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)#