Converting Parametric form equations to rectangular form
0
$begingroup$
I have these parametric equations given by the line of intersection of two planes:
z = 1 + 2x + 3 and y = 2x.
Which gives the line of intersection
L:
x = t
y = 2t
z = 1 + 8t
I am supposed to find the slope of this line, in the direction of increasing t. The way that I'm thinking of getting this is converting the parametric equation to rectangular form and get the slope from there. I tried solving for t but I don't think it works that way. How do I convert the parametric equation to rectangular form? Is there an easier way of finding the slope? Thanks!
multivariable-calculus
asked Oct 5 '15 at 0:32
user2989964user2989964
1339
$endgroup$
add a comment |
0
$begingroup$
I have these parametric equations given by the line of intersection of two planes:
z = 1 + 2x + 3 and y = 2x.
Which gives the line of intersection
L:
x = t
y = 2t
z = 1 + 8t
I am supposed to find the slope of this line, in the direction of increasing t. The way that I'm thinking of getting this is converting the parametric equation to rectangular form and get the slope from there. I tried solving for t but I don't think it works that way. How do I convert the parametric equation to rectangular form? Is there an easier way of finding the slope? Thanks!
multivariable-calculus
asked Oct 5 '15 at 0:32
user2989964user2989964
1339
$endgroup$
add a comment |
0
0
0
$begingroup$
I have these parametric equations given by the line of intersection of two planes:
z = 1 + 2x + 3 and y = 2x.
Which gives the line of intersection
L:
x = t
y = 2t
z = 1 + 8t
I am supposed to find the slope of this line, in the direction of increasing t. The way that I'm thinking of getting this is converting the parametric equation to rectangular form and get the slope from there. I tried solving for t but I don't think it works that way. How do I convert the parametric equation to rectangular form? Is there an easier way of finding the slope? Thanks!
multivariable-calculus
asked Oct 5 '15 at 0:32
user2989964user2989964
1339
$endgroup$
I have these parametric equations given by the line of intersection of two planes:
z = 1 + 2x + 3 and y = 2x.
Which gives the line of intersection
L:
x = t
y = 2t
z = 1 + 8t
I am supposed to find the slope of this line, in the direction of increasing t. The way that I'm thinking of getting this is converting the parametric equation to rectangular form and get the slope from there. I tried solving for t but I don't think it works that way. How do I convert the parametric equation to rectangular form? Is there an easier way of finding the slope? Thanks!
multivariable-calculus
multivariable-calculus
asked Oct 5 '15 at 0:32
user2989964user2989964
1339
asked Oct 5 '15 at 0:32
user2989964user2989964
1339
asked Oct 5 '15 at 0:32
user2989964user2989964
1339
asked Oct 5 '15 at 0:32
user2989964user2989964
1339
asked Oct 5 '15 at 0:32
user2989964user2989964
1339
1339
add a comment |
add a comment |
1 Answer
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oldest
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$x=y/2=(z-1)/8$. We can also write $(x,y,z) =(0,0,1)+(1,2,8)t .$ The line is parallel to the line that pases through $(0,0,0)$ and through $(1,2,8)$.
answered Oct 5 '15 at 0:41
DanielWainfleetDanielWainfleet
36k31648
$endgroup$
$begingroup$
This is what I did initially but I don't know how to get a single equation where it will equal the traditional y = mx + b equation in order to find the slope. Is there any way to do this?
$endgroup$
– user2989964
Oct 5 '15 at 0:45
$begingroup$
The traditional equation form is fine in 2 dimensions but not applicable in 3, where a single linear equation in x,y,z determines a plane, not a line. You could write $(2x-y)^2+(8x-z-1)^2=0$ but it's not much use to do so.The "slope" of a line in 3-D is a vector,not a number.
$endgroup$
– DanielWainfleet
Oct 5 '15 at 0:56
$begingroup$
In dimension $3$ this is impossible: one (cartesian) equation defines a surface. A curve requires at least $2$ equations.
