Find a vector parametrization of the circle contained in the plane x=5 with radius 3 centered at the point...
up vote
0
down vote
favorite
I know the vector parametrization of the circle contained within the yz-plane centered at the origin is:
$vec r$( $theta$ ) = < 0, 3 cos $theta$ , 3 sin $theta$ >
What do I do next?
If the circle was contained in the xy-plane or xz-plane, how would this change the parametrization?
multivariable-calculus circle parametrization
asked Oct 16 '16 at 23:07
sYnChris
1035
add a comment |
up vote
0
down vote
favorite
I know the vector parametrization of the circle contained within the yz-plane centered at the origin is:
$vec r$( $theta$ ) = < 0, 3 cos $theta$ , 3 sin $theta$ >
What do I do next?
If the circle was contained in the xy-plane or xz-plane, how would this change the parametrization?
multivariable-calculus circle parametrization
asked Oct 16 '16 at 23:07
sYnChris
1035
2
It's worth noting that this question as stated doesn't quite make sense; a circle contained in the plane $x=5$ has its center in the same plane, so its center could never be $(3,1,2)$. @AbdallahHammam's answer is one reasonable interpretation, but note that the resulting circle doesn't have radius $3$.
– stewbasic
Oct 17 '16 at 0:48
@stewbasic My bad. I meant to put the point (5,1,2).
– sYnChris
Oct 17 '16 at 1:15
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
I know the vector parametrization of the circle contained within the yz-plane centered at the origin is:
$vec r$( $theta$ ) = < 0, 3 cos $theta$ , 3 sin $theta$ >
What do I do next?
If the circle was contained in the xy-plane or xz-plane, how would this change the parametrization?
multivariable-calculus circle parametrization
asked Oct 16 '16 at 23:07
sYnChris
1035
I know the vector parametrization of the circle contained within the yz-plane centered at the origin is:
$vec r$( $theta$ ) = < 0, 3 cos $theta$ , 3 sin $theta$ >
What do I do next?
If the circle was contained in the xy-plane or xz-plane, how would this change the parametrization?
multivariable-calculus circle parametrization
multivariable-calculus circle parametrization
asked Oct 16 '16 at 23:07
sYnChris
1035
asked Oct 16 '16 at 23:07
sYnChris
1035
edited Oct 17 '16 at 1:20
asked Oct 16 '16 at 23:07
sYnChris
1035
asked Oct 16 '16 at 23:07
sYnChris
1035
asked Oct 16 '16 at 23:07
sYnChris
1035
1035
2
It's worth noting that this question as stated doesn't quite make sense; a circle contained in the plane $x=5$ has its center in the same plane, so its center could never be $(3,1,2)$. @AbdallahHammam's answer is one reasonable interpretation, but note that the resulting circle doesn't have radius $3$.
– stewbasic
Oct 17 '16 at 0:48
@stewbasic My bad. I meant to put the point (5,1,2).
– sYnChris
Oct 17 '16 at 1:15
add a comment |
2
It's worth noting that this question as stated doesn't quite make sense; a circle contained in the plane $x=5$ has its center in the same plane, so its center could never be $(3,1,2)$. @AbdallahHammam's answer is one reasonable interpretation, but note that the resulting circle doesn't have radius $3$.
– stewbasic
Oct 17 '16 at 0:48
@stewbasic My bad. I meant to put the point (5,1,2).
– sYnChris
Oct 17 '16 at 1:15
2
2
It's worth noting that this question as stated doesn't quite make sense; a circle contained in the plane $x=5$ has its center in the same plane, so its center could never be $(3,1,2)$. @AbdallahHammam's answer is one reasonable interpretation, but note that the resulting circle doesn't have radius $3$.
– stewbasic
Oct 17 '16 at 0:48
It's worth noting that this question as stated doesn't quite make sense; a circle contained in the plane $x=5$ has its center in the same plane, so its center could never be $(3,1,2)$. @AbdallahHammam's answer is one reasonable interpretation, but note that the resulting circle doesn't have radius $3$.
– stewbasic
Oct 17 '16 at 0:48
@stewbasic My bad. I meant to put the point (5,1,2).
– sYnChris
Oct 17 '16 at 1:15
@stewbasic My bad. I meant to put the point (5,1,2).
