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I do know how to take the dot product but if its a simple matter with y=6x^2 then I must be missing something here. do you take the derivative of y and put it in dy and then solve for x and do the same thing? and then the dot product?Charles Link said:## dr=\hat{i} dx+\hat{j} dy ## . With ## y=6x^2 ## it should be a simple matter to evaluate ## \int_C F \cdot dr ## over the path. Do you know how to take a dot product?(This one is in two dimensions instead of three).
alright i got it to work, its 1179/2.Charles Link said:## \int_C F \cdot dr=\int 9xy \, dx+\int 8y^2 \, dy ##. It doesn't get much easier. You substitute in ## y=6x^2 ## in the first integral. In both integrals, be sure and use the x-limits for the x-integral, and the y-limits for the y.
And you'd get a different answer so its path-dependent. so we could actually use this information for BCharles Link said:Additional problem: You can see that the line ## y=6x ## also connects the two endpoints. Try computing ## \int F \cdot dr ## over this straight line path between the same two points. Do you get the same answer? ## \\ ## If not, the integral is clearly path dependent. And if the answer is yes, the integral could still be path dependent for some other paths. The way to prove path dependence or independence is to look at ## \nabla \times F ##.
That was a logical step, because it then made the x integral completely a function of ## x ##. ## \\ ## You could also have substituted that into the ## y ## integral, (## y=6x^2 ## ==>> ## dy=12x \, dx ##), and then solved the y-integral as a dx integral, using x-limits. That would have made extra unnecessary work though.jonathanm111 said:last quick question: how did you know to substitute the y into the x-integral
A non conservative vector field is a vector field in which the line integral of its vector components depends on the path taken, rather than just the initial and final points. This means that the total work done by the vector field on an object moving along a closed path is not zero.
A conservative vector field is one in which the line integral of its vector components is path-independent, meaning that the work done by the vector field on an object moving along a closed path is always zero. This is not the case for a non conservative vector field.
Non conservative vector fields can represent physical systems that involve energy loss or gain, such as friction or a force that is not conservative. They are also important in understanding fluid dynamics and electromagnetism.
A non conservative vector field is represented by a vector function, where each component of the vector depends on the position in space. In other words, it is a function of x, y, and z coordinates.
One way to determine if a vector field is non conservative is to calculate the line integral along two different closed paths and compare the results. If the values are not equal, the vector field is non conservative.