![]() The vector field had a potential, which meant that the work done did not depend on the path traveled. In the exercise above, the path we took to get from one point to another did not matter. ![]() Denoteīy $\vc \, d\als$ as $d\lis$.\newcommand\) Find the work done by the force F ( x, y) x 2, x y in moving a particle from ( 1, 0) to ( 0, 1) along the unit circle. Example Okay, let’s look at an example and apply our steps to obtain our solution. Find the work done on the particle as it moves along the straight line from (2, 0, 0) to (2, 1. Integrate the work along the section of the path from t a to t b. The direction given by the velocity $\dllp'(t)$. Evaluate the line integral where C is the given curve. For example, when the movement is 90 degrees from theĭirection of the force, the magnetic field does no work.įirst, what is the direction of movement? It is The work isĬalculated from the component of the force in the direction of It is constrained to move along the slinky. We will see that the work done by a force moving a body along a path is naturally computed as a line integral. 3 Work done by a force along a curve Having seen that line integrals are not unpleasant to compute, we will now try to motivate our interest in doing so. However, the bead does not move in the direction of the magneticįield. This is exactly the same integral as in method (i). The direction of movement, calculating the work would be simple. Let d be the displacement vector from the initial to the final position. We'll ignore any effects of the magnetic force on the position of the bead.Īs you change $t$ to move the bead, work is done by the magnetic field on the bead. constant and the particle move along a straight line. 6.2.4 Describe the flux and circulation of a vector field. However, we'll assume the bead's position is constrained be at $\dllp(t)$. 6.2.3 Use a line integral to compute the work done in moving an object along a curve in a vector field. Since the bead is magnetized, the magnetic field exerts a force on theīead. The magnetic field does work on the bead as it moves along the curve. To compute actual work, we need to parameterize Cwith another parameter tvia a vector-valued function r(t). The green rectangle represents a large magnet, which induces the constant magnetic field represented by the vector field $\dlvf(x,y,z) = (-1/2, 0, 0)$ and illustrated with the green arrows. This line integral is beautiful in its simplicity, yet is not so useful in making actual computations (largely because the arc length parameter is so difficult to work with). For a given value of $t$ (changed by the cyan point on the slider), the red point represents a magnetic bead at point $\dllp(t)$. The blue helix is parametrized by $\dllp(t) = (\cos t, \sin t, t/(3\pi))$, for $0 \le t \le 6\pi$. Work done by a force on an object moving along a curve C is given by the line integral where is the vector force field acting on the object, is the unit tangent vector (Figure 1). Will induce a magnetic field $\dlvf(x,y,z)$, shown by the green ![]() over the curve C parametrized by r ( t) á cos ( t) ,sin ( t) ñ for t in 0, p. Thus, line integrals are evaluated by pulling back to an ordinary integral in the parameter used in parameterizing the curve. be represented in terms of line integrals in continuous and multi-dimensional cases, for example, s R vdt: Another example includes the formula for calculating the work done by the force F (possibly an electric or gravitational eld) in moving the particle along the curve C W Z C Fdr Example 3. Left of the slink, as shown by the large green square in the below That is, a line integral is an integral over a curve which has a well-defined orientation. ![]() Next, imagine that you put a large magnet to the To illustrate this concept, we return to theīy the function $\dllp(t) = (\cos t, \sin t, t/(3\pi))$, for $0 \le t \le 6\pi$,īead on your slinky (the bead has a small hole in it, so it can slideĪlong the slinky). Is the amount of work that a force field does on a particle as it One interpretation of the line integral of a vector field These vector-valued functions are the ones where the input and output dimensionsĪre the same, and we usually represent them as One can also integrate a certain type of vector-valued functions along A line integral (sometimes called a path integral) is the integral of some function along a curve.įunction along a curve, obtaining for example, Because the dot products in the definition of the line integral C F d r can each be viewed as the work done by F as an object moves along the (very small).
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