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Lisa, welcome back.



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I hope you have solved some of these tasks, let us come to the solutions to our exercises in differential



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calculus.



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So the first two tasks were about calculating the gradient of a scalar function and about calculating



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the curl, the curl and the divergence of a vector function.



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And then in task number three and four, you should show these two identities that the curl of the gradient



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of any scareder function is zero and that the divergence of the curve and the vector function is zero.



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So let's get started with task number one.



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So here you can see we have a function F of our and we should calculate the gradient.



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So R corresponds to X, Y and Z.



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As you can see, I have already highlighted here the X, we have another X here and we have a Y square.



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And of course as to the power of three.



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What we have to calculate now is to calculate, first of all, the partial derivatives.



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Let's go ahead to we have the partial derivative with respect to X of X of F, F of our.



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And here this is already the most difficult task because we have X appearing twice in the product.



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And this means we have to apply the product rule.



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So we calculate the function, the first function, which is X times, the derivative of the second



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function, and this is times minus one for this one here and then our exponential function.



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OK, and then the other term here, we have to differentiate the first function, which is one, and



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then we multiply by E to the power, as you can see here now for the Y derivative, we only get one



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term because here of a Y only appears one.



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And so we just get, um, we had our function basically because it's an exponential function.



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And then we have to account for the inner derivative, which is the derivative of minus Y square, which



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is minus to Y and then for the Z derivative.



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We get like here, the function



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minus C to the power of three.



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And then again, the inner derivative, which in this case is minus three C Square.



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OK, and now we can write down the gradient, so, you know, the gradient of F is, of course, the



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vector of the partial derivatives.



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F f f and then we can just write it down, and as you may notice, we have this X times E to do something



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in every single term so we can write down X times E to the minus X minus Y square minus C to the power



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of three.



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And now the only difference for the vectors is that here in the first term we get minus um.



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Sorry, we get.



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OK, yes, small mistake, OK?



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So we cannot pull out this one here, we get minus X.



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And then plus one, and for the Y component, we get mine as to why X and for the Z component, we get



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minus three C, square, X and yeah, that's that's basically all you have to do.



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And yeah, this is where we are finished.



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OK, so now let's go ahead and calculate the divergence and the curl of such a vector function.



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So here we have the position vector R which is X, Y, Z, divided by the absolute value of R to the



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power of three, which is one of a square root to the power of three.



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So first of all, I would say it's good to calculate all of the partial derivatives so that we have



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it easy to calculate divergence and the curve.



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And for these two terms, we we need all of these partial derivatives.



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So let's go ahead and calculate, first of all, the X derivative of our function H.



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And so we have here, of course, three components and let's start with the X components for the X component



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we have here an X and we have here an X, so we have a product of two functions.



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So this means we must use the product rule and get to terms.



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And for the first term, we have to first calculated operative X, which is of course one.



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And then we get the other function, which is one of our AR to the power of three.



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So this is this one over square root to the power of three expression here.



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And then we get the other term, which is X as our first function, and then we must calculate the derivative



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of one over the square root of X X squared plus discrepancy squared to the power of three.



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So this one over square root to the power of three is just the same as this expression to the power



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of minus three 1/2.



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So the derivative is minus three 1/2.



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Then we must decrease the exponent, which gives us minus five 1/2.



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So this is just our one of our to the power of five.



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And then we must not forget the inner derivative, which in this case is to X..



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Now for the Y component, we only have an X here, but not here.



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So we just got one term.



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And this is why times the derivative of one of the square root.



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So this is just the same as here.



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So minus three 1/2 to X of our five and here the same thing, Z minus three 1/2 to X are to the par



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five.



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And so we can write this down as one of our two, the power of three one zero zero plus or better.



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But I use use minus here.



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Let me let me delete this.



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OK, so minus three of our to the power of five and then we get here X square x y.



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X, Y, Z.



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Now we can do the same thing for the other derivatives, and here we just get partial derivative of



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the way with respect to why, and here we will get when you when you do it carefully, of course, we



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get here one of our to the power three zero one zero.



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And here we get minus three are to the power of five and then we get here why x, y, square, y, z.



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And of course partial derivative of each back to Z gives us one of our to depart three zero zero one



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minus three of to the power five z, x, y, z, y, z square.



