 Good morning, towards the end of the previous lecture, we saw the del or nabla operator operated on ordinary scalar and vector functions. Now, we will consider composite operations involving the del operator, always keep in mind that the operator del is a linear operator and therefore, whether applied as gradient divergence or as curl, they distribute over a sum of functions. Next, we consider the del operator applied over products of field functions. Four cases may arise, the product of two scalar fields is again a scalar field, the product of a scalar field and a vector field is a vector field and the product of two vector fields can be in two ways, one is the dot product which is a scalar and the other is a cross product which is a vector. When we apply the del operator on these four composite functions, these are the way to apply the operator. So, the gradient of phi into psi turns out to be this, the divergence of phi into v turns out to be gradient of phi dot v plus phi into divergence of v, the cross product the curl of phi v turns out to be gradient of phi cross v plus phi into curl of v and so on. So, these representations, these expressions you can work out, if you open, if you expand these expressions term by term and then simplify. When we go further and operate the gradient divergence and curl of a scalar or a vector function by the del operator once more, then we get what we call as the second order differential operators. Now, grad phi is a vector quantity, it is a vector function. So, you can apply del in two different ways through the dot product or cross product and accordingly, you get divergence of grad phi and curl of grad phi. When you consider the curl of v and consider applying the del operator over that curl of v, then again you can apply it in two ways, one is through dot product and the cross product. Therefore, you get two further second order operators, one is div curl and the other is curl curl. On the other hand, the divergence of v happens to be a scalar function and the only way you can apply del over that is through gradient. So, that gives you these five second order differential operators and two of them give us very important information, that is curl grad phi is identically 0 whatever may be phi. Similarly, div curl v turn out to be identically 0 whatever is the vector function v. So, what these two mean is that curl of a gradient is always 0 and divergence of a curl is always 0, then divergence of the gradient function grad phi turns out to be the Laplacian del 2 phi. So, curl curl and grad div these two have this relationship between them. So, these give you some of the relationships that the second order differential operators always the first is the line integral along a curve. So, if you have a vector function v, then along a curve if you take its integral that means take its integral that means that you have this curve and from this point to this point if you want to take the line integral of a vector function that means you take v dot a small length element a small vector element displacement element in along this curve. So, v dot dr such v dot dr components if you keep on adding from the starting point to the end point then you get the line integral integral along a curve of this particular vector function along curve t. So, that is defined like this v dot dr which will be v x d x plus v y d y plus v z d z and this will be integrated all over the curve continuously. So, if the curve is parameterized in this manner over an interval a to b for t then this line integral reduces to an ordinary definite integral from t equal to a to b like this. So, v d dot dr becomes v dot dr by dt into d t. So, v dot dr by dt turns out to be a function of t which you can integrate from t equal to a to b. These are some important statements which mean the same situation. So, all of these are equivalent statements for simple non intersecting curves or paths contained in a simply connected region these all of these are equivalent statements v x d x plus v y d y plus v z d z this quantity here this differential quantity here is an exact differential that will mean the same thing as the that the vector function v is the gradient of some scalar function some scalar field phi and that also means that if this is a perfect differential that means it can be integrated and the integral of this will be something which does not depend on the path along which the integral has been performed. So, for that kind of a function v which is the gradient of some scalar function whether you integrate from this point to this point through this curve or along this curve or along this curve the result will be same. And this also means that if the curve c is a closed curve then the point from where you start the it is the same point where you end and that means the integral line integral along a closed curve circulation represented with this circle on the integral sign that turns out to be 0 around any closed path. It also means that curve v is equal to 0 that is clear because v itself is the gradient of some scalar field. So, it is curve must be 0 this in that in terms of relevant physics means