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

Question

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...?

Emilia Aguilar · Accepted Answer

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

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)#

Isabella Ackerman · Answer

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...# 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...#

Evelyn Agard · Answer

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)#

Charles Arocha · Answer

undefined To find the arc length of the function $y = \frac{1}{2}(e^x + e^{-x})$ with parameters $0 \leq x \leq 2$, you first need to calculate the derivative of the function, which is $\frac{dy}{dx} = \frac{1}{2}(e^x - e^{-x})$. Then, you apply the formula for arc length:

$$L = \int_{a}^{b} \sqrt{1 + \left(\frac{dy}{dx}\right)^2} dx$$

Substitute the derivative into the formula and integrate from $x = 0$ to $x = 2$.

$$L = \int_{0}^{2} \sqrt{1 + \left(\frac{1}{2}(e^x - e^{-x})\right)^2} dx$$

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...?

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...?