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Line integral workdone12/24/2022 But instead of being limited to an interval,, along the x-axis, we can explain more general curves along any path in the plane.īy “straightening” out the path using parameterization and arc length. Interestingly, a line integral can help us map out where we benefited from the wind or the current and where those same forces hindered our progress.Ī line integral, called a curve integral or a path integral, is a generalized form of the basic integral we remember from calculus 1. The path of the wind or the flow of the current might make it easier in one direction than the other, right? The work, W, performed moving an object from xa to xb by a force F(x) may be attained by the following: Example: A spring exerts a restoring force of 0. To illustrate the idea, think about how it feels to run on a track on a windy day or to row a boat across a lake with a noticeable current? I believe my answer is thus the negative integral from 3pie/2 to pie/2 of -12cos^3t dt.Jenn, Founder Calcworkshop ®, 15+ Years Experience (Licensed & Certified Teacher) My integral is going in the clockwise direction instead of the counterclockwise, so I will place a negative in front of it. In the line integral of work done by conservative forces, should we care about the direction of force and differential displacement if we choose different coordinate system The formula of potential energy comes out to be different if we choose different coordinate system. I'm left with -12cos^3t if I did my math correctly. Lots of stuff cancels! I wound up with -18costsint + 18cos^2t + 18costsint - 18cos^2t - 12cos^3t. My integral will take the dot product of F(r(t)) and r'(t). I found that F(r(t)) is 3costi + (-6sint+6cost)j - 6cos^2tk. I also found F(r(t)) by sticking the components of r(t) into those of F. I proceeded to find the derivative of r(t) and found that r'(t) = (-6sint + 6cost)i -3costj + 2costk. 15.2.042 - Find work done by the force field on a. Thus I believe my lower limit will be 3 pie / 2. AKPotW: Using line integrals to find the work done by a force along a curve. It follows then that this also satisfies y = 3cost = 0 and x = 6cost + 6sint = -6. Thank you very much for the help! That really helps me out, and I appreciate it.įor the lower limit, I noticed that z = sint = -1 is satisfied by 3 pie / 2. I don't think that will be that difficult, especially if I can simplify some (for instance, I see I could simplify the i component of my hpothetical curve to 6(cost+sint)i.Īny assistance would be much appreciated! I know that once I get an r(t), I need to stick it into my function F, and then dot itself with the derivative of r(t). By doing so, the line integral becomes a measure of how much work the force field does upon the particle on its journey across the path. More than likely simple, but I'm stumped. That's where I come to an abrupt stop.because I need r(t) going from (6,0,2) to (-6,0,-2), and I'm not quite sure how to do this. answered by AmreshRoy (69.9k points) selected by Vikash Kumar. vector integration jee jee mains Share It On Facebook Twitter Email 1 Answer. I then stuck in the y and z components I had (3cost and 2 sint, respectively) and came up with this curve: (6cost+6sint)i + 3costj + 2sintk. Use the line integral, compute work done by a force vector F (2y + 3)i + xzj + (yz + x)k when it moves a particle from the point (0, 0, 0) to the point (2, 1, 1) along the curve x 2t 2, y t, z t 3. In the field of classical mechanics, line integrals are used to calculate the work done. In order to find a curve formed by the intersection, I used 3costj + 2sintk and then, for my i component, I noted that the plane equation gave me x = 2y + 3z. Line integrals can be used to find the three-dimensional surface areas. My main stumping point is coming up with the correct parameterization for r(t). I just need to set the integral up correctly, because I'm going to be letting the computer do the integration. I'm going to need to find the integral of F dr which is actually going to be the integral of Fr(t) dot with r'(t) dt. Once we are comfortable finding potentials, well show that the work done by such a vector field is the. Find the work done on it by the vector field F(x,y,z) = -yi + xj + yzk. This function is called a potential for a vector field. The particle moving along the curve goes from (6,0,2) to (-6,0,-2). A curve is formed by the intersection of y^2/9 + z^2/4 = 1 and the plane x-2y-3z = 0.
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