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3 f Cornpute - Simplily cosO and y rsino. 0r2Let f(x,y) and zuppoee V2+y answer as rnuch a2 possible: your...

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3 f Cornpute - Simplily cosO and y rsino. 0r2Let f(x,y) and zuppoee V2+y answer as rnuch a2 possible: your

3 f Cornpute - Simplily cosO and y rsino. 0r2 Let f(x,y) and zuppoee V2+y answer as rnuch a2 possible: your



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$3-10$ Determine whether or not $\mathbf{F}$ is a conservative vector field. If it is, find a function $f$ such that $\mathbf{F}=\nabla f .$

$\mathbf{F}(x, y)=\left(2 x y+y^{-2}\right) \mathbf{i}+\left(x^{2}-2 x y^{-3}\right) \mathbf{j}, \quad y>0$

Okay. So we'd like to determine whether or not this function, the specter field big F. Of X and Y is equal to. Yeah, that's why Plus Y squared I had plus X squared mhm. Plus two Xy J had is conservative or not. So we know that if this is conservative. Uh So if conservatives we are able to write um the specter of your breath as the gradients Some two variable function. Little F. Yeah. So that means that um we can use the property of the second derivatives in particular the property that for some function F. That's why That it's 2 2nd derivatives uh should be equal. So F X, Y and F. My ex should both be equal. So knowing assuming that Fsn vector field. Yeah. Hello, that describes ingredients. Mhm. We can assume that uh you can do this. So if F it's some vector gradient field, we know that physical a partial negative with respect to X. I had us. The partial derivative would talk to Y. J. Hi. And so then we can take each of these components of big tickets, other popular videos. So if we take um the partial derivative with respect to Y. Of the X. Component of big F. And we compare it to the partial the sex of the white compliance. We can see if they're equal and if they're not then you can say that this is not conservative. So let's look at what our two terms are. We know that apathetic uh Rx component of X is called to great. Uh X. Y plus Y squared. And we know that our white component of a physical too. Um X squared plus two. Xy. Now we can go ahead and take their partial. So if you take the posture from inspector why? Uh the X. Component we get simply X. And if we take um Yeah sorry. We're yet X. Plus to I. And if you take the um the other derivative, the pressure divert to expect the X. Of this black component. We did uh two X plus two Y. And therefore we see that the second that these derivatives are not equal. Since the party with respect to Y. Of the X component, it's not equal parts sugar. Video expect tax the Y component. We can say that this vector field F is not conservative and therefore there is no this is not the gradient of the function that we can solve for.

Were given a victor feel Big f and whereas to determine whether or not the gift is a conservative vector field and if it is, were asked to find a function little left such that Big F is equal to the radiant of little left. The function big F is three x squared minus two y squared I plus four x y plus three j. We have that the partial derivative of the X component three X squared minus two y squared with respect to Why is negative for why, while the partial derivative of the Y component for X Y plus three with respect to X is for why and so these partial derivatives air not equal. Therefore, it follows that our vector field the gaffe is not conservative.

Okay, so we're looking at the vector field F f x Y is equal to why squared minus two X. I had plus two X. Y. Yeah. J hash You want to see if this is a conservative vector field? So so we can do that pipe checking to see if the second derivatives of equal, because if this was a vector field that was a gradient field and this would be of the form uh huh. X derivative of X F times I had plus the wide review of that time she had. So if we go ahead and find The 2nd derivative which uh F X Y. The second derivative F I X after equal. This means that this is in fact a great field that we can solve for. Let's go ahead and do that. So we have um you can find that's sorry the partial with respect to Y. Yeah, of Y squared minus two X. And see if that's equal to the partial perspective tax of two x. Y. We'll see that. This is just equal to why and that this is also equal to Y. So that yes, this is a conservative vector field. Yeah. So basically now what we can do is we can take uh we can write this of the form if it's just a crazy an event. So like I said earlier this is just a derivative. Mhm. So we know that um DF over dx is just equal to um Y squared minus two X. Yeah, This is why I spared -2 x. And that partial F. Partial line is just equal to X. Y. So if we integrate both sides we'll see that um this side leaves us with um partial F is equal to y squared minus two X. Partial X. And then we can integrate to be left with F is equal to x. Y squared minus two Xy plus some uh G function of why? Because if we take the partial respect to X. Uh this is our integration constant. Our integration integration term. Okay so that's our first equation. If you do the second equation we'll see that partial F. Is equal to two X. Y. Sorry to Yeah two x. y. Partial. Why leaving us with integration? And so F. Is equal to um X. Y squared plus some H function will say H. H. Of X. And so we wanted to see how these are equal. Yeah and so that we can see that um dysfunction in this age of X must be equal to -2 x. by yeah and then plus some constant. And we'll see that G. Of Y. It's just the constant. So they are. Yeah but that's why I just equal to um X. Y. Sorry xy squared uh minus two Xy But um Constancy. Oh

