How Do You Calculate the Arc Length of \( y = e^x \) from 0 to 1?

[Vadim]

my problem is to find the arc length of y=e^x between 0 and 1

what I've got is \int_{0}^{1} \sqrt{1+(e^x)^2}dx which i then substitute u=e^x giving \int_{1}^{e} \frac{\sqrt{1+u^2}}{u^2}u du which i then sub in v=\sqrt{1+u^2} giving \int_{\sqrt{2}}^{\sqrt{1+e^2}} \frac{v^2}{(v^2-1)^2} dv = \int_{\sqrt{2}}^{\sqrt{1+e^2}} \frac{v}{(v^2-1)} dv = \int_{\sqrt{2}}^{\sqrt{1+e^2}} \frac{v}{(v-1)(v+1)} dv which then needs to be integrated using partial fractions, and this is where i run into problems

i get \frac{1}{2} \int_{\sqrt{2}}^{\sqrt{1+e^2}} \frac{1}{v-1}+\frac{1}{v+1}dv which results in my getting an answer of 1.

the book on the other hand gets \int_{\sqrt{2}}^{\sqrt{1+e^2}}1+ \frac{\frac{1}{2}}{v-1}-\frac{\frac{1}{2}}{v+1}dv but i can't see how they got the 1, or the negative sign on the second fraction

 

[dextercioby]

my problem is to find the arc length of y=e^x between 0 and 1

what I've got is \int_{0}^{1} \sqrt{1+(e^x)^2}dx which i then substitute u=e^x giving \int_{1}^{e} \frac{\sqrt{1+u^2}}{u^2}u du which i then sub in v=\sqrt{1+u^2} giving \int_{\sqrt{2}}^{\sqrt{1+e^2}} \frac{v^2}{(v^2-1)^2} dv = \int_{\sqrt{2}}^{\sqrt{1+e^2}} \frac{v}{(v^2-1)} dv= \int_{\sqrt{2}}^{\sqrt{1+e^2}} \frac{v}{(v-1)(v+1)} dv which then needs to be integrated using partial fractions, and this is where i run into problems

i get \frac{1}{2} \int_{\sqrt{2}}^{\sqrt{1+e^2}} \frac{1}{v-1}+\frac{1}{v+1}dv which results in my getting an answer of 1.

the book on the other hand gets \int_{\sqrt{2}}^{\sqrt{1+e^2}}1+ \frac{\frac{1}{2}}{v-1}-\frac{\frac{1}{2}}{v+1}dv but i can't see how they got the 1, or the negative sign on the second fraction

I think you have a problem there.EDIT;It doesn't work.Well,when u got rid of the square in the variable "v".

Daniel.

 

[arildno]

my problem is to find the arc length of y=e^x between 0 and 1

what I've got is \int_{0}^{1} \sqrt{1+(e^x)^2}dx which i then substitute u=e^x giving \int_{1}^{e} \frac{\sqrt{1+u^2}}{u^2}u du which i then sub in v=\sqrt{1+u^2} giving \int_{\sqrt{2}}^{\sqrt{1+e^2}} \frac{v^2}{(v^2-1)^2} dv = \int_{\sqrt{2}}^{\sqrt{1+e^2}} \frac{v}{(v^2-1)} dv = \int_{\sqrt{2}}^{\sqrt{1+e^2}} \frac{v}{(v-1)(v+1)} dv which then needs to be integrated using partial fractions, and this is where i run into problems

i get \frac{1}{2} \int_{\sqrt{2}}^{\sqrt{1+e^2}} \frac{1}{v-1}+\frac{1}{v+1}dv which results in my getting an answer of 1.

the book on the other hand gets \int_{\sqrt{2}}^{\sqrt{1+e^2}}1+ \frac{\frac{1}{2}}{v-1}-\frac{\frac{1}{2}}{v+1}dv but i can't see how they got the 1, or the negative sign on the second fraction

