
796311
answered Sep 29 '13 at 2:32
Matt RMatt R
233116
$endgroup$
When you choose your curve, $z$ can take any value, so just choose $f(t)=(t, 7-t^4,-3)$. Clearly, the curve is parallel to the xy-plane (a curve is parallel to a plane if it lies in a plane parallel to the plane)...no reparametrization is necessary and certainly we don't need to think about lines through the point $(0,7,-3)$.
edited Oct 3 '14 at 13:40
John C
796311
answered Sep 29 '13 at 2:32
Matt RMatt R
233116
edited Oct 3 '14 at 13:40
John C
796311
edited Oct 3 '14 at 13:40
John C
796311
edited Oct 3 '14 at 13:40
John C
796311
796311
answered Sep 29 '13 at 2:32
Matt RMatt R
233116
answered Sep 29 '13 at 2:32
Matt RMatt R
233116
answered Sep 29 '13 at 2:32
Matt RMatt R
233116
233116
add a comment |
add a comment |
$begingroup$
Required parametrization of curve parallel to base curve defined by first two coordinates is
$$ x= t , y = 7 -t^4 , z = -3 $$
$x,y $ relation shows that it is a fourth order parabola. Also no $x,y$ is involved in $z$. So it is a surface obtained by extruding the higher order parabola parallelly drawn along $z.$ Such independent two coordinates satisfy not a curve but the full surface, which are called cylinders.
The base parabola curve is shown in green, and the surface generated upto $z=-3$ is yellow.The upper edge $z=-3$ is the curve your question.
answered Mar 9 '16 at 16:06
NarasimhamNarasimham
20.6k52158
$endgroup$
add a comment |
$begingroup$
Required parametrization of curve parallel to base curve defined by first two coordinates is
$$ x= t , y = 7 -t^4 , z = -3 $$
$x,y $ relation shows that it is a fourth order parabola. Also no $x,y$ is involved in $z$. So it is a surface obtained by extruding the higher order parabola parallelly drawn along $z.$ Such independent two coordinates satisfy not a curve but the full surface, which are called cylinders.
The base parabola curve is shown in green, and the surface generated upto $z=-3$ is yellow.The upper edge $z=-3$ is the curve your question.
answered Mar 9 '16 at 16:06
NarasimhamNarasimham
20.6k52158
$endgroup$
add a comment |
$begingroup$
Required parametrization of curve parallel to base curve defined by first two coordinates is
$$ x= t , y = 7 -t^4 , z = -3 $$
$x,y $ relation shows that it is a fourth order parabola. Also no $x,y$ is involved in $z$. So it is a surface obtained by extruding the higher order parabola parallelly drawn along $z.$ Such independent two coordinates satisfy not a curve but the full surface, which are called cylinders.
The base parabola curve is shown in green, and the surface generated upto $z=-3$ is yellow.The upper edge $z=-3$ is the curve your question.
answered Mar 9 '16 at 16:06
NarasimhamNarasimham
20.6k52158
$endgroup$
Required parametrization of curve parallel to base curve defined by first two coordinates is
$$ x= t , y = 7 -t^4 , z = -3 $$
$x,y $ relation shows that it is a fourth order parabola. Also no $x,y$ is involved in $z$. So it is a surface obtained by extruding the higher order parabola parallelly drawn along $z.$ Such independent two coordinates satisfy not a curve but the full surface, which are called cylinders.
The base parabola curve is shown in green, and the surface generated upto $z=-3$ is yellow.The upper edge $z=-3$ is the curve your question.
answered Mar 9 '16 at 16:06
NarasimhamNarasimham
20.6k52158
edited May 23 '17 at 18:00
answered Mar 9 '16 at 16:06
NarasimhamNarasimham
20.6k52158
answered Mar 9 '16 at 16:06
NarasimhamNarasimham
20.6k52158
answered Mar 9 '16 at 16:06
NarasimhamNarasimham
20.6k52158
20.6k52158
add a comment |
add a comment |
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$begingroup$
$y=7-x^4$ defines a surface, not a curve, given the implication that this is all in $mathbb{R}^3$.
$endgroup$
– alex.jordan
Mar 14 '14 at 3:30
$begingroup$
@alex.jordan yes, yet he is clearly asking for the curve defined by intersecting said surface with the plane $z=-3$
$endgroup$
– John C
Oct 3 '14 at 13:22