that the corresponding field v is a conservative field. The second integral that we define is a surface integral this is defined over a surface element. Now, d s is a differential surface element and its magnitude is the area of that small surface element and the direction is along the normal to that surface element. So, if you effect this dot product and then integrate that all over a surface patch then you get the surface integral of the vector field v over that surface patch s. And for this the surface patch s must be orientable that is it should be clear which side of the surface we are talking about that is only those surfaces for which one side of the surface and the other side of the surface are clearly identifiable. Now, if there is a parameterization of the surface in terms of two parameters u and w then this small surface area area element d s can be found from this and we can work out the normal also unit normal which is r u cross r w divided by this magnitude. And therefore, when we insert the relationship between in this expression then v dot n d s in place of n we get this cross product divided by its magnitude. And this d s the magnitude of that small differential area element we get this magnitude into d u d w. So, that magnitude gets cancelled and finally, we get this as the integrand and we can integrate over the region r in the u w plane the parametric plane. Finally, we have got the volume integral which operates over a differential volume. So, volume integral you can evaluate for scalar field functions as well as vector field functions. So, in terms of these integrals we have quite a few important theorems which are called the integral theorem. The first among them is the Green's theorem in the plane. Consider r as a closed bounded region in the x y plane like this region is r bounded by a closed curve c c is the boundary of r this curve a c b d is the curve c. Now, f 1 and f 2 are two first order continuous functions first order continuous means that f 1 and f 2 are continuous functions of x and y and their first order derivatives are also continuous. Then Green's theorem in the plane states that the line integral of the vector function f having f 1 and f 2 as component turns out to be the same as the double integral of this quantity on the entire region r. Now, this gives you a relationship between the double integral over a region and a line integral over along its boundary. Now, among the three theorems that we are going to discuss the line of proof for the first of these the Green's theorem in the plane we will discuss in detail to give you the clever of the proofs of these theorems and for the rest of them we will summarize the theme of the proof only. So, the way you try to prove the way we tend to prove Green's theorem in the plane is first by considering a simple domain in which any line parallel to the coordinate axis parallel to y axis or parallel to x axis cuts the boundary of the region or cuts the curve c only at x axis. Most two points if you draw a vertical line like this it will cut here and here two points only at most two points because if you draw this line then you get a single point if you draw like this you get no point. So, because we consider such a simple region first so that will mean that the entire curve c can be split into two parts one a c b and the other a d b first one can be called the lower half and the second one can be called upper half because any line parallel to y axis cuts it only at two points one is a lower point the other is the upper point. Similarly, since a line parallel to x axis like this also cuts it at at most two points. So, in another way we can subdivide the boundary into two parts one is c b d the right half and the other is c a d left half. Now, if we can do that then consider the one of the double integrals from here say del f 1 by del y d x d y. So, first we consider that and del f 1 by del y d x d y this double integral over the region. So, what we can say is that first we will integrate it with with respect to y and then x. So, we interchange the order of these differentials d y d x and as we do that first integral is with respect to y and y varies from the lower part to the upper part. So, let us represent this lower part as a function y of x call it y 1 of x the lower part of this boundary is a curve that can be represented as y 1 of x the upper part as y 2 of x. So, that means for the first integral with respect to y the lower limit is y 1 of x the upper limit is y 2 of x. So, that is this y 1 x to y 2 x and this integral will be next integrated with respect to x from x equal to a to x equal to b. That means all these vertical strips will be then added together from this end to that end. If we do that then del f 1 by del y integrated with respect to y. So, that will simply give us f 1 f 1 at the upper limit minus f 1 at the lower limit. So, f 1 at the upper limit is f 1 of x and y 2 and f 1 at the lower limit is f 1 of x and y 1. Now, note this that this has to be this has to be integrated with respect to x from a to b over along y 2 the first one. So, first one along y 2 has to be integrated from a to b. We can