Saturday
April 30, 2016

# Homework Help: calc check: curve length

Posted by COFFEE on Monday, July 2, 2007 at 2:31pm.

Find the length of the curve y=(1/(x^2)) from ( 1, 1 ) to ( 2, 1/4 ) [set up the problem only, don't integrate/evaluate]

this is what i did.. let me know asap if i did it right..

y = (1/(x^2))
dy/dx = (-2/(x^3))

L = integral from a to b for: sqrt(1+(dy/dx)^2)dx
L = integral from 1 to 2 for: sqrt(1+(-2/(x^3))^2)dx
L = integral from 1 to 2 for: sqrt(1+(-2/(x^3))(-2/(x^3)))dx
L = integral from 1 to 2 for: sqrt(1+(4/(x^6))dx

a=1
b=2
n=1-
deltaX=0.1
f(x)=sqrt(1+(4/x^6))

L = integral from 1 to 2 for: sqrt(1+(4/(x^6))dx
L = (deltaX/3)[ f(1) + 4f(1.1) + 2f(1.2) + 4f(1.3) + ... + 2f(1.8) + 4f(1.9) + f(2) ]
L = (0.1/3)[ sqrt(1+(4/1)^6) + 4sqrt(1+(4/1.1)^6) + 2sqrt(1+(4/1.2)^6) + 4sqrt(1+(4/1.3)^6) + 2sqrt(1+(4/1.4)^6) + 4sqrt(1+(4/1.5)^6) + 2sqrt(1+(4/1.6)^6) + 4sqrt(1+(4/1.7)^6) + 2sqrt(1+(4/1.8)^6) + 4sqrt(1+(4/1.9)^6) + sqrt(1+(4/2)^6) ]
L = (0.1/3)[720.937]
L = 24.031