Q on hyperbolic equation.

[Sonal7]

I cant work out how they have integrated the (cosh 2u -1)/2. I know there is a mistake where it says sinh^2u, it should be sinh u.

 

[joshuaa]

[MATH]x = \cosh u[/MATH][MATH]dx = \sinh u \ du[/MATH]
you forgot that [MATH]dx[/MATH] will be changed to [MATH]\sinh u \ du[/MATH]
then

you will have [MATH]\sinh^2 u \ du[/MATH]

 

[Sonal7]

Thank you I forgot about that. I got it now. But how did they get 2 root 3. I know what cosh arcosh 2 is 2, but I cant see how you might know sinsh arcosh 2 is root 3. Any idea? Thank you v much.

 

[Dr.Peterson]

Thank you I forgot about that. I got it now. But how did they get 2 root 3. I know what cosh arcosh 2 is 2, but I cant see how you might know sinsh arcosh 2 is root 3. Any idea? Thank you v much.

So you know the "Pythagorean" identity for sinh and cosh?

 

[Sonal7]

yes but still not sure how its root 3. cosh^2 -sinh^2=1. I hope thats right. let me think for a second.

 

[Sonal7]

So you know the "Pythagorean" identity for sinh and cosh?

Ok i got it. Its root 3 as if you rearranged cosh(arcosh 2) (ie =2) then sinh^2 (arcosh 2) is 1-4. Got it, thank you so much.

 

[joshuaa]

[MATH]\frac{\sinh 2u}{2} = \frac{2\sinh u \cosh u}{2} = \sinh u \cosh u[/MATH]
remember this

if you have something and its inverse, they cancel each other

then

[MATH]\cosh \cosh^{-1} 2 = 2[/MATH] (they cancel each other)

And

[MATH]\sinh u = \sqrt{\cosh^2 u - 1}[/MATH]
[MATH]\sinh \cosh^{-1} 2 = \sqrt{(\cosh \cosh^{-1} 2)^2 - 1} = \sqrt{2^2 - 1} = \sqrt{4 - 1} = \sqrt{3}[/MATH]
then

[MATH]\sinh \cosh^{-1} 2 \cosh \cosh^{-1} 2 = \sqrt{3} \cdot 2 = 2\sqrt{3}[/MATH]

 

[Sonal7]

[MATH]\frac{\sinh 2u}{2} = \frac{2\sinh u \cosh u}{2} = \sinh u \cosh u[/MATH]
remember this

if you have something and its inverse, they cancel each other

then

[MATH]\cosh \cosh^{-1} 2 = 2[/MATH] (they cancel each other)

And

[MATH]\sinh u = \sqrt{\cosh^2 u - 1}[/MATH]
[MATH]\sinh \cosh^{-1} 2 = \sqrt{(\cosh \cosh^{-1} 2)^2 - 1} = \sqrt{2^2 - 1} = \sqrt{4 - 1} = \sqrt{3}[/MATH]
then

[MATH]\sinh \cosh^{-1} 2 \cosh \cosh^{-1} 2 = \sqrt{3} \cdot 2 = 2\sqrt{3}[/MATH]

I did that in my head, but it is good of you to do it for the rest of the public to learn from. I hadnt realised i made a typing error I meant to write sinh^2 (arcos 2) =4-1.

 

[joshuaa]

also, one more thing

the limit of integration should be [MATH]\int_{1}^{2}[/MATH] with respect to [MATH] x[/MATH]
the book did it [MATH]\int_{1}^{2\sqrt{3}}[/MATH]

 

[joshuaa]

exact area is

[MATH]\int_{0}^{2\sqrt{3}} \left(10 - \frac{4y}{\sqrt{3}} - \sqrt{\frac{y^2}{4} + 1} \right) \ dy \ = \ 10\sqrt{3} - \sinh^{-1} \sqrt{3} \ = \ 10\sqrt{3} - \cosh^{-1} 2 [/MATH]
this is the beauty when you can find the whole area with only one integral

 

[Sonal7]

also, one more thing

the limit of integration should be [MATH]\int_{1}^{2}[/MATH] with respect to [MATH] x[/MATH]
the book did it [MATH]\int_{1}^{2\sqrt{3}}[/MATH]

Yes that was an obvious mistake. I noticed that. I did wish to ask why did you think they chose such as awkward u sub, why not cos u, why cosh u. I know its hyperbolic parametric values, is that why? would it matter ? bit of a dumb question, i m guessing its to do with value of the area we are trying to derive-its has an arcosh in it.

 

[joshuaa]

Yes that was an obvious mistake. I noticed that. I did wish to ask why did you think they chose such as awkward u sub, why not cos u, why cosh u. I know its hyperbolic parametric values, is that why? would it matter ? bit of a dumb question, i m guessing its to do with value of the area we are trying to derive-its has an arcosh in it.

you don't have to do it with hyperbolic [MATH]\cosh u[/MATH]
you can instead do it with trigonometric [MATH]\sec u[/MATH]
why [MATH]\cosh u[/MATH] or [MATH]\sec u[/MATH] ?

because you have to look for something similar to [MATH]\sqrt{x^2 - 1}[/MATH]
[MATH]\sinh u = \sqrt{\cosh^2 u - 1}[/MATH]
[MATH]\tan u = \sqrt{\sec^2 u - 1}[/MATH]
do you notice that they look like the square root?

but

[MATH]\sin u = \sqrt{1 - \cos^2 u}[/MATH]
i cannot use [MATH]\cos u[/MATH] because i want the integral to be [MATH] \sqrt{1 - x^2} [/MATH] to use it

you can also use [MATH]\csc u[/MATH]
because [MATH]\cot u = \sqrt{\csc^2 u - 1}[/MATH] which is similar to [MATH]\sqrt{x^2 - 1}[/MATH]

 

[Sonal7]

you don't have to do it with hyperbolic [MATH]\cosh u[/MATH]
you can instead do it with trigonometric [MATH]\sec u[/MATH]
why [MATH]\cosh u[/MATH] or [MATH]\sec u[/MATH] ?

because you have to look for something similar to [MATH]\sqrt{x^2 - 1}[/MATH]
[MATH]\sinh u = \sqrt{\cosh^2 u - 1}[/MATH]
[MATH]\tan u = \sqrt{\sec^2 u - 1}[/MATH]
do you notice that they look like the square root?

but

[MATH]\sin u = \sqrt{1 - \cos^2 u}[/MATH]
i cannot use [MATH]\cos u[/MATH] because i want the integral to be [MATH] \sqrt{1 - x^2} [/MATH] to use it

Thank you very much