say that we will integrate its negative from b to a that means x equal to b to x equal to a. If we do that then this gets changed to negative of this with the corresponding swapping of the limits of the integral. Now, see f 1 of x y 2 and here f 1 of x y 1 both signs negative. So, you find that b to a is this integral and a to b is this integral this is along y 2 this is along y 1. So, what this is going to mean from a to b along y 1 and then from b to a along y 2. So, you get the first part gives you this line integral second part gives you this line integral. So, you have got a closed line integral over the entire boundary. So, that means that the these two terms together will mean minus common minus sign the line integral of f 1 over the entire curve c that shows that del f 1 by del y double integral over r turns out to be minus the cyclic integral or circulation of f 1 f 1 d x. So, the second term here turns out to be equal to the first term here. Similarly, by dividing this curve into two parts the left half and the right half as x 1 y from here to here and x 2 y from here to here. You can establish the equality of this part with this part that is the first double integral from here as the same as the first line integral from this side the second line integral from this side. So, as you do that in this manner first integrated with a set to x from this limit to that limit and continue to get the next integral with a set to y and then you add them together then you find them that this turns out to be the same as this and as you take the difference of the two you get the final result which is this. Now, if you carefully evaluate if you carefully check this you will also find that this turns out in alternative form the this one is the line integral f dot d r f is f 1 i plus f 2 j in the plane there is no k component. So, then this is f dot d r and on this side you find that this turns out to be the magnitude of the curl f and its direction is k because i and j are both in the x y plane. So, curl will turn out to be in the direction k and if d x d y is an area element in the x y plane then its direction as a as area is a vector quantity the corresponding direction will be again in the k direction. So, the magnitude will remain and k cross k dot k will turn out to be unity. So, curl f dot k you will find turns out to be the same thing. So, in alternative form this same relationship the result of the Green's theorem in the plane means that the line integral of a vector function turns out to be along a closed curve c turns out to be same as the surface integral of the curl of f over a surface element bounded by this same closed curve c. Later we will see that this in a more general form turns out to be the statement of Stokes theorem. So, that way Green's theorem in the plane is actually a special case of Stokes theorem which is more general. Now, recall that we consider this entire curve for a simple region. Simple region in the sense simple domain in the sense that any line parallel to one of the any of the coordinate axis cuts the curve c in at most two points. Now, if that is not the domain if the domain is like this not only general, but also multiply connected there is a hole also this part is not included in the domain. So, for that also we can have Green's theorem and the proof is not very complicated because for this kind of a domain we can always decompose this domain into simpler regions simpler domains in such a manner that each of the component domains satisfies this kind of a requirement. And then over every component over every component domain we can prove this and then we just sum up all these components 1, 2, 3, 4, 5, 6, 7 components we sum up together. The double integrals are directly additive they add up totally and we get the complete double integral by the simple sum. For the line integrals here what we find is that as we add them up then the actual boundary of the original domain both the outer boundary and the inner boundary is covered only once. On the other hand the spurious boundaries which was due to our subdivision of the domain gets actually circulated twice as a part of this sub domain it got circulated once from lower end to upper end as a part of this region it got circulated once it got included once from upper point to the lower point. That means this inner boundary this spurious boundary between two sub domains which was not part of the original boundary gets included in the line integral twice once along this way and the next time along this way. And in the algebraic sum they get cancelled for each of the inner boundaries the spurious boundaries that is the same thing that is going to happen. So, in the final sum the line integral that remains caters to the actual original boundary of the given domain and removes all the contributions from the spurious boundaries because they are traversed twice in opposite senses. So, they cancel each other. So, this way Green's theorem in the plane can be applied to this kind of domains also. Now, the next important theorem is the Gauss theorem which has a lot of application lot of fundamental interpretation in physics. So, for