Find the arc length of the function y=1/2(e^x+e^-x) with parameters 0\lex\le2?

What I have so far:
ds=\sqrt(1+(dy/dx)^2)dx with a=0 and b=2

dy/dx=1/2(e^x-e^-x)^2
(dy/dx)^2=1/4(e^x-e^-x)

ds=\sqrt(1+[1/4(e^x-e^-x)^2])dx...

So how do I proceed from here...?

3 Answers
Apr 25, 2018

L=1/2(e^2-1/e^2)

Explanation:

So, we have y=1/2(e^x+e^-x)

This is in the form y=f(x), so we know arc length L on [0,2] is given by

L=int_0^2sqrt(1+(dy/dx)^2)dx

Let's find the derivative:

dy/dx=1/2(e^x-e^-x)

Before jumping into the integration, we should realize the following hyperbolic identities:

y=1/2(e^x+e^-x)=coshx

dy/dx=1/2(e^x-e^-x)=sinhx

Then, squaring gives us

(dy/dx)^2=sinh^2x

So, applying the hyperbolic identity, we get

L=int_0^2sqrt(1+sinh^2x)dx

There is a hyperbolic Pythagorean identity we apply here, however, it looks slightly different from the normal trigonometric Pythagorean identity.

Deriving it is a pretty algebraically messy process, so I'll just give the identity:

cosh^2x-sinh^2x=1

This tells us that

1+sinh^2x=cosh^2x

Thus,

L=int_0^2sqrt(cosh^2x)dx=int_0^2coshxdx

In general,

intcoshxdx=sinhx+C.

Since the derivatives hyperbolic sine and hyperbolic sine are simply one another, IE, d/dxsinhx=coshx, d/dxcoshx=sinhx, so their integrals also hold this property.

Thus,

L=sinhx|_0^2

L=1/2(e^x-e^-x)|_0^2

L=1/2(e^2-e^-2-(e^0-e^0))

L=1/2(e^2-1/e^2-(1-1))

L=1/2(e^2-1/e^2)

Apr 25, 2018

note: this answer is for the question: Find the arc length of the function y=ln(secx) with parameters 0 ≤ x ≤ pi/4?

ln(sqrt(2)+1)=0.881374...

Explanation:

arclength =int_0^(pi/4)sqrt(1+(y'(x))^2)dx

[side note:
y'(x)=1/secx*(secxtanx)=tanx]

arclength =int_0^(pi/4)sqrt(1+tan^2x)dx

=int_0^(pi/4)sqrt(sec^2x)dx

=int_0^(pi/4)secxdx
[side note:
you can do this because secx is always positive on 0<=x<=pi/4]

[another side note: intsecxdx=ln|secx+tanx|+C see this]

=ln|sec(pi/4)+tan(pi/4)|-ln|sec0+tan0|

=ln|sqrt(2)+1|-ln|1+0|

=ln|sqrt(2)+1|

=0.881374...

Apr 25, 2018

As you stated,

y=1/2(e^x+e^-x)

dy/dx=1/2(e^x-e^-x)

(dy/dx)^2=1/4(e^x-e^-x)^2

But now let's expand (e^x-e^-x)^2 as (e^x-e^-x)(e^x-e^-x)=e^(2x)+e^(-2x)-2. Recall that e^xe^-x=e^0=1.

Then, we see that

ds=sqrt(1+1/4(e^(2x)+e^(-2x)-2))dx=sqrt((e^(2x)+e^(-2x)+2)/4)dx

Let's try to clear up this numerator (get rid of negative exponents) by multiplying the fraction through by e^(2x)//e^(2x):

ds=sqrt((e^(2x)(e^(2x)+e^(-2x)+2))/(4e^(2x)))dx=sqrt((e^(4x)+2e^(2x)+1)/(4e^(2x)))dx

Now we recognize that our numerator is actually factorable, and in a very favorable way!

ds=sqrt((e^(2x)+1)^2/((2e^x)^2))dx=(e^(2x)+1)/(2e^x)dx=1/2(e^x+e^-x)dx

What a remarkable amount of work to end up... exactly where we started?

Amazing and interesting coincidence aside, we can calculate the arc length now by:

L=int_0^2ds=1/2int_0^2(e^x+e^-x)dx=1/2(e^x-e^-x)|_0^2

Evaluating:

L=1/2[(e^2-1)-(e^-2-1)]=1/2(e^2-1/e^2)=(e^4-1)/(2e^2)