 Hello everyone, myself, Mrs. Mayuri Kangre, Assistant Professor of Mathematics from the Department of Humanities and Sciences, Vulture Institute of Technology, Solaapur. Today we are going to see vector differentiation. The learning outcome is, at the end of this session, the students will be able to compute the curl of vectors. Let us see the definition of curl of vector f bar. Let f bar equals to f1i plus f2j plus f3k be any vector point function, then curl of f bar is given by curl of f bar equals to del cross f bar, where del is the vector differential operator defined by i dou by dou x plus j dou by dou y plus k dou by dou z. So the curl of f bar is given by the cross product of i into dou by dou x plus j into dou by dou y plus k into dou by dou z with f1i plus f2j plus f3k. We know that the cross product is obtained by the determinant, so by this way the del cross f bar is given by the determinant whose first row entries are i, j, k, the second row entries are dou by dou x, dou by dou y, dou by dou z and the third row entries are f1, f2, f3, then the determinant value is called as the curl of f bar. Note that if a vector f bar is a vector point function such that curl of f bar that is del cross f bar is a zero vector, then f bar is called as irrotational. As we know that the cross product is a vector product, so its output is a vector, that is why the curl of f bar is a vector valued function. Now let us go for examples. First example, find the curl of f bar equals to xyz i plus 3x square yj plus xz square minus y square zk at the point 2 minus 1, 1. Here f bar is given as xyz i plus 3x square yj plus xz square minus y square zk, comparing this f bar with f1f2f3, comparing this f bar with f1i plus f2j plus f3k which gives us f1 equals to xyz, f2 equals to 3x square y and f3 equals to xz square minus y square z. As the curl of f bar is defined as del cross f bar equals to the determinant whose first row entries i, j, k, second row entries dou by dou x, dou by dou y, dou by dou z, third row entries f1, f2, f3. In this definition, let us put the values of f1, f2, f3, we get curl of f bar equals to the determinant with first row entries i, j, k, second row entries dou by dou x, dou by dou y, dou by dou z, third row entries xyz 3x square y, xz square minus y square z. Now let us evaluate this determinant. So curl of f bar equals to i into bracket dou by dou y of xz square minus y square z minus dou by dou z of 3x square y minus j into bracket dou by dou x of xz square minus y square z minus dou by dou z of xyz plus plus k into bracket dou by dou x of 3x square y minus dou by dou y of xyz into k, therefore curl of f bar equals to. Now let us evaluate the brackets. Let us see here. Here the differentiation is partially with respect to y, so x and z are treated as constant. So the derivative of xz square will be 0 and minus y square z, its derivative will be minus 2 yz. For the second term, the differentiation is with respect to z, so x and y are treated as constant, so the derivative of 3x square y will be 0. So i will have the coefficient minus 2 yz. Now let us go for the second bracket. Here the differentiation is with respect to x, so y and z are treated as constant. Therefore we get the derivative of xz square as z square minus this y and z are treated as constant, so its derivative will be 0. For this term, the differentiation is with respect to z, so the derivative of xyz will be xy, so we get the coefficient of j as z square minus xy. Now let us go for the third bracket, the differentiation is with respect to x, so y and z are treated as constant and its derivative will be 6xy. Similarly here the differentiation with respect to y, treating x and z as constant, we get the differentiation as xz. So we get the curl of f bar equals to i into bracket minus 2 yz minus j into bracket z square minus xy plus k into bracket 6xy minus xz. Now let us evaluate the curl of f bar at the point 2 minus 1 1, that means here we substitute x as 2, y as minus 1 and z as 1. So we get i into bracket minus 2 as it is, y is replaced by minus 1 and z is replaced by 1, so i into bracket minus 2 minus 1 into 1 minus j into bracket z square, so 1 square minus xy is there, so 2 into minus 1 plus k into bracket 6 into xy, so 6 into 2 into minus 1 minus xz, so 2 into 1. As we know that let us evaluate these brackets, so we get the curl of f bar at the point 2 minus 1 1 as minus 2 into minus 1 into 1 is 2, so 2i minus 1 square minus of 2 into minus 1 will be 1 plus 2, so we get 3j and 6 into 2 into minus 1 is minus 12 minus 2, so we get minus 14k. So the curl of f bar at the point 2 minus 1 1 is 2i minus 3j minus 14k. Now let us go for the second example, show that the vector f bar equals to x square minus yz into i plus y square minus zx into j plus z square minus xy into k is irrotational. Here we have to show that the vector is irrotational, so pause the video for a minute and give the answer of this question when we say that a vector is irrotational. I hope everyone knows the answer, the vector is said to be irrotational if curl of f bar that is del cross f bar is equal to 0 vector. That is here we have to show that the determinant with the first row entries ijk, second row entries doh by doh x, doh by doh y and doh by doh z and third row entries f1, f2, f3 is having the value 0 vector. Now f bar is given as x square minus yz into i plus y square minus zx into j plus z square minus xy into k comparing this f bar with f1i plus f2j plus f3k which gives us the values of f1, f2 and f3. Now replacing the values of f1, f2, f3 in the definition of curl of f bar the curl of f bar is given as the determinant with the first row entries ijk, second row entries doh by doh x, doh by doh by doh z and third row entries by replacing f1, f2, f3 with x square minus yz, y square minus zx and z square minus xy respectively. Now let us evaluate this determinant therefore del cross f bar equals to i into bracket doh by doh y of z square minus xy minus doh by doh z of y square minus zx minus j into bracket doh by doh x of z square minus xy minus doh by doh z of x square minus yz plus k into bracket doh by doh x of y square minus zx minus doh by doh y of x square minus yz. Now let us evaluate this coefficient of i, let us see the first bracket here the differentiation is with respect to y, so x and z are treated as constant, so the derivative with respect to y for z square will be 0 and the differentiation of minus xy with respect to y will be minus x. Similarly for the second term the differentiation of y square with respect to z will be 0 and minus zx its derivative with respect to z will be minus x, so we get i into bracket minus x minus of minus x. Similarly for the other two brackets doh by doh x of z square minus xy will be minus y and doh by doh z of x square minus yz will be minus y, so we get minus j into bracket minus y minus of minus y. Similarly for the third bracket the derivative with respect to x of y square minus zx will be minus z and the derivative of x square minus yz with respect to y will be minus z, so we get del cross f bar equals to i into bracket minus x minus of minus x minus j into bracket minus y minus of minus y plus k into bracket minus z minus of minus z. Now here the minus minus sign becomes plus similarly here also and here also, so we get i into bracket minus x plus x minus j into bracket minus y plus y plus k into bracket minus z plus z which cancels each other and we get del cross f bar equals to 0 i minus 0 j plus 0 k that is a 0 vector which shows that f bar is an irrotational vector. Thank you.