Question

Find the work done by the force field \(\mathbf{F}\) on a particle moving along the given path. \(\mathbf{F}(x, y, z)=y z \mathbf{i}+x z \mathbf{j}+x y \mathbf{k}\) \(C:\) line from (0,0,0) to (5,3,2)

Short answer

The work done by the force field on the particle along the given path is 45.

Step-by-step solution

Parameterize the line segment

The line can be parameterized as \(r(t) = \langle 5t, 3t, 2t \rangle\) for \(0 \leq t \leq 1\).

Compute the derivative of r(t)

The derivative \(dr(t)/dt\) of \(r(t)\) is \(\langle 5, 3, 2 \rangle\).

Substitute r(t) into force field

Substitute the parameterization r(t) into the force field F to get \(F(r(t)) = \langle 5t*3t, 5t*2t, 5t*3t \rangle = \langle 15t^2, 10t^2, 15t^2 \rangle\).

Calculate the dot product of F(r(t)) and dr(t)/dt

The dot product \(\mathbf{F}(r(t)) \cdot dr(t)/dt\) is \(75t^2 + 30t^2 + 30t^2 = 135t^2\).

Calculate the line integral

The work done by the force field is calculated as the line integral of \(\mathbf{F} \cdot dr\), which is equivalent to the integral of \(\mathbf{F}(r(t)) \cdot dr(t)/dt\) with respect to \(t\) from \(0\) to \(1\). Hence the work is \( \int_{0}^{1} 135t^2 dt\).

Evaluate the integral

By evaluating the integral, we get \( \int_{0}^{1} 135t^2 dt = [45t^3]_0^1 = 45\).