Geometric interpretation of tanh












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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!










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    0












    $begingroup$


    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!










    share|cite|improve this question









    $endgroup$















      0












      0








      0





      $begingroup$


      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!










      share|cite|improve this question









      $endgroup$




      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






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      asked Dec 3 '18 at 20:09









      JBuckJBuck

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      566






















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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 $mptanh a$.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            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











          Your Answer





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          1 Answer
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          1 Answer
          1






          active

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          active

          oldest

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          active

          oldest

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          1












          $begingroup$

          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 $mptanh a$.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            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
















          1












          $begingroup$

          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 $mptanh a$.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            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














          1












          1








          1





          $begingroup$

          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 $mptanh a$.






          share|cite|improve this answer









          $endgroup$



          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 $mptanh a$.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          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


















          • $begingroup$
            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


















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