10_ If F: R' _ R} is a diffeomorphism and M is a surface in R' prove that the image F(M) is also a surface in R?. (Hint: If x is a patch in M then the com...

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

$\mathbf{F}(x, y)=e^{x} \sin y \mathbf{i}+e^{x} \cos y \mathbf{j}$

Were given a vector field big F, and were asked to determine whether or not this is a conservative vector field If it is, were asked to find a function little left such that big F is equal to the ingredient of little left big f is the function e to the X cosine y I plus e to the x sign Why j So we have that the partial derivative of the X component e to the x co sign Why, with respect to why is negative e to the X sign Why you notice that this is the same as the partial derivative of the y component of big If eat the X sign Why? Oh, sorry. This is not equal to the partial derivative of the y component e to the X sign y With respect to X, this is simply eat the X sign y. So we have that these partial derivatives are not equal. Therefore it follows that our vector field big f is not conservative and so no potential function for they give

Okay. So we want to go ahead and analyze this vector field F of X. Y is equal to why eat it. X. I had plus E to the X plus E. To the Y. Y. Had jihad. So we know that if we call this key and if we call this Q. This vector field F is conservative. Yes. Mhm. If um partial was P with respect to lie is equal to the partial of key would respect that. And this also implies that F is difficult to some gradient of other function. Little F. And so if we go ahead and take these derivatives, if we do de do I of P, it's the same as D. D. Y of Y to the X. And in this case since why is the variable X. Dx is just constant. So this is just a linear term leaving us simply with E to the X. If you go ahead and do partial or partial execute you, that's just a partial partial X. Of E to the X plus Y. And in this case, why is our constant? So this is simply just a constant term goes zero and it's the ex derivatives. Just pcs. So in this case, yes, it's this deep. You it's the dP over D. Y is equal to the partial key over partial X. So a big F his conservative actually. And and in this case we know that there is some function that this is a gradient of. So we go ahead and look at it. We have we can say that since f uh f of X Y is equal to the gradient of some function F. We know that it's must equal the X derivative and I had direction and the wide derivative J had direction. So in this case we know that um F this is just equal to why eats the X. I had. Let's get to the X. Let's eat them. All right. J hat. In this case we know um that's derivative is equal to Y. E. To the X. And then the wider groups F. Y. Is just equal to see X plus two Y. So we can go ahead and integrate both of these functions with in this case respect to X and expect to lie. And this will leave us with the following. So if we say that this is partial F partial X is equal to like the X. We can separate this. Yeah. And then integrate both sides. Leaving us simply with F is equal to um derivative of disrespect to X. We know why it's constant. So this is just Y. E. To the X. Since it's just sort of a constant times GTX and then plus some G. Function of why? Since um instead of to have an integration council we have this integration function. And then on the other side we have partial partial Y. Is E. To the X plus two. Y. And if we separate this we get the following, we can integrate leaving us with F. Is equal to and then E. T. X. The constant. So we just get Y to the X. Let me go ahead and not keep parentheses if Y. E. To the X. And then we know that he to the Y is simply P. Two, Y plus some age function of X. And so now we want to compare these two and we can see that this G. M. Y. It's just this E. To the Y. Here. So we know that um comparing these they must be what we know by each the X plus G of Y is equal to Y. E T X plus Y plus some function H of X. And since there's no other ex uh ex functions we know this is zero and that this G o Y must be equal to E to the Y. And then always we have to include the constant term C. Because in any case either it's a function of X and Y. This constant term is still here, so, and finally we know that this gradient, uh this vector field F is a function, is the uh the gradient of this function, Little F, which in this case is just equal to Y. E to the X. You go ahead and get rid of crisis. He actually, this is just why into the X plus you to the by plus some constant C.

Were given a vector field biggest and were asked to determine whether or not Big F is a conservative vector field, if it is, were asked to find a scaler function. Little left such that Big F is equal to the ingredients of little F Big F is the Vector field X Y hyperbolic cosine of X y, plus the hyperbolic sign of X Y I plus X Square hyperbolic co sin of X times Y j. Well, we have the partial derivative of the X component with respect to why we have that this is equal to using the product rule. Mhm. We have X times. Why times X so X squared. Why times hyperbolic sign? Because this is hyperbolic and not regular. Tricking a metric of x y plus x times hyperbolic cosine of X y plus hyperbolic co sign of X y times X or X times. The hyperbolic co sign. That's why, which, of course, could be written as X squared. Why hyperbolic sign of X Y, plus two times x times the hyperbolic co sign of X Y, and we have the partial derivative of the Y component X squared, hyperbolic co sign of X Y yeah. With respect to X again by the product rule, This is two X times the hyperbolic co sign of X Y, plus X squared. Why times a hyperbolic sign of X y, which is the same as the other partial derivative was. We also have that the vector field f has a domain, which is all of our two, which of course we know is open and simply connected. Therefore, by a fear, um, from this section, we have that big F is a conservative vector field. So by another theory, um, there exists a scaler function little f such that big death is equal to the Grady int of little F. This implies that the partial derivative of little Left with respect to X is the X component of big F x y hyperbolic cosine, X y plus hyperbolic sign of X y and the partial derivative of little left with respect to why is the y component of big F, which is X squared, hyperbolic cosine of X y now taking the anti derivative of the top equation. With respect to X, we have that f x Y is equal to here. I'm going to use a little bit of intuitive thinking, but you might guests using the product rule that this would be something like X times the hyperbolic sign of X y. So if we where to do this in our head here, taking the derivative with respect to X product rule tells us this would be x times y times, the hyperbolic co sign of X Y plus and then one times hyperbolic sign of X y, which is what we have appear. Plus, we also need to account for some function g of why which may disappear when we take the derivative and taking the derivative with respect to why we have that F y is equal to x squared, hyperbolic co sign of X y plus g prime of y. Comparing this to the other expression for F y. This is equal to X squared, hyperbolic co sign of X Y, and solving we have the G prime of y is equal to zero or at G of y itself is simply some constant. In putting this together, we have that our potential function Little F is equal to X times the hyperbolic sign of X y, plus K

Were given a vector field big s and were asked to determine whether or not the gift is a conservative vector field if it is whereas to find the function little left such that big F is three ingredient of little left vector field. Big F is why Times e the X plus sign y I plus e to the X plus x cosine wine j. So we have the partial derivative of the X component. Why times e X plus sign. Why, with respect to why this is e to the X plus co sign of why and likewise we have the partial derivative of R Y component e to the X plus x cosine y with respect to X is also eat the X plus cosign. Why notice also that our vector field f has a domain which is our to which, of course, we know are too is open and simply connected therefore, by a theorem from this section we have that the vector field f is conservative therefore, by another, the're um we had that there exists a scaler function little left such that big af is equal to the ingredients of little left. This implies that the partial derivative of Little Left with respect to X is equal to the X component of big F. Why e to the X Plus sign? Why and the partial derivative of F with respect to why is the y component of big F E to the X plus x Kassian? Why this first equation implies that f of X y taking the anti derivative with respect to X is equal to y times e to the X plus X times sign of why, plus some function of why call it G and taking the derivative with respect to why we have that partial derivative of little left with respect to Why is each of the X plus x Co sign of why plus g prime of why? Comparing this to our other expression for the partial derivative of little left with respect to why this is equal to eat to the X plus x co sign why and therefore we have the G prime of y is equal to zero or, in other words, that g of y. Taking the anti derivative is some constant K putting this all together we have that are function. Little left is equal to why times e to the X plus x times sign of why plus okay