Find the work done by the force field
F(x, y, z) = y + z, x + z, x + y
on a particle that moves along the line segment from (1, 0, 0) to (5, 2, 3).
To find the work done by a force field on a particle moving along a line segment, we use the line integral. The formula for the line integral of a force field F along a curve C is given by:
∫ F · dr
where F is the force field and dr represents the differential displacement vector along the curve C.
In this case, we need to find the work done by the force field F(x, y, z) = y + z, x + z, x + y on a particle that moves along the line segment from (1, 0, 0) to (5, 2, 3).
Step 1: Parametrize the line segment
To compute the line integral, we need to parametrize the line segment. Let's denote the starting point (1, 0, 0) as A and the ending point (5, 2, 3) as B.
We can define a parameter t to represent the position of the particle along the line segment. For example, when t = 0, the particle is at point A, and when t = 1, the particle is at point B.
Let's find the parametric equations for the line segment AB:
x = 1 + (5 - 1) t
y = 0 + (2 - 0) t
z = 0 + (3 - 0) t
Simplifying these equations gives:
x = 1 + 4t
y = 2t
z = 3t
Step 2: Compute the differential displacement vector dr
To compute the differential displacement vector dr, we take the derivative of the parametric equations with respect to t.
dx/dt = 4
dy/dt = 2
dz/dt = 3
So, the differential displacement vector dr is:
dr = dx i + dy j + dz k = 4 dt i + 2 dt j + 3 dt k
Step 3: Compute the dot product F · dr
Substitute the parametric equations into the force field F(x, y, z) = y + z, x + z, x + y and evaluate the dot product F · dr.
F(x(t), y(t), z(t)) = (2t + 3t) i + (1 + 3t) j + (1 + 2t) k
= (5t) i + (1 + 3t) j + (1 + 2t) k
Now, compute the dot product:
F · dr = (5t)(4 dt) + (1 + 3t)(2 dt) + (1 + 2t)(3 dt)
= 20t dt + 2(1 + 3t) dt + 3(1 + 2t) dt
= (20t + 2 + 6t + 3 + 6t) dt
= (32t + 5) dt
Step 4: Integrate the dot product over the parameter range
To find the work done, we need to integrate the dot product F · dr along the curve from t = 0 to t = 1.
The line integral becomes:
∫ F · dr = ∫ (32t + 5) dt
Integrating the expression, we get:
= 16t^2 + 5t
Now, calculate the work done by evaluating the integrand at the limits of integration:
Work = ∫ F · dr from 0 to 1
= (16(1)^2 + 5(1)) - (16(0)^2 + 5(0))
= 16 + 5
= 21
Therefore, the work done by the force field on the particle moving along the line segment from (1, 0, 0) to (5, 2, 3) is 21 units of work.