Could you solve this?

[Guest]

x=cosh2t
y=2t
0 [tex]\le[/tex] t
t [tex]\le[/tex] ln2 find the length of parabol by using these equations.

[Guest]

What parabola? x= cosh(2t), y= 2t which is equivalent to x= cosh(y) or [tex]y= cosh^{-1}(x)[/tex], is NOT a parabola!

[HallsofIvy]

This has been here a while now so: for the non-parabola (its graph is close to a parabola),
x= cosh(2t) so dx/dt= 2sinh(2t)
y= 2t so dy/dt= 2.

The "differential of arc-length" is [tex]ds= \sqrt{\left(\frac{dx}{dt}\right)^2+ \left(\frac{dy}{dt}\right)^2}dt[/tex].
[tex]= \sqrt{4sinh^2(2t)+4}dt= 2\sqrt{sinh^2(2t)+ 1}= 2 cosh(2t)dt[/tex].

Since t goes from 0 to ln(2) the arclength is given by [tex]2\int_0^{ln(2)} cosh(2t)dt= \left[sinh(2t)\right]_0^{ln(2)}[/tex].

[Guest]

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