Suppose F⃗ (x, y)=(x+6)i⃗ +(6y+3)j⃗ . Use the fundamental theorem of line integrals to calculate the following. (a) The line integral of F⃗ along the line segment C from the point P=(1,0) to the point Q=(3,3).

52 units

Step-by-step explanation:

We require ∫꜀ F.Tds along the line segment given

We obtain the equation of the line in parameterized vector form

r(t) = (1 - t) (1,0) + t (3,3) (where 0≤t≤1)

r(t) = (1 - t, 0) + (3t, 3t)

r(t) = (1 + 2t, 3t) = (1 + 2t)î + 3tj

x = (1 + 2t) and y = 3t

∫꜀ F.Tds = ∫¹₀ F(r(t)).r'(t) dt

r'(t) = 2î + 3j

F(x,y) = (x+6)î +(6y+3)j

x = (1 + 2t) and y = 3t

F(r(t)) = (1 + 2t + 6)î + (6(3t) + 3)j

F(r(t)) = (2t + 7)î + (18t + 3)j

∫꜀ F.Tds = ∫¹₀ F(r(t)).r'(t) dt = ∫¹₀ [(2t + 7)î + (18t + 3)j] . [2î + 3j] dt = ∫¹₀ [(4t + 14) + (54t + 9)] dt

∫꜀ F.ds = ∫¹₀ (58t + 23) dt = [29t² + 23t]¹₀ = (29 + 23) = 52 units.

i'm assuming it's looking for the missing length, which should be 4cm

hope this !

ur mom

step-by-step explanation: