How would you solve this vector force function calculus problem.?

Question

Let the vector force function F(x, y, z)= (2x - y) i + (2z) j + (y - z) k . Find the work done by the force F on an object moving along the straight line from the point (0, 0, 0) to the point (1, 1, 1) .

these could be possible solutions...


a. 3/2
b. 5/2
c. 4
d. 9/2
e. 11/2
f. 17/2
g. 27/2
h. 33/2
or it is none of these

Answer 1

In this instance, dr would be (√3/3 dx) i + (√3/3 dy) j + (√3/3 dz) k for all points along the line. The integral of F • dr in this case would therefore be

∫ (2x - y)(√3/3 dx) + (2z)(√3/3 dy) + (y - z)(√3/3 dz)
= √3/3( ∫ (2x - y)dx + ∫ 2z dy + ∫ (y - z) dz

Because everywhere along the line x = y = z, this becomes
√3/3( ∫ (2x - x)dx + ∫ 2y dy + ∫ (z - z) dz
= √3/3( ∫ x dx + ∫ 2y dy + ∫ 0 dz)
= √3/3 (x^2/2 + y^2 + 0)
= √3/3 ((1^2/2 - 0^2/2) + (1^2 - 0^2) + 0)
= √3/3 (3/2)
= √3/2
which is none of the given choices.

Answer 2

I have seen the solution presented above analyzing this as a right triangle configuration. I beg to disagree with the solution using the Pythagorean theorem as the outright procedure. This problem is not as simple as you may realize. Certain laws of physics need to be considered in the solution of this problem. The working equations for this are the following: Y = V(sin 45)T - (1/2)gT^2 and X = V(cos 45)T where Y = vertical distance of the bullet from the ground = 1000 ft (given) V = initial velocity at which bullet was fired T = travel time g = acceleration due to gravity = 32.2 ft/sec^2 (constant) X = horizontal distance travelled by the bullet At the onset, you have three unknowns ---- V (initial velocity), T = time for bullet to be 1000 feet above the ground and X = horizontal distance travelled by the bullet. AND you only have two equations. Do you really want to go this route?

Answer 3

The work is given by the dot product of the force and the vector
v=i+j+k
W = 2x-y +2z+y-z = 2x+z
My former answer is totally wrong
W = Int along the line x=y=z of F*ds where ds = dx*i+dy*j+dz*k
here dx=dy =dz and the work is
Int(0,1)( (2x-x)dx+2xdx+0dx =Int3xdx =3/2x^2 (0,1) = 3/2 (a)