$endgroup$
– Bernard
Oct 5 '15 at 0:57
$begingroup$
So the slope is pretty much the equation of the line itself? If I am told to find the slope, what would be an acceptable answer?
$endgroup$
– user2989964
Oct 5 '15 at 1:55
$begingroup$
The slope specifies the orientation of the line , You need more than 1 number to describe it for a pole stuck in the ground (not necessarily vertically) .If you want to specify the orientation with respect to a "ground plane" (e.g. the plane z=0) two numbers will do. There is however another meaning: When a line L. not in a plane P ,.intersects P, there is a unique plane Q which contains L and is orthogonal to P, and intersects P in a line M. Sometimes the angle x (or pi-x) between L and M is called the angle between . L and P . Ask your teacher to clarify what is asked for.
$endgroup$
– DanielWainfleet
Oct 5 '15 at 3:52
add a comment |
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1 Answer
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1 Answer
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0
$begingroup$
$x=y/2=(z-1)/8$. We can also write $(x,y,z) =(0,0,1)+(1,2,8)t .$ The line is parallel to the line that pases through $(0,0,0)$ and through $(1,2,8)$.
answered Oct 5 '15 at 0:41
DanielWainfleetDanielWainfleet
36k31648
$endgroup$
$begingroup$
This is what I did initially but I don't know how to get a single equation where it will equal the traditional y = mx + b equation in order to find the slope. Is there any way to do this?
$endgroup$
– user2989964
Oct 5 '15 at 0:45
$begingroup$
The traditional equation form is fine in 2 dimensions but not applicable in 3, where a single linear equation in x,y,z determines a plane, not a line. You could write $(2x-y)^2+(8x-z-1)^2=0$ but it's not much use to do so.The "slope" of a line in 3-D is a vector,not a number.
$endgroup$
– DanielWainfleet
Oct 5 '15 at 0:56
$begingroup$
In dimension $3$ this is impossible: one (cartesian) equation defines a surface. A curve requires at least $2$ equations.
$endgroup$
– Bernard
Oct 5 '15 at 0:57
$begingroup$
So the slope is pretty much the equation of the line itself? If I am told to find the slope, what would be an acceptable answer?
$endgroup$
– user2989964
Oct 5 '15 at 1:55
$begingroup$
The slope specifies the orientation of the line , You need more than 1 number to describe it for a pole stuck in the ground (not necessarily vertically) .If you want to specify the orientation with respect to a "ground plane" (e.g. the plane z=0) two numbers will do. There is however another meaning: When a line L. not in a plane P ,.intersects P, there is a unique plane Q which contains L and is orthogonal to P, and intersects P in a line M. Sometimes the angle x (or pi-x) between L and M is called the angle between . L and P . Ask your teacher to clarify what is asked for.
$endgroup$
– DanielWainfleet
Oct 5 '15 at 3:52
add a comment |
0
$begingroup$
$x=y/2=(z-1)/8$. We can also write $(x,y,z) =(0,0,1)+(1,2,8)t .$ The line is parallel to the line that pases through $(0,0,0)$ and through $(1,2,8)$.
answered Oct 5 '15 at 0:41
DanielWainfleetDanielWainfleet
36k31648
$endgroup$
$begingroup$
This is what I did initially but I don't know how to get a single equation where it will equal the traditional y = mx + b equation in order to find the slope. Is there any way to do this?
$endgroup$
– user2989964
Oct 5 '15 at 0:45
$begingroup$
The traditional equation form is fine in 2 dimensions but not applicable in 3, where a single linear equation in x,y,z determines a plane, not a line. You could write $(2x-y)^2+(8x-z-1)^2=0$ but it's not much use to do so.The "slope" of a line in 3-D is a vector,not a number.
$endgroup$
– DanielWainfleet
Oct 5 '15 at 0:56
$begingroup$
In dimension $3$ this is impossible: one (cartesian) equation defines a surface. A curve requires at least $2$ equations.
$endgroup$
– Bernard
Oct 5 '15 at 0:57
$begingroup$
So the slope is pretty much the equation of the line itself? If I am told to find the slope, what would be an acceptable answer?
