Geometric interpretation of tanh
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
JBuck
366
add a comment |
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
JBuck
366
add a comment |
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
JBuck
366
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
JBuck
366
asked Dec 3 '18 at 20:09
JBuck
366
asked Dec 3 '18 at 20:09
JBuck
366
asked Dec 3 '18 at 20:09
JBuck
366
asked Dec 3 '18 at 20:09
JBuck
366
366
add a comment |
add a comment |
1 Answer
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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.
23k22137
Thanks for that, but what do you mean by the ''normal'' of the curve?
– JBuck
Dec 3 '18 at 20:51
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 $-1$.
– J.G.
Dec 3 '18 at 20:54
Now it make sense, thank you!
– JBuck
Dec 3 '18 at 20:57
add a comment |
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1 Answer
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1 Answer
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oldest
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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.
23k22137
Thanks for that, but what do you mean by the ''normal'' of the curve?
– JBuck
Dec 3 '18 at 20:51
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 $-1$.
– J.G.
Dec 3 '18 at 20:54
Now it make sense, thank you!
– JBuck
Dec 3 '18 at 20:57
add a comment |
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.
23k22137
Thanks for that, but what do you mean by the ''normal'' of the curve?
– JBuck
Dec 3 '18 at 20:51
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 $-1$.
– J.G.
Dec 3 '18 at 20:54
Now it make sense, thank you!
– JBuck
Dec 3 '18 at 20:57
add a comment |
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.
23k22137
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.
23k22137
answered Dec 3 '18 at 20:30
J.G.
23k22137
answered Dec 3 '18 at 20:30
J.G.
23k22137
answered Dec 3 '18 at 20:30
J.G.
23k22137
23k22137
Thanks for that, but what do you mean by the ''normal'' of the curve?
– JBuck
Dec 3 '18 at 20:51
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 $-1$.
– J.G.
Dec 3 '18 at 20:54
Now it make sense, thank you!
– JBuck
Dec 3 '18 at 20:57
add a comment |
Thanks for that, but what do you mean by the ''normal'' of the curve?
– JBuck
Dec 3 '18 at 20:51
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 $-1$.
– J.G.
Dec 3 '18 at 20:54
Now it make sense, thank you!
– JBuck
Dec 3 '18 at 20:57
Thanks for that, but what do you mean by the ''normal'' of the curve?
– JBuck
Dec 3 '18 at 20:51
Thanks for that, but what do you mean by the ''normal'' of the curve?
– JBuck
Dec 3 '18 at 20:51
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 $-1$.
– J.G.
Dec 3 '18 at 20:54
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 $-1$.
– J.G.
Dec 3 '18 at 20:54
Now it make sense, thank you!
– JBuck
Dec 3 '18 at 20:57
Now it make sense, thank you!
– JBuck
Dec 3 '18 at 20:57
add a comment |
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