VC1.15

From Example Problems
Jump to: navigation, search

Here \phi f=xy^{2}z(xzi-xyj+yz^{2}k)=x^{2}y^{2}z^{2}i-x^{2}y^{3}zj+xy^{3}z^{3}k\,

Now {\frac  {\partial }{\partial z}}(\phi f)={\frac  {\partial }{\partial z}}(x^{2}y^{2}z^{2})i-{\frac  {\partial }{\partial z}}(x^{2}y^{3}z)j+{\frac  {\partial }{\partial z}}(xy^{3}z^{3})k\, which is equal to

2x^{2}y^{2}zi-x^{2}y^{3}j+3xy^{3}z^{2}k\,

Now differentiating the above,we get

{\frac  {\partial ^{2}(\phi f)}{\partial x\partial z}}={\frac  {\partial }{\partial x}}[{\frac  {\partial (\phi f)}{\partial z}}]\, which is equal to

={\frac  {\partial }{\partial x}}(2x^{2}y^{2}z)i-{\frac  {\partial }{\partial x}}(x^{2}y^{3})j+{\frac  {\partial }{\partial x}}(3xy^{3}z^{2})k\,

=4xy^{2}zi-2xy^{3}j+3y^{3}z^{2}k\,

Again partially differentiating the above result,

{\frac  {\partial ^{3}(\phi f)}{\partial x^{2}\partial z}}={\frac  {\partial }{\partial x}}[{\frac  {\partial ^{2}(\phi f)}{\partial x\partial z}}]\,

={\frac  {\partial }{\partial x}}(4xy^{2}zi-2xy^{3}j+3y^{3}z^{2}k)\,

={\frac  {\partial }{\partial x}}(4xy^{2}z)i-{\frac  {\partial }{\partial x}}(2xy^{3})j+{\frac  {\partial }{\partial x}}(3y^{3}z^{2})k\,

=4y^{2}zi-2y^{3}j\,

Therefore at A(2,-1,1),

{\frac  {\partial ^{3}(\phi f)}{\partial x^{2}\partial z}}=4(-1)^{2}(1)i-2(-1)^{3}j=4i+2j\,

Main Page