What is the arc length of #r(t)=(t,t,t)# on #tin [1,2]#?
1 Answer
Apr 12, 2018
#sqrt(3) #
Explanation:
We seek the arc length of the vector function:
# bb( ul r(t)) = << t,t,t >> # for#t in [1,2]#
Which we can readily evaluated using:
# L = int_alpha^beta \ || bb( ul (r')(t)) || \ dt #
So we calculate the derivative,
# bb( ul r'(t)) = << 1,1,1 >> #
Thus we gain the arc length:
# L = int_1^2 \ || <<1,1,1>> || \ dt #
# \ \ = int_1^2 \ sqrt(1^1+1^2+1^2) \ dt #
# \ \ = int_1^2 \ sqrt(3) \ dt #
# \ \ = [sqrt(3)t]_1^2 #
# \ \ = sqrt(3)(2-1) #
# \ \ = sqrt(3) #
This trivial result should come as no surprise as the given original equation is that of a straight line.
