Answer:
Explanation:
Given the integral
∫ (5x + 9y) ds
C is the line segment in xy plane with the end points
P = (2,0)
Q = (0, 6)
Let compute the direction of the vector
d = (a, b) = Q - P
(a, b) = (0, 6) - (2,0)
(a, b) = (-2, 6).
A. Now, the equation of line passing through point P(2,0) and Q(0,6) is and have direction d(-2,6) is written as
(x - 2) / -2 = (y - 0) / 6 = t
(x - 2) / -2 = t
x - 2 = -2t
x = 2 - 2t
Or
(y - 0) / 6 = t
y = 6t
Then, the parametric equations of the curve C is
r(t) = (2 - 2t, 6t)
Given that t ranges from t =0 to t = 1
f(x, y) = 5x + 9y
Let compute F(r(t))
From r(t), x = 2-2t and y = 6t
f(r(t)) = 5(2-2t) + 9(6t)
f(r(t) = 10 - 10t + 54t.
f(r(t)) = 10 - 44t.
Thus, the line integral becomes
∫ (5x+9y)ds = ∫f(r(t))•|r'(t)| dt t=0 to 1
Let find |r'(t)|
r(t) = (2 - 2t, 6t)
r'(t) = (-2, 6)
|r'(t)| = √(-2)²+6²
|r'(t)| =√40
So,
∫ (5x+9y)ds = ∫f(r(t))•|r'(t)|
∫f(r(t))•|r'(t)| = ∫(10 - 44t)•√40 dt
∫f(r(t))•|r'(t)| = √40 ∫(10 - 44t) dt
∫f(r(t))•|r'(t)| = √40 (10t - 44t²/2)
∫f(r(t))•|r'(t)| = √40 (10t - 22t²)
t ranges from 0 to 1
Then,
∫f(r(t))•|r'(t)| = √40 (10(1) - 22(1²) - 0 - 0)
∫f(r(t))•|r'(t)| = √40 (10 - 22)
∫f(r(t))•|r'(t)| = √40 × -12
∫f(r(t))•|r'(t)| = -12√40
∫f(r(t))•|r'(t)| = -75.89
So, the line integral of (5x+9y)ds is -75.89
