Math Problem Statement

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 (3, 2, 4).

Solution

To find the work done by the force field F(x,y,z)=(y+z,x+z,x+y)\mathbf{F}(x, y, z) = (y + z, x + z, x + y) on a particle moving along the line segment from (1,0,0)(1, 0, 0) to (3,2,4)(3, 2, 4), we use the formula:

W=CFdr,W = \int_C \mathbf{F} \cdot d\mathbf{r}, where CC is the path of the particle and drd\mathbf{r} is the differential displacement vector.


1. Parameterize the Path

The line segment from (1,0,0)(1, 0, 0) to (3,2,4)(3, 2, 4) can be parameterized as: r(t)=(1+2t,2t,4t),0t1.\mathbf{r}(t) = (1 + 2t, 2t, 4t), \quad 0 \leq t \leq 1.

The velocity vector is: r(t)=ddtr(t)=(2,2,4).\mathbf{r}'(t) = \frac{d}{dt}\mathbf{r}(t) = (2, 2, 4).


2. Force Field Along the Path

Substitute the parameterized path r(t)\mathbf{r}(t) into F\mathbf{F}: F(r(t))=(y+z,x+z,x+y).\mathbf{F}(\mathbf{r}(t)) = (y + z, x + z, x + y). Using r(t)=(1+2t,2t,4t)\mathbf{r}(t) = (1 + 2t, 2t, 4t): y+z=2t+4t=6t,x+z=(1+2t)+4t=1+6t,x+y=(1+2t)+2t=1+4t.y + z = 2t + 4t = 6t, \quad x + z = (1 + 2t) + 4t = 1 + 6t, \quad x + y = (1 + 2t) + 2t = 1 + 4t. Thus: F(r(t))=(6t,1+6t,1+4t).\mathbf{F}(\mathbf{r}(t)) = (6t, 1 + 6t, 1 + 4t).


3. Dot Product Fdr\mathbf{F} \cdot d\mathbf{r}

The differential displacement vector is: dr=r(t)dt=(2,2,4)dt.d\mathbf{r} = \mathbf{r}'(t) dt = (2, 2, 4) dt.

The dot product F(r(t))dr\mathbf{F}(\mathbf{r}(t)) \cdot d\mathbf{r} is: F(r(t))r(t)=(6t)(2)+(1+6t)(2)+(1+4t)(4).\mathbf{F}(\mathbf{r}(t)) \cdot \mathbf{r}'(t) = (6t)(2) + (1 + 6t)(2) + (1 + 4t)(4). Simplify: F(r(t))r(t)=12t+2+12t+4+16t=40t+6.\mathbf{F}(\mathbf{r}(t)) \cdot \mathbf{r}'(t) = 12t + 2 + 12t + 4 + 16t = 40t + 6.


4. Integrate Over tt

The work done is: W=01(40t+6)dt.W = \int_0^1 (40t + 6) \, dt.

Compute the integral: W=[20t2+6t]01.W = \left[ 20t^2 + 6t \right]_0^1. At t=1t = 1: 20(1)2+6(1)=20+6=26.20(1)^2 + 6(1) = 20 + 6 = 26. At t=0t = 0: 20(0)2+6(0)=0.20(0)^2 + 6(0) = 0. Thus: W=260=26.W = 26 - 0 = 26.


Final Answer:

The work done by the force field is: 26.\boxed{26}.

Would you like a breakdown of specific steps or clarification on any part of the solution?


Related Questions:

  1. How is the work formula CFdr\int_C \mathbf{F} \cdot d\mathbf{r} derived?
  2. What if the force field was conservative—how would the work calculation change?
  3. Can you explain why parameterizing the path is essential in such problems?
  4. How do you determine if a force field is path-independent?
  5. What is the geometric interpretation of the dot product in this context?

Tip:

Always confirm the parameterization of your path aligns with the problem's constraints—it ensures accurate integration.

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Math Problem Analysis

Mathematical Concepts

Vector Calculus
Line Integrals
Vector Fields
Dot Product

Formulas

Work = ∫C F · dr
Parameterized line segment r(t) = (1 + 2t, 2t, 4t)
Dot product of F(r(t)) and dr/dt

Theorems

Fundamental Theorem of Line Integrals

Suitable Grade Level

College level (Calculus III or Vector Calculus)