$endgroup$
– user2989964
Oct 5 '15 at 1:55
$begingroup$
The slope specifies the orientation of the line , You need more than 1 number to describe it for a pole stuck in the ground (not necessarily vertically) .If you want to specify the orientation with respect to a "ground plane" (e.g. the plane z=0) two numbers will do. There is however another meaning: When a line L. not in a plane P ,.intersects P, there is a unique plane Q which contains L and is orthogonal to P, and intersects P in a line M. Sometimes the angle x (or pi-x) between L and M is called the angle between . L and P . Ask your teacher to clarify what is asked for.
$endgroup$
– DanielWainfleet
Oct 5 '15 at 3:52
add a comment |
0
0
0
$begingroup$
$x=y/2=(z-1)/8$. We can also write $(x,y,z) =(0,0,1)+(1,2,8)t .$ The line is parallel to the line that pases through $(0,0,0)$ and through $(1,2,8)$.
answered Oct 5 '15 at 0:41
DanielWainfleetDanielWainfleet
36k31648
$endgroup$
$x=y/2=(z-1)/8$. We can also write $(x,y,z) =(0,0,1)+(1,2,8)t .$ The line is parallel to the line that pases through $(0,0,0)$ and through $(1,2,8)$.
answered Oct 5 '15 at 0:41
DanielWainfleetDanielWainfleet
36k31648
edited Oct 5 '15 at 0:50
answered Oct 5 '15 at 0:41
DanielWainfleetDanielWainfleet
36k31648
answered Oct 5 '15 at 0:41
DanielWainfleetDanielWainfleet
36k31648
answered Oct 5 '15 at 0:41
DanielWainfleetDanielWainfleet
36k31648
36k31648
$begingroup$
This is what I did initially but I don't know how to get a single equation where it will equal the traditional y = mx + b equation in order to find the slope. Is there any way to do this?
$endgroup$
– user2989964
Oct 5 '15 at 0:45
$begingroup$
The traditional equation form is fine in 2 dimensions but not applicable in 3, where a single linear equation in x,y,z determines a plane, not a line. You could write $(2x-y)^2+(8x-z-1)^2=0$ but it's not much use to do so.The "slope" of a line in 3-D is a vector,not a number.
$endgroup$
– DanielWainfleet
Oct 5 '15 at 0:56
$begingroup$
In dimension $3$ this is impossible: one (cartesian) equation defines a surface. A curve requires at least $2$ equations.
$endgroup$
– Bernard
Oct 5 '15 at 0:57
$begingroup$
So the slope is pretty much the equation of the line itself? If I am told to find the slope, what would be an acceptable answer?
$endgroup$
– user2989964
Oct 5 '15 at 1:55
$begingroup$
The slope specifies the orientation of the line , You need more than 1 number to describe it for a pole stuck in the ground (not necessarily vertically) .If you want to specify the orientation with respect to a "ground plane" (e.g. the plane z=0) two numbers will do. There is however another meaning: When a line L. not in a plane P ,.intersects P, there is a unique plane Q which contains L and is orthogonal to P, and intersects P in a line M. Sometimes the angle x (or pi-x) between L and M is called the angle between . L and P . Ask your teacher to clarify what is asked for.
$endgroup$
– DanielWainfleet
Oct 5 '15 at 3:52
add a comment |
$begingroup$
This is what I did initially but I don't know how to get a single equation where it will equal the traditional y = mx + b equation in order to find the slope. Is there any way to do this?
$endgroup$
– user2989964
Oct 5 '15 at 0:45
$begingroup$
The traditional equation form is fine in 2 dimensions but not applicable in 3, where a single linear equation in x,y,z determines a plane, not a line. You could write $(2x-y)^2+(8x-z-1)^2=0$ but it's not much use to do so.The "slope" of a line in 3-D is a vector,not a number.
$endgroup$
– DanielWainfleet
Oct 5 '15 at 0:56
$begingroup$
In dimension $3$ this is impossible: one (cartesian) equation defines a surface. A curve requires at least $2$ equations.
