# length of line

• Apr 19th 2010, 07:42 PM
gralla55
length of line
Calculate the length of the line segment determined by the path:

x(t) = (a1t + b1, a2t + b2)

as t varies from t0 to t1.

It should be really easy, I'm just not quite sure what is what in this formula

L = (integral from a to b)(root)(1+(f'(x))^2) dx

Thanks!
• Apr 20th 2010, 03:54 AM
HallsofIvy
Quote:

Originally Posted by gralla55
Calculate the length of the line segment determined by the path:

x(t) = (a1t + b1, a2t + b2)

as t varies from t0 to t1.

It should be really easy, I'm just not quite sure what is what in this formula

L = (integral from a to b)(root)(1+(f'(x))^2) dx

Thanks!

When t= t0, x= a1t0+ b1 and y= a2t0+ b2. When t= x1, x= a1t1+ b1 and y= a2t1+ b2. Those x values are the limits, a and b, of your integral.

Since x= a1t+ b1, t= (x- b1)/a1 so y= a2t+ b2= a2(x- b1)/a1+ b2. That is your f(x)= (a2/a1)x- (a2b1/a1) +b1. f'= (a2/a1). Your integration is just $\displaystyle \int_{a1t0+ b1}^{a1t1+ b1} \sqrt{1+ (a2/a1)^2} dx$

That is, of course, a very easy integral because the integrand is a constant. In fact, you don't need to integrate at all: x goes from a1t0+ b1 to a1t1+ b1, a difference of a1(t1- t0). y goes from a2t0+ b2 to a2t1+ b2, a difference of a2(t1- t0). Use the Pythagorean theorem to find the distance: $\displaystyle (t1- t0)\sqrt{(a1)^2+ (a2)^2}$.