Hyperbolic Sine

  • Thread starter Xkaliber
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  • #1
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Hi everyone,

I was in the middle of solving a physics problem and came across a math term I am having trouble solving. It is the hyperbolic sine term in this equation:

[tex]T=\frac{2.0x10^{-8}}{sinh^{2}[\frac{\sqrt{130(2.6x10^{10}-130)}}{1.05x10^{-34}}6x10^{11}]+2.0x10^{-8}}[/tex]

When plugging the sinh term into the calculator, I receive an error due to the massive size of the solution. Is there an equivalent method of taking a hyperbolic sine that can be done by hand since the calulator cannot handle the large exponent?
 

Answers and Replies

  • #2
You have the high road or the low road.

For the low road, get yourself a copy of the CRC Standard Math Tables. It has tables for Hyperbolic Functions.

For the high road there is Mathematica (my favorite) or MatLab.

For myself, even though I have copies of Mathematica and Matlab (both) at my easy disposal, I still like my book of standard math tables at times.
 
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  • #3
arildno
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Why bother?

HypSine(Humungous) equals 1/2*e^Humungous.

And 1/Superbig is just about zero.
 
  • #4
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Thanks for the replies. I suppose I will just use Mathematica this evening to obtain some number besides zero since it's a probability. Basically, it is the likelihood of a human launching from the Earth's surface at 4 m/s and reaching Jupiter.
 
  • #6
HallsofIvy
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Xkaliber said:
Thanks for the replies. I suppose I will just use Mathematica this evening to obtain some number besides zero since it's a probability. Basically, it is the likelihood of a human launching from the Earth's surface at 4 m/s and reaching Jupiter.
Why would you want an answer "besides zero"? Since 4 m/s is well below escape speed from earth, the "probabilty" of reaching Jupiter (or any thing except earth itself) is zero.
 
  • #7
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Well, quantum mechanically the probability is not exacly zero. Through the process known as "tunneling", a particle or object can travel through a potential energy barrier, in this case gravity, with its total energy less than the potential energy. I did actually put zero as my final answer in the end since I included my steps for solving the problem. The final probability ended up being something along the lines of [tex]e^{10^{44}}[/tex]
 
  • #8
arildno
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I think you forgot a minus sign somewhere in your exponent. :smile:
 
  • #9
Office_Shredder
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Actually, thanks to the many worlds variety of quantum mechanics there are [tex]e^{10^{5000}}[/tex] ways of a person launching from earth at 4 m/s. So [tex]e^{10^{44}}[/tex] people actually escape earth for each person that attempts to launch.

*nod*
 
  • #10
HallsofIvy
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Office_Shredder said:
Actually, thanks to the many worlds variety of quantum mechanics there are [tex]e^{10^{5000}}[/tex] ways of a person launching from earth at 4 m/s. So [tex]e^{10^{44}}[/tex] people actually escape earth for each person that attempts to launch.

*nod*
Are you sure there shouldn't be a negative sign in the exponent? If not, I jump up and down a few times and Jupiter is overrun with people!
 
  • #11
Office_Shredder
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Jupiters are overrun with people. But imagine how many Jupiters there are at that point
 

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