VC1.1

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\frac{\partial A}{\partial y}=\frac{\partial}{\partial y}(xyz) i + \frac{\partial}{\partial y}(xz^2) j-\frac{\partial}{\partial y} (y^3) k

Therefore, \frac{\partial ^2 A}{\partial y^2}=\frac{\partial}{\partial y}(xz) i-\frac{\partial}{\partial y} (3y^2)k = -6yk

Again, \frac{\partial B}{\partial x}=\frac{\partial}{\partial x}(x^3)i-\frac{\partial}{\partial x}(xyz) j+\frac{\partial}{\partial x} (x^2 z) k = 2x^3 i-yz j+2xz k

Therefore, \frac{\partial^2 B}{\partial x^2}=\frac{\partial}{\partial x}(3x^2)i-\frac{\partial}{\partial x}(yz) j+\frac{\partial}{\partial x} (2xz) k

Now, \frac {\partial ^2 A}{\partial y^2}\times \frac{\partial ^2 B}{\partial x^2} = -6yk \times (6xi+2zk)=-36xy k\times i (As k \times k =0 )

Hence \frac {\partial ^2 A}{\partial y^2}\times \frac{\partial ^2 B}{\partial x^2}=-36xy j (As k \times i =j )

Which is equal to − 36j at the point (1,1,0).

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