Consider the parametric equation: #x = 15(cos(theta) + (theta)sin(theta))# and #y = 15(sin(theta) - (theta)cos(theta))#, What is the length of the curve for #theta = 0# to #theta = pi/8#?
1 Answer
Explanation:
The Arc Length for a Parametric Curve is given by
# L=int_alpha^beta sqrt((dx/dt)^2+(dy/dt)^2) \ dt #
So in this problem we have (using the product rule):
# \ \ \ \ \ \ \ \ x= 15(cos theta + theta sin theta) #
# \ \ \ \ \ \ \ \ \ \ = 15 cos theta+15 theta sin theta #
# :. dx/(d theta) = -15 sin theta + (15 theta)(cos theta) + (15)(sin theta)#
# \ \ \ \ \ \ \ \ \ \ = 15 theta cos theta #
# \ \ \ \ \ \ \ \ y= 15(sin theta - theta cos theta) #
# \ \ \ \ \ \ \ \ \ \ = 15 sin theta - 15 theta cos theta) #
# :. dy/(d theta) = 15 cos theta - (15 theta)(sin theta) - (15)(cos theta) #
# \ \ \ \ \ \ \ \ \ \ = - 15 theta sin theta #
So the Arc Length is;
# L= int_0^(pi/8) sqrt( (15 theta cos theta)^2 + (- 15 theta sin theta)^2) \ d theta #
# \ \ = int_0^(pi/8) sqrt( 15^2 theta^2 cos^2 theta + 15^2 theta^2 sin^2 theta) \ d theta #
# \ \ = int_0^(pi/8) sqrt( 15^2 theta^2 (cos^2 theta + sin^2 theta) \ d theta #
# \ \ = int_0^(pi/8) sqrt( 15^2 theta^2 ) \ d theta #
# \ \ = int_0^(pi/8) 15 theta \ d theta #
# \ \ = [15/2 theta^2]_0^(pi/8) #
# \ \ = 15/2 (pi^2/64-0) #
# \ \ = (15pi^2)/128 #
