Let f(x, y,z)=x+yz and let C be the line segment from P=(0,0,0) to (6,3,1). (a) Calculate f(c(t)) and ds=∥c′(t)∥dt for the parametrization c(t)=(6t,3t,1t) for 0≤t≤1. (b) Evaluate ∫Cf(x, y,z)ds.

step-by-step explanation:

place r′ so that qr′ is parallel to the tangent.

now we get that ∠r′py=∠pr′q, where y is on the tangent, on the opposite side of p from x.

also, because of the symmetry, ∠r′py=∠qpx.

finally, note that no matter where we place r, as long as it's on the correct arc, we have ∠prq=∠pr′q.

stringing all these equalities together, we end up with ∠prq=∠qpx.

this picture will show how its going to look like.

hope this by jack monhen