Question

if u=x²-2y,v=x+y+,w=x-2y+3z, find d(u,v,w)/d(x,y,z)​

Answer 1

$\frac{d(u,v,w)}{d(x,y,z)} = 14x + 4$

Step-by-step explanation:

Given,

$u=x^{2} -2y \\v=x+y+z\\w=x-2y+3z$

To find :

$\frac{d(u,v,w)}{d(x,y,z)}$

The above expression is called Jacobian of u,v,w with respect to x,y,z.

It can also be expressed as  $J\frac{(u,v,w)}{(x,y,z)}$

It can be calculated by the determinant ,

$\left[\begin{array}{ccc}\frac{du}{dx} &\frac{du}{dy} &\frac{du}{dz} \\\frac{dv}{dx}&\frac{dv}{dy}&\frac{dv}{dz}\\\frac{dw}{dx}&\frac{dw}{dy}&\frac{dw}{dz}\end{array}\right]$

Finding the values of determinant elements we get ,

$(i) \frac{du}{dx} = \frac{d}{dx}(x^{2} -2y) \\\\\frac{du}{dx}= 2x$

$(ii) \frac{du}{dy} = \frac{d}{dy}(x^{2} -2y) \\\\\frac{du}{dy}= -2\\\\(iii) \frac{du}{dz} = \frac{d}{dz}(x^{2} -2y) \\\\\frac{du}{dz}= 0\\\\\\(iv) \frac{dv}{dx} = \frac{d}{dx}(x+y+z) \\\\\frac{dv}{dx}= 1\\\\(v) \frac{dv}{dy} = \frac{d}{dy}(x+y+z) \\\\\frac{dv}{dy}= 1\\\\(vi) \frac{dv}{dz} = \frac{d}{dz}(x+y+z) \\\\\frac{dv}{dz}= 1\\\\(vii) \frac{dw}{dx} = \frac{d}{dx}(x-2y+3z) \\\\\frac{dw}{dx}= 1\\\\(viii) \frac{dw}{dy} = \frac{d}{dy}(x-2y+3z) \\\\\frac{dw}{dy}= -2$

$(ix) \frac{dw}{dz} = \frac{d}{dz}(x-2y+3z) \\\\\frac{dw}{dz}= 3$

=> Substituting the values in the determinant,

$\left[\begin{array}{ccc}2x&-2&0\\1&1&1\\1&-2&3\end{array}\right]$

=> Solving the determinant,

$2x(3-(-2)) -(-2)(3-1)+0(-2-1)\\2x(3+4)+2(2)+0\\2x(7)+4\\14x+4$

Hence the value of $\frac{d(u,v,w)}{d(x,y,z)} = 14x + 4$

Answer 2

$\displaystyle \sf{ \frac{ \partial (u,v,w)}{ \partial (x,y,z)} = 10x + 4}$

Given :

u = x² - 2y , v = x + y + z , w = x - 2y + 3z

To find :

$\displaystyle \sf{ \frac{ \partial (u,v,w)}{ \partial (x,y,z)} }$

Solution :

Step 1 of 3 :

Write down the given functions

Here it is given that

u = x² - 2y , v = x + y + z , w = x - 2y + 3z

Step 2 of 3 :

Find the partial derivatives

$\displaystyle \sf{ \frac{ \partial u}{ \partial x} = 2x }$

$\displaystyle \sf{ \frac{ \partial u}{ \partial y} = - 2 }$

$\displaystyle \sf{ \frac{ \partial u}{ \partial z} = 0}$

$\displaystyle \sf{ \frac{ \partial v}{ \partial x} } = 1$

$\displaystyle \sf{ \frac{ \partial v}{ \partial y} } = 1$

$\displaystyle \sf{ \frac{ \partial v}{ \partial z} } = 1$

$\displaystyle \sf{ \frac{ \partial w}{ \partial x} } = 1$

$\displaystyle \sf{ \frac{ \partial w}{ \partial y} } = - 2$

$\displaystyle \sf{ \frac{ \partial w}{ \partial z} } = 3$

Step 3 of 3 :

Find the required value

$\displaystyle \sf{ \frac{ \partial (u,v,w)}{ \partial (x,y,z)} }$

$= \displaystyle \begin{vmatrix} \frac{ \partial \sf u}{ \partial \sf x} & \frac{ \partial \sf u}{ \partial \sf y} & \frac{ \partial \sf u}{ \partial \sf z}\\ \\ \frac{ \partial \sf v}{ \partial \sf x} & \frac{ \partial \sf v}{ \partial \sf y} & \frac{ \partial \sf v}{ \partial \sf z} \\ \\ \frac{ \partial \sf w}{ \partial \sf x} & \frac{ \partial \sf w}{ \partial \sf y} & \frac{ \partial \sf w}{ \partial \sf z} \end{vmatrix}$

$= \displaystyle \begin{vmatrix} \sf 2x & \sf - 2 & 0\\ \\ 1 & 1 & 1 \\ \\ 1 & - 2 & 3 \end{vmatrix}$

$\displaystyle \sf{ = 2x(3 + 2) + 2(3 - 1) + 0}$

$\displaystyle \sf{ = 10x + 4}$

Correct question : If u = x² - 2y , v = x + y + z , w = x - 2y + 3z find ∂(u,v,w)/∂(x,y,z)

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