Lisa, welcome back.
I hope you have solved some of these tasks, let us come to the solutions to our exercises in differential
calculus.
So the first two tasks were about calculating the gradient of a scalar function and about calculating
the curl, the curl and the divergence of a vector function.
And then in task number three and four, you should show these two identities that the curl of the gradient
of any scareder function is zero and that the divergence of the curve and the vector function is zero.
So let's get started with task number one.
So here you can see we have a function F of our and we should calculate the gradient.
So R corresponds to X, Y and Z.
As you can see, I have already highlighted here the X, we have another X here and we have a Y square.
And of course as to the power of three.
What we have to calculate now is to calculate, first of all, the partial derivatives.
Let's go ahead to we have the partial derivative with respect to X of X of F, F of our.
And here this is already the most difficult task because we have X appearing twice in the product.
And this means we have to apply the product rule.
So we calculate the function, the first function, which is X times, the derivative of the second
function, and this is times minus one for this one here and then our exponential function.
OK, and then the other term here, we have to differentiate the first function, which is one, and
then we multiply by E to the power, as you can see here now for the Y derivative, we only get one
term because here of a Y only appears one.
And so we just get, um, we had our function basically because it's an exponential function.
And then we have to account for the inner derivative, which is the derivative of minus Y square, which
is minus to Y and then for the Z derivative.
We get like here, the function
minus C to the power of three.
And then again, the inner derivative, which in this case is minus three C Square.
OK, and now we can write down the gradient, so, you know, the gradient of F is, of course, the
vector of the partial derivatives.
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
in every single term so we can write down X times E to the minus X minus Y square minus C to the power
of three.
And now the only difference for the vectors is that here in the first term we get minus um.
Sorry, we get.
OK, yes, small mistake, OK?
So we cannot pull out this one here, we get minus X.
And then plus one, and for the Y component, we get mine as to why X and for the Z component, we get
minus three C, square, X and yeah, that's that's basically all you have to do.
And yeah, this is where we are finished.
OK, so now let's go ahead and calculate the divergence and the curl of such a vector function.
So here we have the position vector R which is X, Y, Z, divided by the absolute value of R to the
power of three, which is one of a square root to the power of three.
So first of all, I would say it's good to calculate all of the partial derivatives so that we have
it easy to calculate divergence and the curve.
And for these two terms, we we need all of these partial derivatives.
So let's go ahead and calculate, first of all, the X derivative of our function H.
And so we have here, of course, three components and let's start with the X components for the X component
we have here an X and we have here an X, so we have a product of two functions.
So this means we must use the product rule and get to terms.
And for the first term, we have to first calculated operative X, which is of course one.
And then we get the other function, which is one of our AR to the power of three.
So this is this one over square root to the power of three expression here.
And then we get the other term, which is X as our first function, and then we must calculate the derivative
of one over the square root of X X squared plus discrepancy squared to the power of three.
So this one over square root to the power of three is just the same as this expression to the power
of minus three 1/2.
So the derivative is minus three 1/2.
Then we must decrease the exponent, which gives us minus five 1/2.
So this is just our one of our to the power of five.
And then we must not forget the inner derivative, which in this case is to X..
Now for the Y component, we only have an X here, but not here.
So we just got one term.
And this is why times the derivative of one of the square root.
So this is just the same as here.
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
five.
And so we can write this down as one of our two, the power of three one zero zero plus or better.
But I use use minus here.
Let me let me delete this.
OK, so minus three of our to the power of five and then we get here X square x y.
X, Y, Z.
Now we can do the same thing for the other derivatives, and here we just get partial derivative of
the way with respect to why, and here we will get when you when you do it carefully, of course, we
get here one of our to the power three zero one zero.
And here we get minus three are to the power of five and then we get here why x, y, square, y, z.
And of course partial derivative of each back to Z gives us one of our to depart three zero zero one
minus three of to the power five z, x, y, z, y, z square.