that we have a closed boundary region represented as t and its boundary s which is a piecewise smooth closed orientable surface. And over the entire region we have got defined a vector function f which is first order continuous that means it is continuous and its first order derivatives are also continuous. So, for that kind of a function the Gauss theorem or the divergence theorem says that the volume integral of the divergence of the vector function f over the entire volume is the same as the surface integral of the function itself of the vector function itself over the boundary of the region that is the surface integral is evaluated over the boundary s. Now, this is actually a direct result of the interpretation of the definition of divergence to the finite domains that is whatever is the meaning or interpretation of the quantity divergence for an infinitesimal domain around a point the same theme when extended to finite domains the corresponding extension turns out to be this divergence theorem. So, when you open these expressions then you get this divergence of f is this the volume integral of that is this over t and f dot n gives you this scalar quantity and you get the surface integral of that with scalar surface area element d s. So, this equality is the result of Gauss divergence theorem. If you want to establish this result what you try to do is that you try to establish the equality of this term by term that is the third triple integral from here is going to be equal to the third double integral from here the z component to z component and so on. So, term by term equality you can establish that is del f by del z triple integral will be equal to the surface integral of this part and so on for all three parts. So, for this also first we consider a region a volumetric region such that the boundary of which is cut by any line parallel to x axis y axis z axis at most in at most two points not more than that first we consider that kind of a region over that we establish the equality and then for extension to general regions we again subdivide the general region into many such simpler regions which satisfy this requirement and sum up the contributions. And in the case of Green's theorem in the plane the direct additive sum was on the double integral here the same thing happens for the volumetric and there the boundaries were segments of lines or curves which were traverse the inner courier boundaries were traverse twice in opposite senses here the boundaries will be surface elements and they whatever courier surface element is going to be used from one side in one of the double integral one of the surface integral for one region one sub region it is going to be considered in the other sub region as from the other side. And that is why those courier surface elements in the final sum get cancelled out and you get the actual surface integral over that surface which is the part of the original domain. Now, we will omit the detailed proof of this step by step and you can follow the proof at leisure later and work out the proof in the same lines. There are two further important identities or results called Green's identities which also work with a region volumetric region T with its closed surface boundary S as required in the premises of Gauss theorem. And when we apply Gauss theorem over certain functions in certain manner that is once on phi grad psi and then on psi grad phi then as a direct consequences of Gauss theorem we can establish these relations which have a lot of important applications in many fields. The third important theorem of vector calculus is the Stokes theorem that is that makes reference to not a closed surface, but an open surface so in this manner. So, suppose this is an open surface S and this is the boundary of the open surface. Now, note that with the same boundary of the open surface with the same boundary, boundary being a closed curve you can have many different surfaces. For example, if you have a ring and then here is a net. So, if you have with which you try to catch something. So, now you can go on changing the shape of this net, but the ring which is the boundary of the net remains the same. So, all these surfaces all these surfaces will have the same boundary and these are all open surfaces. This side it is open. So, that is why you get this boundary. Note that a closed surface will have no boundary. For an open surface there is a boundary. So, even the planar region if the if this curve is a plane curve then the planar region bounded by this curve is also one such surface and that kind of a surface we encountered in the case of Green's theorem in the plane. So, this is a surface S, one of these is taken as S and the boundary of the open surface S is this curve this closed curve C. Now, if the if there is a field function first order continuous field vector function f defined over this entire region. Then Stokes theorem tells that the line integral of the function f over this closed curve like this turns out to be the same as the surface integral of the curl of f over this entire surface S. That is the statement of Green's theorem Stokes theorem in which