Find the line integral with respect to arc length (5x + 3y)ds, where C is the line segment in the xy-plane with endpoints P = (5, 0) and Q = (0,4).
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Given integral
where C is the line segment in x-y plane with the end points P(5,0) and Q(0,4)
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- Find the line integral with respect to arc length ∫C(6x+4y)ds, where C is the line segment in the xy-plane with endpoints P=(8,0) and Q=(0,6). (a) Find a vector parametric equation r⃗ (t) for the line segment C so that points P and Q correspond to t=0 and t=1, respectively. (b) Using the parametrization in part (a), the line integral with respect to arc length is ∫C(6x+4y)ds=∫ba ______dt, with limits of integration a= and b= (c) Evaluate the line integral with respect to arc length in part (b). ∫C(6x+4y)ds= ∫C(6x+4y)ds=∫baThis is a three part problem. Given: Find the line integral with respect to arc length (6x + 8y)ds, where C is the line segment in the xy-plane with endpoints P = (8,0) and Q = (0,9) A. Find a vector parametric equation r(t) for the line segment C so that points P and Q correspond to t = 0 and t = 1, respectively. B. Using the parametrization in part A, what is the line integral with respect to arc length? C. Evaluate the line integral with respect to arc length in part B.Show that the curve = Vti + vt + (2t - 1) k is tangent to the surface x² + y2 -z = 1 when t = 1
- Compute the line integral∫C [2x3y2 dx + x4y dy]where C is the path that travels first from (1, 0) to (0, 1) along the partof the circle x2 + y2 = 1 that lies in the first quadrant and then from(0, 1) to (−1, 0) along the line segment that connects the two points.Evaluate the line integral f(x2+2y2)ds for C: the line segment from (1,2) to (4,6)Find a potential function F for (vector)F = < (1+xy)exy, x2exy >, then evalute the integral (vector )F x dr a long the curve C parametrized by r(t) = < cos t , 2 sin t>