Find the line integral with respect to arc length integral_C (5x +9y)ds, where C is the line segment in the xy-plane with endpoints P = (2,0) and Q = (0,6). (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. r(t) = (b) Using the parametrization in part (a), the line integral with respect to arc length is integral_C (5x + 9y)ds = integral_a^b dt with limits of integration a = and b = (c) Evaluate the line integral with respect to arc length in part (b). integral_C (5x + 9y)ds =

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

answer:

yes

explanation:

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