(1 point) suppose f⃗ (x, y)=−yi⃗ +xj⃗ and c is the line segment from point p=(2,0) to q=(0,5).

(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. r⃗ (t)= < 2-2t,5t>

(b) using the parametrization in part (a), the line integral of f⃗ along c is

r(t) = (2 - 2*t, 5*t)

∫c  F dr = 10

Step-by-step explanation:

Given

F(x,y) = (-y, x)

C is the line segment from point P=(2,0) to Q=(0,5).

we can say that

x(t) = 2 - 2*t

if t = 0  ⇒   x(0) = 2 - 2*0 = 2 = xp

if t = 1   ⇒   x(1) = 2 - 2*1 = 0 = xq

y (t) = 5*t

if t = 0  ⇒   y(0) = 5*0 = 0 = yp

if t = 1   ⇒   y(1) = 5*1 = 5 = yq

Now, we have the 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) = (2 - 2*t, 5*t)    where  0 ≤ t ≤ 1

F₁(x,y) = -y    ⇒  M = -5*t

dx = - 2*dt

F₂(x,y) = x     ⇒  N = 2 - 2*t

dy = 5*dt

then, we apply the formula

∫c  F dr = ∫c (M*dx + N*dy)

⇒ ∫ (-5*t)(- 2*dt) + (2 - 2*t)(5*dt) = (5)(2) ∫ (t + (1 - t)) dt = 10 ∫ dt = 10*t + C

If   0 ≤ t ≤ 1

we have

∫c  F dr = 10* ( 1 - 0) = 10

(5x-1)^2= (5x)^2-2(5x)×1+(1)^2=25x^2-10x+1