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The process of finding a potential function of a conservative vector field is a multi-step procedure that includes both integration and discriminant, while also paying attention to the variables you are integrating or discriminating against. For this reason, for the $dlvf$ vector field, we recommend that you first determine that the $dlvf$ is indeed preserved before starting this procedure. That way you know a potential function exists, so the process resolves eventually. In this page, we focus on finding a potential function of a two-dimensional conservative vector field. We deal with three-dimensional fields in another page. Read: what is a potential function We introduce the process of finding a potential function through an example. Let’s use the vector field begin {align *} dlvf(x, y) = (y cos x + y^2, sin x + 2xy-2y). end {align *} The first step is to check if $dlvf$ is conservative. Since the beginning {align *} pdiff {dlvfc_2} {x} & = pdiff {} {x} (sin x + 2xy-2y) = cos x + 2y pdiff {dlvfc_1} {y} & = pdiff {} {y } (y cos x + y ^ 2) = cos x + 2y, end {align *} we conclude that the scalar curvature of $dlvf$ is zero, because begin {align *} pdiff {dlvfc_2} {x} – pdiff {dlvfc_1 } {y} = 0. end {align *} Next we see that $dlvf$ is defined over all $R^2$, so no tricks to worry about. The vector field $dlvf$ is actually conservative Read more: What is the employer name Since $dlvf$ is conservative, we know that there exists some potential function $f$ so that $nabla f= dlvf $. The first step to find $f$, we notice that the condition $nabla f = dlvf$ means starting {align *} left (pdiff {f} {x}, pdiff {f} {y} right) & = ( dlvfc_1, dlvfc_2) & = (y cos x + y ^ 2, sin x + 2xy-2y). end {align *} This vector equation is two scalar equations, one component each. We need to find a function $f (x, y) $ that satisfies two conditions begin {align} pdiff {f} {x} (x, y) = y cos x + y ^ 2 label {cond1} end {align } and start {align} pdiff {f} {y} (x, y) = sin x + 2xy -2y. label {cond2} end {align} Let’s do these conditions in turn and see if we can find $f(x, y)$ satisfying both of them. (We know this can happen because $dlvf$ is conservative. If $dlvf$ depends on the path, the process will then throw an error somewhere.) Let’s start with the condition eqref {cond1} . We can use the equation begin {align *} pdiff {f} {x} (x, y) = y cos x + y^2, end {align *} and treat $y$ as if it were a number . In other words, we assume that the equation is begin {align *} diff {f} {x} (x) = a cos x + a ^ 2 end {align *} for some $a$. We can integrate the equation with $ x $ and get begin { align * } f (x) = a sin x + a ^ 2x + C. end { align * } But, then we have to remember that $ a $ is actually $y $ variable to start {align *} f(x, y) = y sin x + y^2x + C. end {align *} But actually, that’s not true either. Since we’re looking at $y$ as a constant, the “constant” integration $C$ can be a function of $y$ and it won’t make a difference. The partial derivative of any function of $y$ with respect to $x$ is zero. We can replace $C$ with any function of $y$, assuming $g(y)$ and the condition eqref{cond1} will be satisfied. A new expression for the potential function is begin {align} f(x, y) = y sin x + y ^ 2x + g(y). label {betweenep} end {align} If you are still in doubt, try taking the partial derivative with respect to $x of $f(x, y)$ as determined by the equation eqref {midep}. Since $g(y)$ does not depend on $x$, we can conclude that $displaystyle pdiff {} {x} g(y) = 0$. Indeed, the condition eqref{cond1} is satisfied for f(x, y)$ of the equation eqref {midep}. Now, we need to satisfy the condition eqref {cond2}. We can take f(x, y)$ of the equation eqref{midep} (so we know the condition eqref{cond1} will be satisfied) and take its partial derivative with respect to $y$ , get begin {align *} pdiff {f} {y} (x, y) & = pdiff {} {y} left (y sin x + y ^ 2x + g (y) right) & = sin x + 2yx + diff {g} {y } (y). end {align *} Compare this with the eqref {cond2} condition, we’re in luck. We can easily make this $f(x, y)$ satisfy the condition eqref {cond2} as long as begin {align *} diff {g} {y} (y) = – 2y. end {align *} If the vector field $dlvf$ depends on the path, we will find that both the eqref {cond1} condition and the eqref {cond2} condition cannot be satisfied. We’ll run into trouble at this point, because we find that $diff{g}{y}$ will have to be a function of $x as well as $y. Since $diff{g}{y}$ is a function of its own $y$, our calculation verifies that $dlvf$ is conservative. align *} with some constant $k$, then start {align *} pdiff {f} {y} (x, y) = sin x + 2yx -2y, end {align *} and we are done satisfy both conditions. definition of $g(y)$ with the equation eqref {midep}, we conclude that the function begin {align *} f (x, y) = ysin x + y ^ 2x -y ^ 2 + k end { align *} is a potential function for $dlvf. $ You can verify that indeed begin {align *} nabla f = (ycos x + y^2, sin x + 2xy -2y) = dlvf(x, y). end {align *} Read more: Kinchay: English-Tagalog Translation of “Kinchay” | Top Q & AW With this in hand, integrating begin {align *} dlint end {align *} is simple, no matter what path $dlc$ is. We can apply the gradient theorem to conclude that the integral is simply $f(vc{q}) – f(vc{p})$, where $vc{p}$ is the starting point and $vc {q} $ is the end point of $dlc $. (For this reason, if $dlc $ is a closed curve, the integral is zero.) We can come up with a problem such as find begin {align *} dlint end {align *} where $ dlc $ is the curve given by the following graph.The simple answer is to start {align *} dlint &= f(pi/2, -1) – f(-pi, 2) &= – sin pi/2 + frac {pi} {2} -1 + k – (2 sin (-pi) – 4pi -4 + k) & = – sin pi / 2 + frac {9pi} {2} + 3 = frac {9pi} {2} +2 end {align *} (The constant $k$ is always guaranteed to cancel, so you can just set $k = 0$.) If the $dlc$ curve is complex, one would expect that $dlvf$ to be just right. You should check if $dlvf$ is conservative before calculating its line integral begin {align *} dlint. end {align *} You can save yourself a lot of work. Read more: Square root of 244 | Top Q&A

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