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OK, so these are partial derivatives now that just put all of the parts together.



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So divergence of H is the partial derivative of X, the X components plus partial derivative of Y of



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the Y component plus partial derivative Z of the Z component.



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So what this does give us, it gives us one of our to the power of three minus.



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Three of us are to the power five x square for the first term, then we get one of our to the power



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of three minus three of our to the power of five Y squared plus one over to the power of three minus



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three of our to the power five Z Square.



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And now we can write it down as three of our to the power of three because we have three times this



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term, one over to the power of three and then we can use the other term and write it down like, like



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this.



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So yeah, once again better use a minus sign here.



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Looks better.



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Um OK so we get minus ah to the power five and then we have X squared for the first term, Y squared



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for the second term and Z squared for the third term.



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And now you can see that.



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Yeah.



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And of course here three for all of them and now you can see that X squared plus Y scrupulously square



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is the same thing as our square.



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So we get three over R to the power of three minus three times one of our to the power of three.



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So it gives zero.



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Finally we can calculate the Kearl, all the rotation of H.



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So this is of course a three component vector, as you know, and the components are the partial derivative



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with respect to Y of the C component minus partial derivative of C with of the Y component.



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And then for the others we add A Z, X and Y, Sivarasa, and here we get an X and Y and vice versa.



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And now we just have to yeah.



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Use these six terms and write them down.



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So let's just take them from the top.



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Maybe we start with the wider of a tip of the Z component.



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Here we get zero and here we get minus three of R to the power of five of Y Z.



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And for the other term we get the Z derivative of the Y component, which we have here is also one only



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one term.



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So we get minus three over R to the power of five C, Y.



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And so here you can see already that these two terms are the same.



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So we get zero.



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And for the other two terms components, it's the same thing.



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So here we will get yeah, we will get the X and this one here we will get X, Z and here we will get



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X, Y and he will get Y, X.



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So in all three components we get zero.



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So this means the divergence and the core of our function H is zero.



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Let's now come to the third task, which was to show that the kernel of a gradient of any scalar function



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is always zero.



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This is something that's, for example, very important in electrodynamics and it allows us to introduce



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the electrostatic potential.



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And the solution is quite easy if you just write this Nabala operator down in Cartesian coordinates.



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So you have here to operate at the X divided and here you have the gradient, which is, you know,



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the X derivative of F as the X component, and then here Y and Z as one Z components.



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And then you have to calculate the vector product, which is this derivative acting on this one here,



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minus this derivative acting on this one as the X component.



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And I mean, of course, you know how to calculate a vector product.



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You just write it down.



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And then the only step that you have to take is you have to realize that you can switch around these



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two derivatives so you can switch a partial derivative of Y acting on partial derivative of C, acting



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on F is the same as the other way around.



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So this means these two terms are equal, these will now equal and these are equal.



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So overall we get to zero vector, which means that's really the kernel of the gradient of F is zero.



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Now, also in electrodynamics, we find a problem where we have to show that the divergence of a curl



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of a vector function is always zero.



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So let's do the same thing.



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This derivation is a bit longer, but here I have already given you the hint that you should write down



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this function in terms of the coffees and coordinates.



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So what we have to do is we, first of all have to write down the Nabala vector, of course, here and



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here.



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And now we start by calculating this vector product.



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So once again, as the X component, we have to let this Y derivative act on the Z component of this



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function, minus the Z component is the derivative acting on the Y component of this function.



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So this is this one here.



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And then for the other two components, we have to change these indices here.



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Now we have to calculate the divergence, so we have to calculate the DOT product of this Nabala vector



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and this vector here, so we get this derivative acting on these two terms, plus this derivative acting



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on these two terms, plus this derivative acting on these two terms.



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So of all, we get six different terms and always we have plus minus, plus, minus, plus minus.



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And now what I have done is I have looked at these functions age, X, Y and Z, and I have ordered



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them and use these brackets here.



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And you can see, for example, for the terms that include H X, we have again these two derivatives



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here.



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Here we have a derivative with respect to Y and Z, and here we have it the other way around.



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And like in the third task, also here, it is possible to switch these two relatives around.



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And then you find that these two terms are always equal, these are always equal and these are always



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equal.



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So of all, we get zero.