$endgroup$
– Bernard
Oct 5 '15 at 0:57
$begingroup$
So the slope is pretty much the equation of the line itself? If I am told to find the slope, what would be an acceptable answer?
$endgroup$
– user2989964
Oct 5 '15 at 1:55
$begingroup$
The slope specifies the orientation of the line , You need more than 1 number to describe it for a pole stuck in the ground (not necessarily vertically) .If you want to specify the orientation with respect to a "ground plane" (e.g. the plane z=0) two numbers will do. There is however another meaning: When a line L. not in a plane P ,.intersects P, there is a unique plane Q which contains L and is orthogonal to P, and intersects P in a line M. Sometimes the angle x (or pi-x) between L and M is called the angle between . L and P . Ask your teacher to clarify what is asked for.
$endgroup$
– DanielWainfleet
Oct 5 '15 at 3:52
$begingroup$
This is what I did initially but I don't know how to get a single equation where it will equal the traditional y = mx + b equation in order to find the slope. Is there any way to do this?
$endgroup$
– user2989964
Oct 5 '15 at 0:45
$begingroup$
This is what I did initially but I don't know how to get a single equation where it will equal the traditional y = mx + b equation in order to find the slope. Is there any way to do this?
$endgroup$
– user2989964
Oct 5 '15 at 0:45
$begingroup$
The traditional equation form is fine in 2 dimensions but not applicable in 3, where a single linear equation in x,y,z determines a plane, not a line. You could write $(2x-y)^2+(8x-z-1)^2=0$ but it's not much use to do so.The "slope" of a line in 3-D is a vector,not a number.
$endgroup$
– DanielWainfleet
Oct 5 '15 at 0:56
$begingroup$
The traditional equation form is fine in 2 dimensions but not applicable in 3, where a single linear equation in x,y,z determines a plane, not a line. You could write $(2x-y)^2+(8x-z-1)^2=0$ but it's not much use to do so.The "slope" of a line in 3-D is a vector,not a number.
$endgroup$
– DanielWainfleet
Oct 5 '15 at 0:56
$begingroup$
In dimension $3$ this is impossible: one (cartesian) equation defines a surface. A curve requires at least $2$ equations.
$endgroup$
– Bernard
Oct 5 '15 at 0:57
$begingroup$
In dimension $3$ this is impossible: one (cartesian) equation defines a surface. A curve requires at least $2$ equations.
$endgroup$
– Bernard
Oct 5 '15 at 0:57
$begingroup$
So the slope is pretty much the equation of the line itself? If I am told to find the slope, what would be an acceptable answer?
$endgroup$
– user2989964
Oct 5 '15 at 1:55
$begingroup$
So the slope is pretty much the equation of the line itself? If I am told to find the slope, what would be an acceptable answer?
$endgroup$
– user2989964
Oct 5 '15 at 1:55
$begingroup$
The slope specifies the orientation of the line , You need more than 1 number to describe it for a pole stuck in the ground (not necessarily vertically) .If you want to specify the orientation with respect to a "ground plane" (e.g. the plane z=0) two numbers will do. There is however another meaning: When a line L. not in a plane P ,.intersects P, there is a unique plane Q which contains L and is orthogonal to P, and intersects P in a line M. Sometimes the angle x (or pi-x) between L and M is called the angle between . L and P . Ask your teacher to clarify what is asked for.
$endgroup$
– DanielWainfleet
Oct 5 '15 at 3:52
$begingroup$
The slope specifies the orientation of the line , You need more than 1 number to describe it for a pole stuck in the ground (not necessarily vertically) .If you want to specify the orientation with respect to a "ground plane" (e.g. the plane z=0) two numbers will do. There is however another meaning: When a line L. not in a plane P ,.intersects P, there is a unique plane Q which contains L and is orthogonal to P, and intersects P in a line M. Sometimes the angle x (or pi-x) between L and M is called the angle between . L and P . Ask your teacher to clarify what is asked for.
$endgroup$
– DanielWainfleet
Oct 5 '15 at 3:52
add a comment |
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