So we have this vector field F. Of X. Y. It's equal to um I squared to the X. Y. I. Hat plus one plus x. wide you to the X. Fine shanghai. And so what we want to do here is show whether or not it's concerned it's a conservative vector field. And if so um find its function. So uh determined if F. Is conservative and then if so if so find F. Where. Mhm. This big F is just this gradient field of F. So the first thing you want to do is determine whether this vector field is conservative. And if we rewrite this as P I. Hat plus Q. J. Hat, we can find these derivatives. Um and see if actually ah partial P partial Y, secret of partial Q, partial X. And so partial P over partial Y. In this case it's just partial or partial Y. Of Y squared E. To the X. Y. So Y squared B. T. X. Y. And we'll see that we need to use the product rule in this case, I'm leaving us with two. I eat the X by plus. Um So we have to eat the X. Y. Since wise are are variable that we're integrating with respect to. This leaves us with uh xy squared E to the X. Y. And so now you want to find partial Q partial X. And that's just partial partial X. Of quantity one plus X. Y. You're right, a little bit neater. So one plus X Y meets the X. Y. Um and in this case we can use powerful, we can and to leave us with um we take the derivative of this parentheses term we get one plus Y. Um Sorry? We get Viet um why eat the X. Y. And then adding on the second part, we get one plus X Y times. Um We have in this case E T. X. Y. Is derivative index. Uh with respect to X. Is Y. E. To the X. Y. So this leaves us with um Y. E. To the X. Y. And then distributing gives us Y. E. To the X. Y plus x. Y square. That's why. And we can see that we have the same thing. So combining these like terms leaves us with two Y. E. To the X. Y plus X. Y square to eat to the X. Y. So we can see here that partial P partial wide and partial Q partial X are indeed equal. So this does mean F these conservatives because there is no issue with the domain. This is sufficient to say F is conservative. Yeah. So this means that we can write F as um this gradient function, this gradient field of dysfunction that left. And we want to say now that we know that this means that this is a partial F partial X. I had pledged partial or partial why she had. And if we do some comparisons, if we write these down, we have two equations. This is equal to let's go back to the top. Mhm. Uh Y squared to eat the Xy. So this is while squared E. To the X. Y. And then partial left partial Y is equal to one plus X. Y. It's the X. Y. Yeah. And we'll have to we'll just look here at R. two. Are you expressions? We'll see here. This is we can solve through normal integration and this part we have to use integration by parts. So let's go ahead and do this accident to go first. If we separate both sides will C. D. F is equal to Y squared T. Two X Y partial X. And if we interviewed both sides were left with physical too um noticed since X. Is the integration variable? Why is this constant? We know that this is going to equal Y squared becomes one of Hawaii's Xy or just uh Y. E to X. Y plus some integration constant. Um In this case it's a function of why? So it's G. F. Y. Since we're integrating respect to X. And so next we have um partial F over partial. Why is this? So you can multiply give us one plus X. Y. Eat the xy partial. Why? And if we integrate both sides we're going to have to use integration by parts in this case. Um So we can use um U. Is equal to one plus X. Y. And T. V. Is equal to we can replace the wise with partial or the Ds with partials later but we'll see that this is just um eats the X. Y. Um Given. And if we go ahead and differentiate we C. D. U. Is equal to in this case we're using why is our our integration variable? So we're going to differentiate with respect to Y. Leaving us with ex de vie. And then V. Is just equal to in this case we're using wiser integration variable. So this is one over um X. He's the Xy. And so we know this is just this is um in this case have is equal to U. Times V minus the integral of VD you. And so we see is just um one plus X. Y times one over X. E. To the X. Y minus the integral of X times one over X. So this is X over X. P. T. X. Y. And we can go ahead and cancel that too. Just leave us with the integral of E. To the X. Y. Um Dubai. And so we're going to bring back the car show in this case now we have F. Physical to this. So we've seen one over X. Plus. Um Why? And then multiply all that. It's the X. Y. And then if we go ahead and subtract we have this integral again which we actually evaluated earlier and it's just equal to one over X. Eat the X. Y. And so we can cancel this one and this one leaving us without physical to Y. E. To the X. Y. And we have to add on our integration term H. Fx some function H. Of X. And so we'll see that these are identical are too terms for our function of except we have G. Of Y. H. Of X. Which are clearly both chest. Uh Constance not any actual dependence on X and Y. So we can say that if Y is equal to H. Of X, just equal to the sum constancy leaving us finally, with our F. Of X. Y equal to Y E T. X. Y plus C.


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