variable force given as F→=2yi^+7xj^ moves a particle along a parabolic given as y=3x2 from the origin (0, 0) to a point (3, 3). Calculate the work done.
To calculate the work done by the variable force, we need to integrate the dot product of the force and the differential displacement along the curve.
The differential displacement along the curve can be expressed as dr→ = dx i^ + dy j^.
For the parabolic curve y = 3x^2, we can solve for y in terms of x by substituting the given values of (3, 3) for x and y:
3 = 3(3^2)
3 = 27
This proves that the given point (3, 3) lies on the parabolic curve.
Now, let's calculate the differential displacement:
dx = dx i^
dy = d(3x^2) = 6x dx
dy = 6x dx j^
We can express the force F→ as F→ = 2y i^ + 7x j^.
Taking the dot product between F→ and dr→, we have:
F→ · dr→ = (2y i^ + 7x j^) · (dx i^ + 6x dx j^)
F→ · dr→ = 2y dx + 42x^2 dx
To calculate the work done, we integrate this expression over the limits of the curve.
The limits of the curve are from x = 0 to x = 3.
Work done = ∫[0 to 3] (F→ · dr→)
= ∫[0 to 3] (2y dx + 42x^2 dx)
= ∫[0 to 3] (2(3x^2) dx + 42x^2 dx)
= ∫[0 to 3] (6x^2 dx + 42x^2 dx)
= ∫[0 to 3] (48x^2 dx)
= 48∫[0 to 3] (x^2 dx)
= 48 [(1/3)x^3] [0 to 3]
= 48 [(1/3)(27) - (1/3)(0)]
= 48 (9 - 0)
= 432
Therefore, the work done by the variable force along the parabolic curve from the origin to the point (3, 3) is 432 units of work.