the unit normal that you need to use here is given by the right hand class rule on C. That is since it is an open surface. So, whether to take the normal in this way or in this way has to be decided both are valid, but it will depend on which way we are traversing C. So, as we are traversing C. So, if we put our right hand along the arrow like this, then whichever way the thumb will point out that works out to be the n not this. So, if we point to the arrow by the clasping of the right hand in whichever way the thumb points through the surface that is the direction of n to be used here in this context. Now, we will again omit the proof of this and as a special case the Green's theorem in the plane the proof of that we have already seen. So, here you will find that if we consider the surface region S to be a region in the x y plane itself, then you will find that we will recover the Green's theorem in the plane as we have studied earlier. We will bypass the proof of this. So, the important points to be noted from this lecture or this lesson are the del operator that the del operator applied in three different ways on scalar and vector functions on scalar function you get radiant on vector functions through dot and cross product you get divergence and curl. The way we have the composite and second order operators on field functions, then next the line surface and volume integral and the three important theorems which are Green's Gauss and Stokes theorems and these theorems are important in the physics and engineering of large number of systems through their applications. So, in the exercises of this chapter in the book, you will find several important problems from physics as well as from applied mathematics where some of these theorems are applied directly to break the problem or reduce the problem into much simpler situations. In these two lectures on vector calculus by avoiding some of the long proof we have saved some time. So, let us take some examples on vector calculus. Some examples on physics and engineering problem as well as on applied mathematics problem pertaining to the scalar and vector field are given in the exercises in the text book in chapter 18. I strongly advise you to attempt those exercises particularly the exercise on Maxwell's equation because that takes you through a complete exposure to almost all the important issues discussed in this lesson. Now, we consider one problem on parametric curve. This is actually the problem of 17-6 of the text book that we are following. In the appendix of the book corresponding to this problem three approaches have been outlined. Here I will elaborate one of the approaches. The problem is to verify or check whether these two parametric equations represent the same curve or not. This is the first curve and this is the second curve. Now, you will note very easily that the first curve is evidently a circle with origin at the center and radius free lying in the x y plane that much is very clear. Now, we need to check whether R 2 of t also represents the same curve. So, for that we try to transform R 2 in such a manner that it gets connected get started from the same point where the first curve starts. So, for R 1 for the first curve let us see what is R 1 at t equal to 0 the starting point. So, if you put t equal to 0 here it is an i. If we put t equal to 0 here then we get R 1 0 as 3 i its derivative R 1 t the derivative of the first curve the tangent vector that you will get as 3 sin t i plus 3 sin t i plus cos t j at t equal to 0. If we evaluate this tangent vector then we get this is 0 and from here we get 3 j that means that the curve starts here at 3 0 3 0 0 actually in the 3 d 3 0 0. So, in x y plane this curve is in planar curve in x y plane. So, we are making the diagram only in the x y plane. So, it starts from 3 0 and since it is a circle it must start like this and that is why the tangent vector is found to be in the direction j. So, this is the way the circle proceeds. So, this is the point of our interest right now. Now, if this is the tangent vector then what is the unit tangent. So, unit tangent at the starting point for the first curve is j at the starting point note that you can write u 1 of 0, but to make the notation simple currently we are writing here understood is the fact that it is the starting point which is being analyzed right now. Similarly, we can find out the second derivative of r 1 t and then we put the value t equal to 0 in that second derivative which will give us this as minus 3 i which will mean that the unit principle normal is minus i for the first curve. So, if unit tangent is there unit principle normal is there then from these two we can work out what is the binormal u 1 cross p 1. So, j cross minus i we will get that as k. Now, this much we have in hand now we will try to find out the same quantities from r 2 and then try to work out the transformation which will bring u 2 to u 1 p 2 to p 1 and b 2 to b 1 and r 2 at pi by 4 to r 1 at 0. That means we want to match the starting point and we want to match the select frame at the starting point. So, we want to match the point r 1 and u p b point r u p b