Mixing
Prove: The normal to the involute of a circle is tangent to the circle
$begingroup$
Please refrain from using algebraic equations (in the Cartesian system) to prove it. I was looking for some kind of geometric proof for it. Or using differentitiation of vectors. For using the definition of tangent, let the curve be described by $vec r(t) $ and $theta(t)$ for a parameter $t$. Then the tangent may be given by the direction of $frac{dvec r}{dt}$.
I got a start, let the origin be the center of the circle. Let $vec m$ be the vector given by the radius of the circle pointing radially outwards ending at the point on the circle till where the string is bound to the circle, and $vec n$ be the vector such that $vec m+vec n=vec r$.
Now differentiating the vectors with respect to $t$, and using $vec r=r.hat r$. Now $frac{dm}{dt}=0$.
So $mfrac{dhat m}{dt}+nfrac{dhat n}{dt}+hat nfrac{dn}{dt}=$ tangent vector.
Now $|vec n|=n=mtheta$, where $theta$ is the angle rotated till now by the $vec m$. noLet |$frac{dhat m}{dt}$| be constant. So $frac{dn}{dt}=m|frac{dhat m}{dt}|$.
Now all that is left to show is that $frac{dhat m}{dt}$ is $perp$ to $vec m$. How do I do that?
geometry proof-writing tangent-line
asked Jan 18 at 6:56
Aditya AgarwalAditya Agarwal
2,92911839
$endgroup$
add a comment |
$begingroup$
Please refrain from using algebraic equations (in the Cartesian system) to prove it. I was looking for some kind of geometric proof for it. Or using differentitiation of vectors. For using the definition of tangent, let the curve be described by $vec r(t) $ and $theta(t)$ for a parameter $t$. Then the tangent may be given by the direction of $frac{dvec r}{dt}$.
I got a start, let the origin be the center of the circle. Let $vec m$ be the vector given by the radius of the circle pointing radially outwards ending at the point on the circle till where the string is bound to the circle, and $vec n$ be the vector such that $vec m+vec n=vec r$.
Now differentiating the vectors with respect to $t$, and using $vec r=r.hat r$. Now $frac{dm}{dt}=0$.
So $mfrac{dhat m}{dt}+nfrac{dhat n}{dt}+hat nfrac{dn}{dt}=$ tangent vector.
Now $|vec n|=n=mtheta$, where $theta$ is the angle rotated till now by the $vec m$. noLet |$frac{dhat m}{dt}$| be constant. So $frac{dn}{dt}=m|frac{dhat m}{dt}|$.
Now all that is left to show is that $frac{dhat m}{dt}$ is $perp$ to $vec m$. How do I do that?
geometry proof-writing tangent-line
asked Jan 18 at 6:56
Aditya AgarwalAditya Agarwal
2,92911839
$endgroup$
$begingroup$
Differentiating $hat mcdothat m=1$ you get $2hat mcdot{dhat mover dt}=0$.
$endgroup$
– Aretino
Jan 18 at 10:53
add a comment |
$begingroup$
Please refrain from using algebraic equations (in the Cartesian system) to prove it. I was looking for some kind of geometric proof for it. Or using differentitiation of vectors. For using the definition of tangent, let the curve be described by $vec r(t) $ and $theta(t)$ for a parameter $t$. Then the tangent may be given by the direction of $frac{dvec r}{dt}$.
I got a start, let the origin be the center of the circle. Let $vec m$ be the vector given by the radius of the circle pointing radially outwards ending at the point on the circle till where the string is bound to the circle, and $vec n$ be the vector such that $vec m+vec n=vec r$.
Now differentiating the vectors with respect to $t$, and using $vec r=r.hat r$. Now $frac{dm}{dt}=0$.
So $mfrac{dhat m}{dt}+nfrac{dhat n}{dt}+hat nfrac{dn}{dt}=$ tangent vector.
Now $|vec n|=n=mtheta$, where $theta$ is the angle rotated till now by the $vec m$. noLet |$frac{dhat m}{dt}$| be constant. So $frac{dn}{dt}=m|frac{dhat m}{dt}|$.
Now all that is left to show is that $frac{dhat m}{dt}$ is $perp$ to $vec m$. How do I do that?
geometry proof-writing tangent-line
asked Jan 18 at 6:56
Aditya AgarwalAditya Agarwal
2,92911839
$endgroup$
Please refrain from using algebraic equations (in the Cartesian system) to prove it. I was looking for some kind of geometric proof for it. Or using differentitiation of vectors. For using the definition of tangent, let the curve be described by $vec r(t) $ and $theta(t)$ for a parameter $t$. Then the tangent may be given by the direction of $frac{dvec r}{dt}$.
I got a start, let the origin be the center of the circle. Let $vec m$ be the vector given by the radius of the circle pointing radially outwards ending at the point on the circle till where the string is bound to the circle, and $vec n$ be the vector such that $vec m+vec n=vec r$.
Now differentiating the vectors with respect to $t$, and using $vec r=r.hat r$. Now $frac{dm}{dt}=0$.
So $mfrac{dhat m}{dt}+nfrac{dhat n}{dt}+hat nfrac{dn}{dt}=$ tangent vector.
Now $|vec n|=n=mtheta$, where $theta$ is the angle rotated till now by the $vec m$. noLet |$frac{dhat m}{dt}$| be constant. So $frac{dn}{dt}=m|frac{dhat m}{dt}|$.
Now all that is left to show is that $frac{dhat m}{dt}$ is $perp$ to $vec m$. How do I do that?
geometry proof-writing tangent-line
geometry proof-writing tangent-line
asked Jan 18 at 6:56
Aditya AgarwalAditya Agarwal
2,92911839
asked Jan 18 at 6:56
Aditya AgarwalAditya Agarwal
2,92911839
edited Jan 18 at 7:30
Aditya Agarwal
asked Jan 18 at 6:56
Aditya AgarwalAditya Agarwal
2,92911839
asked Jan 18 at 6:56
Aditya AgarwalAditya Agarwal
2,92911839
asked Jan 18 at 6:56
Aditya AgarwalAditya Agarwal
2,92911839
2,92911839
$begingroup$
Differentiating $hat mcdothat m=1$ you get $2hat mcdot{dhat mover dt}=0$.
$endgroup$
– Aretino
Jan 18 at 10:53
add a comment |
$begingroup$
Differentiating $hat mcdothat m=1$ you get $2hat mcdot{dhat mover dt}=0$.
$endgroup$
– Aretino
Jan 18 at 10:53
$begingroup$
Differentiating $hat mcdothat m=1$ you get $2hat mcdot{dhat mover dt}=0$.
$endgroup$
– Aretino
Jan 18 at 10:53
$begingroup$
Differentiating $hat mcdothat m=1$ you get $2hat mcdot{dhat mover dt}=0$.
$endgroup$
– Aretino
Jan 18 at 10:53
add a comment |
0
active
oldest
votes
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',
autoActivateHeartbeat: false,
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%2f3077917%2fprove-the-normal-to-the-involute-of-a-circle-is-tangent-to-the-circle%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
0
active
oldest
votes
0
active
oldest
votes
active
oldest
votes
active
oldest
votes
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.
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%2f3077917%2fprove-the-normal-to-the-involute-of-a-circle-is-tangent-to-the-circle%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
$begingroup$
Differentiating $hat mcdothat m=1$ you get $2hat mcdot{dhat mover dt}=0$.
$endgroup$
– Aretino
Jan 18 at 10:53