Geometric interpretation of tanh
$begingroup$
Ok so in today's lecture on hyperbolic functions, the lecturer drew the well-known graph of the equilateral hyperbola, which shows sinh(a), cosh(a) and the area which is equal to a/2.
However, when I asked him if the hyperbolic tangent appears anywhere in this graph (a la its circular trigonometric cousin, which is the tangent of the unit circle), he didn't have a response and said he'd never thought about that. Can anyone explain if the tangent actually appears somehow in this graph and its geometric meaning related to it (if there is one at all)? I hope I made my question clear enough, but, if this is not the case, I will try to explain it better. Thanks!
hyperbolic-functions
asked Dec 3 '18 at 20:09
JBuckJBuck
566
$endgroup$
add a comment |
$begingroup$
Ok so in today's lecture on hyperbolic functions, the lecturer drew the well-known graph of the equilateral hyperbola, which shows sinh(a), cosh(a) and the area which is equal to a/2.
However, when I asked him if the hyperbolic tangent appears anywhere in this graph (a la its circular trigonometric cousin, which is the tangent of the unit circle), he didn't have a response and said he'd never thought about that. Can anyone explain if the tangent actually appears somehow in this graph and its geometric meaning related to it (if there is one at all)? I hope I made my question clear enough, but, if this is not the case, I will try to explain it better. Thanks!
hyperbolic-functions
asked Dec 3 '18 at 20:09
JBuckJBuck
566
$endgroup$
add a comment |
$begingroup$
Ok so in today's lecture on hyperbolic functions, the lecturer drew the well-known graph of the equilateral hyperbola, which shows sinh(a), cosh(a) and the area which is equal to a/2.
However, when I asked him if the hyperbolic tangent appears anywhere in this graph (a la its circular trigonometric cousin, which is the tangent of the unit circle), he didn't have a response and said he'd never thought about that. Can anyone explain if the tangent actually appears somehow in this graph and its geometric meaning related to it (if there is one at all)? I hope I made my question clear enough, but, if this is not the case, I will try to explain it better. Thanks!
hyperbolic-functions
asked Dec 3 '18 at 20:09
JBuckJBuck
566
$endgroup$
Ok so in today's lecture on hyperbolic functions, the lecturer drew the well-known graph of the equilateral hyperbola, which shows sinh(a), cosh(a) and the area which is equal to a/2.
However, when I asked him if the hyperbolic tangent appears anywhere in this graph (a la its circular trigonometric cousin, which is the tangent of the unit circle), he didn't have a response and said he'd never thought about that. Can anyone explain if the tangent actually appears somehow in this graph and its geometric meaning related to it (if there is one at all)? I hope I made my question clear enough, but, if this is not the case, I will try to explain it better. Thanks!
hyperbolic-functions
hyperbolic-functions
asked Dec 3 '18 at 20:09
JBuckJBuck
566
asked Dec 3 '18 at 20:09
JBuckJBuck
566
asked Dec 3 '18 at 20:09
JBuckJBuck
566
asked Dec 3 '18 at 20:09
JBuckJBuck
566
asked Dec 3 '18 at 20:09
JBuckJBuck
566
566
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
The hyperbola $x^2-y^2=1,,xge 1$ can be parameterised as $x=cosh a,,y=pmsinh a$ so $dy/dx=pmcoth a$. The normal to the curve then has gradient $mptanh a$.
answered Dec 3 '18 at 20:30
J.G.J.G.
25.2k22539
$endgroup$
$begingroup$
Thanks for that, but what do you mean by the ''normal'' of the curve?
$endgroup$
– JBuck
Dec 3 '18 at 20:51
$begingroup$
Given a point $p$ on a curve $C$, the normal to $C$ at $p$ is a line passing through $p$ orthogonal to the tangent to $C$ at $p$. Because the lines are orthogonal, their gradients have product $-1$.
$endgroup$
– J.G.
Dec 3 '18 at 20:54
$begingroup$
Now it make sense, thank you!
$endgroup$
– JBuck
Dec 3 '18 at 20:57
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',
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%2f3024594%2fgeometric-interpretation-of-tanh%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
$begingroup$
The hyperbola $x^2-y^2=1,,xge 1$ can be parameterised as $x=cosh a,,y=pmsinh a$ so $dy/dx=pmcoth a$. The normal to the curve then has gradient $mptanh a$.
answered Dec 3 '18 at 20:30
J.G.J.G.
25.2k22539
$endgroup$
$begingroup$
Thanks for that, but what do you mean by the ''normal'' of the curve?
