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:
So how do I proceed from here...?
3 Answers
Explanation:
So, we have
This is in the form
Let's find the derivative:
Before jumping into the integration, we should realize the following hyperbolic identities:
Then, squaring gives us
So, applying the hyperbolic identity, we get
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:
This tells us that
Thus,
In general,
Since the derivatives hyperbolic sine and hyperbolic sine are simply one another, IE,
Thus,
note: this answer is for the question: Find the arc length of the function
Explanation:
arclength
[side note:
arclength
[side note:
you can do this because
[another side note:
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
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
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)