We first compute the work along the line segment Now we find the work when the object moves along the curve Example 5. The first two fields in the popup menu are conservative. A field is called conservative if only the starting and ending points matter in a conservative field the work done around a closed curve is zero. The value of the line integral is the sum of values of the field at all points on the curve, weighted by some scalar function on the curve (commonly arc length or, for a vector field, the scalar product of the vector field with a differential vector in the curve). Find the work done by the force field on an object moving from the origin to the point along the path where is the line segment is the curve Solution. The line integral of the vector field along the curve gives the work done by the field on an object moving along the curve through the field. The function to be integrated may be a scalar field or a vector field. The terms path integral, curve integral, and curvilinear integral are also used contour integral is used as well, although that is typically reserved for line integrals in the complex plane. I believe my answer is thus the negative integral from 3pie/2 to pie/2 of -12cos^3t dt.In mathematics, a line integral is an integral where the function to be integrated is evaluated along a curve. My integral is going in the clockwise direction instead of the counterclockwise, so I will place a negative in front of it. 15.2.042 - Find work done by the force field on a. I'm left with -12cos^3t if I did my math correctly. AKPotW: Using line integrals to find the work done by a force along a curve. Line integral helps to calculate the work done by a force on a moving object in a vector field. 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. Thus I believe my lower limit will be 3 pie / 2. 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. Using a line integral to find the work done by a vector field example Khan Academy. 15.2.042 - Find work done by the force field on a particle moving along C. 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! AKPotW: Using line integrals to find the work done by a force along a curve. Work done by force field in moving particle along given path. 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). Interpreting path independent line integrals in terms of work done. More than likely simple, but I'm stumped. Line integral workdone how to#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. Let's apply what we learned in the last video into a concrete example of the work done by a vector field on something going through some type of path. 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. 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. 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. Find the work done on it by the vector field F(x,y,z) = -yi + xj + yzk. 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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