Question

X 16.1.19 Evaluate Sxas x ds, where C is С t a. the straight line segment x =t, y= 5, from (0,0) to (20,4) b. the parabolic curve x =t, y = 262, from (0,0) to (1,2)

Answer 1

Ans: Sx de с a) c is the straight une segment xat, y= t/s from (0,0) to (200 we know that s flx,y) ds - Ś fl alt), y(t)) | (d'at) + City ? at li a Now, *(+)-t , yo t/s from (0,0) to (20,4) +(x(+), y(+)) = X(t) et from (ii), 20 S x d 8 = 2 2 t + at C 0 20 20 26 dt 25 26 25 Stat 0 20 . 26 25 0 2 $26 (400) = 40826 5x2 - b) the pasabouc Cusse, xat, y= 2t², from (0,0) to (1,2). dy da dt - 4t dt Jean) g4 : 51271043 from i xd8 t J 1 + 16t² at - lin) 0 dan

st 1 + 16t² at o Let 1+ 16t2 = 2 c) 327 dt = da =) t dt = (dz 32 17 & Vit 16t2 at √x dz 13:12) 17 3/2 312-1) 32 '16 ti (17 -1) 48 17 STY 48 is seanined solution: 48