Evaluate the line integral ∫5ydx+2xdy where C is the straight line path from (2,4) to (5,9).
the line from (2,4) to (5,9) can be written as
x = 3t+2
y = 5t+4
So, using the path indicated, with 0 <= t <= 1
∫5ydx+2xdy
= ∫(5y dx/dt + 2x dy/dt) dt
= ∫5(5t+4)(3) + 2(3t+2)(5) dt
= ∫(105t+80) dt
= 52.5 t^2 + 80t [0,1]
= 132.5
To evaluate the line integral ∫5ydx+2xdy along the straight line path from (2,4) to (5,9), we can parameterize the path using a scalar parameter t.
Let's find the parametric equations for the path from (2,4) to (5,9).
The x-coordinate can be parametrized as:
x = 2 + (5-2)t = 2 + 3t
The y-coordinate can be parametrized as:
y = 4 + (9-4)t = 4 + 5t
Now, let's find the differentials dx and dy:
dx = d(2 + 3t) = 3dt
dy = d(4 + 5t) = 5dt
Substitute these values into the line integral expression:
∫5ydx+2xdy = ∫5(4 + 5t)(3dt) + 2(2 + 3t)(5dt)
Simplify the expression:
= ∫(20 + 25t)(3dt) + (4 + 6t)(10dt)
= ∫(60 + 75t)dt + ∫(40 + 60t)dt
= 60t + (75/2)t^2 + 40t + (60/2)t^2 + C
Evaluate it from t = 0 to t = 1:
= [60(1) + (75/2)(1)^2 + 40(1) + (60/2)(1)^2] - [60(0) + (75/2)(0)^2 + 40(0) + (60/2)(0)^2]
= 60 + 75/2 + 40 + 60/2 - 0 - 0 - 0 - 0
= 60 + 37.5 + 40 + 30
= 167.5
Therefore, the value of the line integral ∫5ydx+2xdy along the straight line path from (2,4) to (5,9) is 167.5.
To evaluate the line integral ∫5ydx+2xdy along the path C, we need to parameterize the path C.
The equation of the straight line passing through the points (2,4) and (5,9) can be represented as follows:
x = t, and y = 5t - 6, where t represents the parameter along the path C.
Now, we need to express dx and dy in terms of dt to rewrite the line integral in terms of dt:
dx = dt
dy = d(5t - 6) = 5dt
Substituting dx and dy in the line integral expression, we get:
∫5ydx + 2xdy = ∫5(5t - 6)dt + 2(t)5dt
= ∫25t - 30 dt + 10t dt
= ∫35t - 30 dt
Integrating with respect to t will yield:
∫35t - 30 dt = 17.5t^2 - 30t + C
Evaluating the integral along the path C, where t goes from the starting point to the end point:
t = 2 → (17.5(2)^2 - 30(2)) - (17.5(4)^2 - 30(4)) = -125
Therefore, the value of the line integral ∫5ydx + 2xdy along the path C is -125.