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# Find the work required to move an object in the force field F = ex+y &lt;1,1,z&gt; along the straight line from A(0,0,0) to B(-1,2,-5). Also, deternine if the force is conservative.Find the work required to move an object in the fo

Accepted Solution

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Answer:Work = e+24F is not conservative.Step-by-step explanation:To find the work required to move an object in the force field   $$\large F(x,y,z)=(e^{x+y},e^{x+y},ze^{x+y})$$ along the straight line from A(0,0,0) to B(-1,2,-5), we have to parameterize this segment. Given two points P, Q in any euclidean space, you can always parameterize the segment of line that goes from P to Q with r(t) = tQ + (1-t)P with 0 ≤ t ≤ 1 so   r(t) = t(-1,2,-5) + (1-t)(0,0,0) = (-t, 2t, -5t)  with 0≤ t ≤ 1 is a parameterization of the segment. the work W required to move an object in the force field F along the straight line from A to B is the line integral $$\large W=\int_{C}Fdr$$ where C is the segment that goes from A to B. $$\large \int_{C}Fdr =\int_{0}^{1}F(r(t))\circ r'(t)dt=\int_{0}^{1}F(-t,2t,-5t)\circ (-1,2,-5)dt=\\\\=\int_{0}^{1}(e^t,e^t,-5te^t)\circ (-1,2,-5)dt=\int_{0}^{1}(-e^t+2e^t+25te^t)dt=\\\\\int_{0}^{1}e^tdt-25\int_{0}^{1}te^tdt=(e-1)+25\int_{0}^{1}te^tdt$$ Integrating by parts the last integral: $$\large \int_{0}^{1}te^tdt=e-\int_{0}^{1}e^tdt=e-(e-1)=1$$ and   $$\large \boxed{W=\int_{C}Fdr=e+24}$$ To show that F is not conservative, we could find another path D from A to B such that the work to move the particle from A to B along D is different to e+24 Now, let D be the path consisting on the segment that goes from A to (1,0,0) and then the segment from (1,0,0) to B. The segment that goes from A to (1,0,0) can be parameterized as  r(t) = (t,0,0) with 0≤ t ≤ 1so the work required to move the particle from A to (1,0,0) is $$\large \int_{0}^{1}(e^t,e^t,0)\circ (1,0,0)dt =\int_{0}^{1}e^tdt=e-1$$ The segment that goes from (1,0,0) to B can be parameterized as  r(t) = (1-2t,2t,-5t) with 0≤ t ≤ 1so the work required to move the particle from (1,0,0) to B is $$\large \int_{0}^{1}(e,e,-5et)\circ (-2,2,-5)dt =25e\int_{0}^{1}tdt=\frac{25e}{2}$$ Hence, the work required to move the particle from A to B along D is  e - 1 + (25e)/2 = (27e)/2 -1 since this result differs from e+24, the force field F is not conservative.