at the starting point of both the curve. So, we leave the first curve at its place and the second curve we try to bring here. So, for that purpose please make note of this and we have let me mark here unit tangent is here unit principle normal is here and of course, b 1 is perpendicular out of the board because of the cross product. So, now we get rid of this because all the information from here is actually available here as well this is the starting point. Now, the same things we try to evaluate from here. So, for the starting point we put t equal to pi by 4. So, as we put t equal to pi by 4 here. So, pi by 4 into 2 that is pi by 2. So, sin pi by 2 is 1 cos pi by 2 is 0. So, from here we get 3 i from here we get 2 plus 2 4 4 j and from here we get 1 plus 4 5 k. So, this is the starting point for the second curve as it is given. Now, if we develop the first derivative and second derivative that also you can similarly do finding derivative is not very complicated. So, I am omitting that and giving you the result if you evaluate the first derivative of this that is r 2 prime t and then at t equal to pi by 4 you evaluate that then you find that you get this as minus 4 i plus 2 j plus 4 k. You must evaluate this expression for the derivative and then verify that this indeed is what you get. Similarly, differentiating that again for the second derivative and inserting the value pi by 4 in place of t you would find minus 8 i minus 8 j minus 4 k. So, from here you can work out the unit tangent and unit principle normal between these two. So, from here you will get u 2 at the value pi by 4 at the value t equal to pi by 4 that is at the starting point as now this is minus 4 to 4. So, what will be the magnitude? Magnitude will be 6. So, you divide by 6. So, you get minus 2 by 3 i plus 1 by 3 j and plus 2 by 3 k this turns out to be the unit tangent at the starting point for the second curve. Now, this in the first case as you saw it was very clear that the tangent vector came to be in the direction j and the second derivative appeared in the direction minus i which are orthogonal to each other perpendicular to each other anyway. So, we did not have to subtract the component of r double prime from the direction of u. If the same thing happens here also then it is fine otherwise we will have to subtract that. So, whether this r prime and r double prime are perpendicular to each other that you can check otherwise we would have to subtract. So, you get 32 minus 16 minus 16. So, the dot product between these two is 0 that means this is indeed perpendicular to this. So, that makes our life easy. So, the unit vector along this direction itself is the unit principle normal. So, you get the unit vector in this direction and divide by 12. Yes, the magnitude of this is 12. So, you divide by 12 and you get minus 2 by 3 i minus 2 by 3 j and minus 1 by 3 k. So, this you get as the principle unit principle normal and from these two you can work out b 2 as the cross product which turns out to be 1 by 3 i minus 2 by 3 j plus 2 by 3. Now, we have the starting point of the curve r 1 here and its u p b serif and net frame oriented in this manner. And for the second curve also we have got the corresponding pieces of information r 2 at the starting point and u 2 p 2 b 2 at the starting point the point and the serif and net frame. Now, if the two if the two parametric equations these two represent the same curve then if through a rigid body motion a displacement and a rotation. If we can bring this r 2 and this u 2 p 2 b 2 in the location and orientation of the first curve r 1 then the complete curve should match. So, let us try to do that. So, we ask what rotation transforms these three vectors in the direction of these three vectors this is the question we ask. So, suppose that rotation matrix 3 by 3 rotation matrix is r. So, r transforms u 1 p 1 b 1 to u 2 p 2 b 2 we could do the other way also that is what rotation would transform u 2 p 2 b 2 to here. But that is not very different because the two rotation matrices are transposed of each other. So, if we try to do this we are going to do it like this because the matrix that will appear here will be easier to invert. The matrix that would come from there will be more complicated to invert and we must invert one of the two matrices. So, r into u 1 what is the column representation of this vector u 1 u 1 is j that means 0 1 0 0 i plus 1 j plus 0 k. So, u 1 is 0 1 0 0 1 0 0 1 0. So, p is minus 1 0 0 that is minus i minus 1 0 0 and b is the z vector k that is 0 0 1. We find that we want the rotation matrix which transforms u 1 to u 2 p 1 to p 2 and b 1 to b 2. So, u 1 to u 2 r into first column of this should be here that is the vector u 2 there. Similarly, r into p 1 should be p 2 and finally, r into b 1 should be b 2 that is this. Now, we want to determine r. So, we need to post multiply the matrix on the right side with the inverse of this which is quite simple. So, let me give you the expression for that r which you can verify later. Now, it is actually quite simple what we need to do is to interchange these two columns on both sides. So, as we interchange these two columns