This equality is wrong:
\int_{\sqrt{2}}^{\sqrt{1+e^2}} \frac{v^2}{(v^2-1)^2} dv = \int_{\sqrt{2}}^{\sqrt{1+e^2}} \frac{v}{(v^2-1)} dv

 

[Vadim]

yeah, i am seeing that...the book jumps from \int_{1}^{e} \frac{\sqrt{1+u^2}}{u^2}u du to \int_{\sqrt{2}}^{\sqrt{1+e^2}} \frac{v}{(v^2-1)} dv and (without checking) i figured \frac{x^2}{y^2}=\frac{x}{y} my mistake. now I'm really confused, because i see how they get \int_{1}^{e} \frac{\sqrt{1+u^2}}{u^2}u du but don't know where to go from there now.

 

[dextercioby]

\int_{1}^{e} \sqrt{\frac{1}{u^{2}}+1} \ du

Make the sub.

\frac{1}{u} =\sinh t

Daniel.

 

[GCT]

From here, \int_{1}^{e} \frac{\sqrt{1+u^2}}{u} du[/quote]

try the substitution u=1/t

 

[Vadim]

can anyone see what the book did(i would like to get it solved the same way they did it.) up to the point where it subs in v i can see what they did, but when i sub in v i get \int_{\sqrt{2}}^{\sqrt{1+e^2}} \frac{v^2}{(v^2-1)^2} dv and don't know how they got the roots of it.
and even then i don't see how they got the 1 and the negative in the partial fraction.

 

[dextercioby]

GCT,you're not being original...

Daniel.

 

[GCT]

I don't now what you're talking about;) nope, "not being original" as in I got this solution from another discussion, thought it would work here

https://www.physicsforums.com/showthread.php?t=76156

 

[OlderDan]

yeah, i am seeing that...the book jumps from \int_{1}^{e} \frac{\sqrt{1+u^2}}{u^2}u du to \int_{\sqrt{2}}^{\sqrt{1+e^2}} \frac{v}{(v^2-1)} dv and (without checking) i figured \frac{x^2}{y^2}=\frac{x}{y} my mistake. now I'm really confused, because i see how they get \int_{1}^{e} \frac{\sqrt{1+u^2}}{u^2}u du but don't know where to go from there now.

If the book makes that jump, it is a typographical error. It should be

\int_{1}^{e} \frac{\sqrt{1+u^2}}{u^2}u du = \int_{\sqrt{2}}^{\sqrt{1+e^2}} \frac{v^2}{(v^2-1)} dv

And their result follows. The u in this equation is just the original y, and v is as you defined it. Taking it from the top

y = e^x

dy = e^x dx = ydx \Rightarrow dx = \frac{{dy}}{y}

ds = \sqrt {dx^2 + dy^2 } = \left( {\sqrt {\frac{1}{{y^2 }} + 1} } \right)dy = \frac{{\sqrt {1 + y^2 } }}{y}dy\left( { = \frac{{\sqrt {1 + y^2 } }}{{y^2 }}ydy} \right)

v = \sqrt {1 + y^2 } \Rightarrow dv = \frac{1}{{2v}}2ydy = \frac{{ydy}}{v} \Rightarrow dy = \frac{v}{y}dv

v^2 = 1 + y^2 \Rightarrow y^2 = v^2 - 1

ds = \frac{{\sqrt {1 + y^2 } }}{y}dy = \frac{v}{y} \cdot \frac{v}{y}dv = \frac{{v^2 }}{{y^2 }}dv = \frac{{v^2 }}{{v^2 - 1}}dv

ds = \left( {1 + \frac{1}{{v^2 - 1}}} \right)dv = \left( {1 + \frac{1}{{v - 1}} \cdot \frac{1}{{v + 1}}} \right)dv = \left( {1 + \frac{{\frac{1}{2}}}{{v - 1}} - \frac{{\frac{1}{2}}}{{v + 1}}} \right)dv