Find the exact length of the curve:
y=3+(1/2)cosh(2x) 0<=x<=1
Cannot figure out this problem. Is y'=sinh(2x)?
If it is I have L=integral from 0 to 1 of (sqrt(1+(sinh(2x))^2)). Don't know where to go from here if I even did it right. I've never worked with cosh before. Help would be great. Thanks!
y=3+(1/2)cosh(2x) 0<=x<=1
Cannot figure out this problem. Is y'=sinh(2x)?
If it is I have L=integral from 0 to 1 of (sqrt(1+(sinh(2x))^2)). Don't know where to go from here if I even did it right. I've never worked with cosh before. Help would be great. Thanks!
-
You are right on!
1
∫ √[(dx)² + (dy) ²]
0
1
∫ √[1 + (y)' ²] dx
0
1
∫ √[1 + sinh²2x] dx
0
1
∫ cosh2x dx ------- cosh²2x - sinh²2x = 1, cosh2x is always positive
0
= 1/2 sinh2x, x from 0 to 1
=1/2 sinh2
=(e⁴ - 1)/(4e²)
1
∫ √[(dx)² + (dy) ²]
0
1
∫ √[1 + (y)' ²] dx
0
1
∫ √[1 + sinh²2x] dx
0
1
∫ cosh2x dx ------- cosh²2x - sinh²2x = 1, cosh2x is always positive
0
= 1/2 sinh2x, x from 0 to 1
=1/2 sinh2
=(e⁴ - 1)/(4e²)