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Dot Product - Calculus 3

  1. A constant force F⃗=⟨8,−6,4⟩\vec{F} = \langle 8, -6, 4 \rangle moves a particle along three consecutive displacement segments: d1⃗=⟨2,0,−1⟩\vec{d_1} = \langle 2, 0, -1 \rangle, d2⃗=⟨0,3,2⟩\vec{d_2} = \langle 0, 3, 2 \rangle, and d3⃗=⟨−1,1,0⟩\vec{d_3} = \langle -1, 1, 0 \rangle. If the total work done is WtotalW_{total} and the work done along the straight-line path from start to finish is WdirectW_{direct}, what is Wtotal−WdirectW_{total} - W_{direct}?
    Answer: 00 For a conservative force (constant force), work depends only on displacement, not path. The total displacement is dtotal⃗=d1⃗+d2⃗+d3⃗=⟨2,0,−1⟩+⟨0,3,2⟩+⟨−1,1,0⟩=⟨1,4,1⟩\vec{d_{total}} = \vec{d_1} + \vec{d_2} + \vec{d_3} = \langle 2, 0, -1 \rangle + \langle 0, 3, 2 \rangle + \langle -1, 1, 0 \rangle = \langle 1, 4, 1 \rangle. Both WtotalW_{total} and WdirectW_{direct} equal F⃗⋅dtotal⃗=8(1)+(−6)(4)+4(1)=8−24+4=−12\vec{F} \cdot \vec{d_{total}} = 8(1) + (-6)(4) + 4(1) = 8 - 24 + 4 = -12. Therefore Wtotal−Wdirect=0W_{total} - W_{direct} = 0.

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