– sYnChris
Oct 17 '16 at 1:15
add a comment |
1 Answer
1
active
oldest
votes
up vote
2
down vote
accepted
The circle we want the parametrization for, which is the intersection of the sphere of equation
$$(x-5)^2+(y-1)^2+(z-2)^2=9$$
with the plane $x=5$ has equation
$$(y-1)^2+(z-2)^2=9.$$
Thus, we can parametrize the circle in the following way:
begin{align}
x&=5\
y&=1+3cos(t)\
z&=2+3sin(t).\
end{align}
edited Nov 23 at 4:41
Alex D
508219
answered Oct 16 '16 at 23:24
hamam_Abdallah
37.5k21634
add a comment |
Your Answer
StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");
StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "69"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});
function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});
}
});
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f1971809%2ffind-a-vector-parametrization-of-the-circle-contained-in-the-plane-x-5-with-radi%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
accepted
The circle we want the parametrization for, which is the intersection of the sphere of equation
$$(x-5)^2+(y-1)^2+(z-2)^2=9$$
with the plane $x=5$ has equation
$$(y-1)^2+(z-2)^2=9.$$
Thus, we can parametrize the circle in the following way:
begin{align}
x&=5\
y&=1+3cos(t)\
z&=2+3sin(t).\
end{align}
edited Nov 23 at 4:41
Alex D
508219
answered Oct 16 '16 at 23:24
hamam_Abdallah
37.5k21634
add a comment |
up vote
2
down vote
accepted
The circle we want the parametrization for, which is the intersection of the sphere of equation
$$(x-5)^2+(y-1)^2+(z-2)^2=9$$
with the plane $x=5$ has equation
$$(y-1)^2+(z-2)^2=9.$$
Thus, we can parametrize the circle in the following way:
begin{align}
x&=5\
y&=1+3cos(t)\
z&=2+3sin(t).\
end{align}
edited Nov 23 at 4:41
Alex D
508219
answered Oct 16 '16 at 23:24
hamam_Abdallah
37.5k21634
add a comment |
up vote
2
down vote
accepted
up vote
2
down vote
accepted
The circle we want the parametrization for, which is the intersection of the sphere of equation
$$(x-5)^2+(y-1)^2+(z-2)^2=9$$
with the plane $x=5$ has equation
$$(y-1)^2+(z-2)^2=9.$$
Thus, we can parametrize the circle in the following way:
begin{align}
x&=5\
y&=1+3cos(t)\
z&=2+3sin(t).\
end{align}
edited Nov 23 at 4:41
Alex D
508219
answered Oct 16 '16 at 23:24
hamam_Abdallah
37.5k21634
The circle we want the parametrization for, which is the intersection of the sphere of equation
$$(x-5)^2+(y-1)^2+(z-2)^2=9$$
with the plane $x=5$ has equation
$$(y-1)^2+(z-2)^2=9.$$
Thus, we can parametrize the circle in the following way:
begin{align}
x&=5\
y&=1+3cos(t)\
z&=2+3sin(t).\
end{align}
edited Nov 23 at 4:41
Alex D
508219
answered Oct 16 '16 at 23:24
hamam_Abdallah
37.5k21634
edited Nov 23 at 4:41
Alex D
508219
edited Nov 23 at 4:41
Alex D
508219
edited Nov 23 at 4:41
Alex D
508219
508219
answered Oct 16 '16 at 23:24
hamam_Abdallah
37.5k21634
answered Oct 16 '16 at 23:24
hamam_Abdallah
37.5k21634
answered Oct 16 '16 at 23:24
hamam_Abdallah
37.5k21634
37.5k21634
add a comment |
add a comment |
Thanks for contributing an answer to Mathematics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Some of your past answers have not been well-received, and you're in danger of being blocked from answering.
Please pay close attention to the following guidance:
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f1971809%2ffind-a-vector-parametrization-of-the-circle-contained-in-the-plane-x-5-with-radi%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
2
It's worth noting that this question as stated doesn't quite make sense; a circle contained in the plane $x=5$ has its center in the same plane, so its center could never be $(3,1,2)$. @AbdallahHammam's answer is one reasonable interpretation, but note that the resulting circle doesn't have radius $3$.
– stewbasic
Oct 17 '16 at 0:48
@stewbasic My bad. I meant to put the point (5,1,2).
– sYnChris
Oct 17 '16 at 1:15