$endgroup$
– JBuck
Dec 3 '18 at 20:51
$begingroup$
Given a point $p$ on a curve $C$, the normal to $C$ at $p$ is a line passing through $p$ orthogonal to the tangent to $C$ at $p$. Because the lines are orthogonal, their gradients have product $-1$.
$endgroup$
– J.G.
Dec 3 '18 at 20:54
$begingroup$
Now it make sense, thank you!
$endgroup$
– JBuck
Dec 3 '18 at 20:57
add a comment |
$begingroup$
The hyperbola $x^2-y^2=1,,xge 1$ can be parameterised as $x=cosh a,,y=pmsinh a$ so $dy/dx=pmcoth a$. The normal to the curve then has gradient $mptanh a$.
answered Dec 3 '18 at 20:30
J.G.J.G.
25.2k22539
$endgroup$
$begingroup$
Thanks for that, but what do you mean by the ''normal'' of the curve?
$endgroup$
– JBuck
Dec 3 '18 at 20:51
$begingroup$
Given a point $p$ on a curve $C$, the normal to $C$ at $p$ is a line passing through $p$ orthogonal to the tangent to $C$ at $p$. Because the lines are orthogonal, their gradients have product $-1$.
$endgroup$
– J.G.
Dec 3 '18 at 20:54
$begingroup$
Now it make sense, thank you!
$endgroup$
– JBuck
Dec 3 '18 at 20:57
add a comment |
$begingroup$
The hyperbola $x^2-y^2=1,,xge 1$ can be parameterised as $x=cosh a,,y=pmsinh a$ so $dy/dx=pmcoth a$. The normal to the curve then has gradient $mptanh a$.
answered Dec 3 '18 at 20:30
J.G.J.G.
25.2k22539
$endgroup$
The hyperbola $x^2-y^2=1,,xge 1$ can be parameterised as $x=cosh a,,y=pmsinh a$ so $dy/dx=pmcoth a$. The normal to the curve then has gradient $mptanh a$.
answered Dec 3 '18 at 20:30
J.G.J.G.
25.2k22539
answered Dec 3 '18 at 20:30
J.G.J.G.
25.2k22539
answered Dec 3 '18 at 20:30
J.G.J.G.
25.2k22539
answered Dec 3 '18 at 20:30
J.G.J.G.
25.2k22539
25.2k22539
$begingroup$
Thanks for that, but what do you mean by the ''normal'' of the curve?
$endgroup$
– JBuck
Dec 3 '18 at 20:51
$begingroup$
Given a point $p$ on a curve $C$, the normal to $C$ at $p$ is a line passing through $p$ orthogonal to the tangent to $C$ at $p$. Because the lines are orthogonal, their gradients have product $-1$.
$endgroup$
– J.G.
Dec 3 '18 at 20:54
$begingroup$
Now it make sense, thank you!
$endgroup$
– JBuck
Dec 3 '18 at 20:57
add a comment |
$begingroup$
Thanks for that, but what do you mean by the ''normal'' of the curve?
$endgroup$
– JBuck
Dec 3 '18 at 20:51
$begingroup$
Given a point $p$ on a curve $C$, the normal to $C$ at $p$ is a line passing through $p$ orthogonal to the tangent to $C$ at $p$. Because the lines are orthogonal, their gradients have product $-1$.
$endgroup$
– J.G.
Dec 3 '18 at 20:54
$begingroup$
Now it make sense, thank you!
$endgroup$
– JBuck
Dec 3 '18 at 20:57
$begingroup$
Thanks for that, but what do you mean by the ''normal'' of the curve?
$endgroup$
– JBuck
Dec 3 '18 at 20:51
$begingroup$
Thanks for that, but what do you mean by the ''normal'' of the curve?
$endgroup$
– JBuck
Dec 3 '18 at 20:51
$begingroup$
Given a point $p$ on a curve $C$, the normal to $C$ at $p$ is a line passing through $p$ orthogonal to the tangent to $C$ at $p$. Because the lines are orthogonal, their gradients have product $-1$.
$endgroup$
– J.G.
Dec 3 '18 at 20:54
$begingroup$
Given a point $p$ on a curve $C$, the normal to $C$ at $p$ is a line passing through $p$ orthogonal to the tangent to $C$ at $p$. Because the lines are orthogonal, their gradients have product $-1$.
$endgroup$
– J.G.
Dec 3 '18 at 20:54
$begingroup$
Now it make sense, thank you!
$endgroup$
– JBuck
Dec 3 '18 at 20:57
$begingroup$
Now it make sense, thank you!
$endgroup$
– JBuck
Dec 3 '18 at 20:57
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.
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%2f3024594%2fgeometric-interpretation-of-tanh%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
