Question

If u=x+3y^2-z^3, v=4x^2 yz, w=2z^2-xy then \(\frac{∂(u,v,w)}{∂(x,y,z)} \) at (1,1,1).

(a) -184

(b) -90

(c) 20

(d) 40

This question was addressed to me in class test.

The origin of the question is Jacobians in division Partial Differentiation of Engineering Mathematics

Answer 1

The correct option is (a) -184

Best explanation: Given that u=x+3y^2-z^3, v=4x^2 yz, w=2z^2-xy

\(\frac{∂(u,v,w)}{∂(x,y,z)} = \begin{vmatrix}

\frac{∂u}{∂x} & \frac{∂u}{∂y} &\frac{∂u}{∂z}\\

\frac{∂v}{∂x} &\frac{∂v}{∂y} &\frac{∂v}{∂z}\\

\frac{∂w}{∂x} &\frac{∂w}{∂y} &\frac{∂w}{∂z}\\

\end{vmatrix} = \begin{vmatrix}

1 & 6y & -3z^2\\

8xyz &4x^2 z &4x^2 y\\

-y &-x &4z\\

\end{vmatrix}\)

\(\frac{∂(u,v,w)}{∂(x,y,z)} \,at\, (1,1,1) = \begin{vmatrix}

1 & 6 & -3\\

8 &4&4\\

-1 &-1 &4\\

\end{vmatrix}\)

=1(16+4) – 6(32+4) – 3(-8+4) = -184.