OK, so these are partial derivatives now that just put all of the parts together.
So divergence of H is the partial derivative of X, the X components plus partial derivative of Y of
the Y component plus partial derivative Z of the Z component.
So what this does give us, it gives us one of our to the power of three minus.
Three of us are to the power five x square for the first term, then we get one of our to the power
of three minus three of our to the power of five Y squared plus one over to the power of three minus
three of our to the power five Z Square.
And now we can write it down as three of our to the power of three because we have three times this
term, one over to the power of three and then we can use the other term and write it down like, like
this.
So yeah, once again better use a minus sign here.
Looks better.
Um OK so we get minus ah to the power five and then we have X squared for the first term, Y squared
for the second term and Z squared for the third term.
And now you can see that.
Yeah.
And of course here three for all of them and now you can see that X squared plus Y scrupulously square
is the same thing as our square.
So we get three over R to the power of three minus three times one of our to the power of three.
So it gives zero.
Finally we can calculate the Kearl, all the rotation of H.
So this is of course a three component vector, as you know, and the components are the partial derivative
with respect to Y of the C component minus partial derivative of C with of the Y component.
And then for the others we add A Z, X and Y, Sivarasa, and here we get an X and Y and vice versa.
And now we just have to yeah.
Use these six terms and write them down.
So let's just take them from the top.
Maybe we start with the wider of a tip of the Z component.
Here we get zero and here we get minus three of R to the power of five of Y Z.
And for the other term we get the Z derivative of the Y component, which we have here is also one only
one term.
So we get minus three over R to the power of five C, Y.
And so here you can see already that these two terms are the same.
So we get zero.
And for the other two terms components, it's the same thing.
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
X, Y and he will get Y, X.
So in all three components we get zero.
So this means the divergence and the core of our function H is zero.
Let's now come to the third task, which was to show that the kernel of a gradient of any scalar function
is always zero.
This is something that's, for example, very important in electrodynamics and it allows us to introduce
the electrostatic potential.
And the solution is quite easy if you just write this Nabala operator down in Cartesian coordinates.
So you have here to operate at the X divided and here you have the gradient, which is, you know,
the X derivative of F as the X component, and then here Y and Z as one Z components.
And then you have to calculate the vector product, which is this derivative acting on this one here,
minus this derivative acting on this one as the X component.
And I mean, of course, you know how to calculate a vector product.
You just write it down.
And then the only step that you have to take is you have to realize that you can switch around these
two derivatives so you can switch a partial derivative of Y acting on partial derivative of C, acting
on F is the same as the other way around.
So this means these two terms are equal, these will now equal and these are equal.
So overall we get to zero vector, which means that's really the kernel of the gradient of F is zero.
Now, also in electrodynamics, we find a problem where we have to show that the divergence of a curl
of a vector function is always zero.
So let's do the same thing.
This derivation is a bit longer, but here I have already given you the hint that you should write down
this function in terms of the coffees and coordinates.
So what we have to do is we, first of all have to write down the Nabala vector, of course, here and
here.
And now we start by calculating this vector product.
So once again, as the X component, we have to let this Y derivative act on the Z component of this
function, minus the Z component is the derivative acting on the Y component of this function.
So this is this one here.
And then for the other two components, we have to change these indices here.
Now we have to calculate the divergence, so we have to calculate the DOT product of this Nabala vector
and this vector here, so we get this derivative acting on these two terms, plus this derivative acting
on these two terms, plus this derivative acting on these two terms.
So of all, we get six different terms and always we have plus minus, plus, minus, plus minus.
And now what I have done is I have looked at these functions age, X, Y and Z, and I have ordered
them and use these brackets here.
And you can see, for example, for the terms that include H X, we have again these two derivatives
here.
Here we have a derivative with respect to Y and Z, and here we have it the other way around.
And like in the third task, also here, it is possible to switch these two relatives around.
And then you find that these two terms are always equal, these are always equal and these are always
equal.
So of all, we get zero.