we get 1 here and minus 1 here then the first column we should make negative because this is minus 1. So, that will immediately make this identity matrix and whatever is there will be the product of this with its inverse from the right side. So, this tells us that the matrix r is the negative of this column comes first this column goes next and the third column remains in its own place because this is 0 0 1. So, those column operations which will make it identity the corresponding column operations done here give the value of the rotation matrix r. Now, this is the matrix r which operated over u 1 p 1 b 1 takes it to u 2 p 2 b 2 and what we are looking for is actually the inverse transformation because we want the second curve to come and merge here if they turn out to be the same curve. So, we need the inverse of this rotation matrix and inverse of rotation matrix is just transpose. So, we transpose this matrix. So, these two exchange their places these two exchange their places in this case we do not have to make any change because they are same and these two exchange their places. So, this transformation applied on the second curve will bring it to the first curve if they turn out to be the same curve. So, that is the check that we have to perform. So, the second curve transformed through now first orientation only till now we have not given the position transformation. So, the orientation of the second curve is changed in order to match the orientation at the starting point. So, let us call it r 2 tilde that will be r transpose into what is r 2 given. So, that means that this matrix multiplied to the vector now this i j k the way it is written the first entry will be here the second entry the j entry will come here and the k entry will come here. If we multiply this matrix with this completely then we get r 2 t r 2 tilde t this one as 3 sin 2 t plus 10 by 3 minus 3 cos 2 t plus 8 by 3 3 this is after turning it. So, now the orientation of the second curve at the starting at its starting point is the same as the orientation of the first curve at its own starting point. That means now if they are the same curve then this curve and that curve now have taken parallel cross trails. Now to get the second curve here we need to apply a displacement. So, for displacement we will need the displacement that is needed say delta r will be this r 2 tilde at its starting point which is pi by 4 minus this r 1 at its own starting point and earlier we saw the starting point of this r 2 at pi by 4 and that was 3 i plus 4 j plus 5 k from that we subtract this which is 3 i. So, the rest of it we get 4 j plus 5 k r 2 has been transformed r 2 has been transformed so better we evaluate this. We evaluate r 2 tilde at pi by 4 from here fresh because during rotation its starting point has gone to some other locations. So, put pi by 4 here so that means sin pi by 2 is 1 so 3 plus 10 by 3 from which we subtract this 3 i so only 10 by 3 will remain and then minus 3 into cos pi which is 0. So, this 8 by 3 minus 0 8 by 3 remains and finally the k part which is 3 minus 0. So, we have got 3. So, this is the displacement which we have to give in a negative direction to r 2 tilde to bring it here. So, now let us call this now this r 2 tilde was the rotated version of r 2 now r 2 bar will be the translated further translated. So, this r 2 bar will be this r 2 tilde minus this delta r. So, what we will find from here if we subtract this then we will get 3 sin 2 t i minus 3 cos 2 t j and 0. So, now we find that this is also a circle in the x y plane starting from here no question about all this. But still you find that the equation of this and the equation of this do not exactly match because there is a change of the parameterization. If you now try to reparameterize this same curve with a little different parameter say with tau equal to 2 t minus pi by 2 then you will find that the equation of this transforms exactly to the equation of this in terms of tau. So, whatever is this equation in terms of t the same equation here you will get in terms of tau you will get 3 sin 3 cos tau 2 t minus pi by 2 you will get 3 cos tau i plus 3 sin tau j and that will also transform the starting point in the first curve the starting point was 0 in the second curve till this point the starting point is pi by 4. When you put that pi by 4 here you will find that the starting point of tau will be 0. So, that way you will find that through these changes rotation to make the set the difference of the two curves parallel then translation to make them merge at the initial point and then reparameterization will show that the two curves exactly have the same equation. So, up to this stage the curve has actually come here, but its parameterization is different. So, at different values of the parameter it is going to different points rather than tracing the same curve at the same parameter values. So, this reparameterization will convince that the two equations are found to be exactly same. So, far in the discussion on vector calculus and next lecture onwards we will start our next module on numerical analysis starting from polynomial equations. Thank you.