Chapter 2.7, Problem 1E

Find the jacobian of the functions

a. $F(u,v)=\left(u^3,u{v}^3\right)$

b. $F(u,v)=\left(\sin uv,e^{uw}\right)$

c. $F(u,v)=\left(u^2+v^2-1,{\left(u-1\right)}^2+v^2-1\right)$

d. $F(u,v,w)=\left(u^2+v-w^2,\sin uvw,uv{w}^4\right).$

To determine

To find: The Jacobian of the function.

Answer to Problem 1E

TheJacobian of the function is $\left[\begin{array}{cc}3u^2& 0\\ v^3& 3u{v}^2\end{array}\right]$ .

Explanation of Solution

Given information:

The given function is,

  $F\left(u,v\right)=\left(u^3,u{v}^3\right)$

Concept used:

The Jacobian of the function is calculated as,

  $DF\left(u,v\right)=\left[\begin{array}{cc}\frac{\partial f_1}{\partial u}& \frac{\partial f_1}{\partial v}\\ \frac{\partial f_2}{\partial u}& \frac{\partial f_2}{\partial v}\end{array}\right]$

The given function is,

  $\begin{array}{l}F\left(u,v\right)=\left(u^3,u{v}^3\right)\\ F\left(u,v\right)=\left(f_1\left(u,v\right),f_2\left(u,v\right)\right)\\ f_1\left(u,v\right)=u^3\\ f_2\left(u,v\right)=u{v}^3\end{array}$

The Jacobian of the function is calculated as,

  $DF\left(u,v\right)=\left[\begin{array}{cc}\frac{\partial f_1}{\partial u}& \frac{\partial f_1}{\partial v}\\ \frac{\partial f_2}{\partial u}& \frac{\partial f_2}{\partial v}\end{array}\right]$

Substitute $u^3$ for $f_1$ and $u{v}^3$ for $f_2$ .

  $\begin{array}{c}DF\left(u,v\right)=\left[\begin{array}{cc}\frac{\partial \left(u^3\right)}{\partial u}& \frac{\partial \left(u^3\right)}{\partial v}\\ \frac{\partial \left(u{v}^3\right)}{\partial u}& \frac{\partial \left(u{v}^3\right)}{\partial v}\end{array}\right]\\ =\left[\begin{array}{cc}3u^2& 0\\ v^3& 3u{v}^2\end{array}\right]\end{array}$

Therefore, the Jacobian of the functionis $\left[\begin{array}{cc}3u^2& 0\\ v^3& 3u{v}^2\end{array}\right]$ .

To determine

To find: The Jacobian of the function.

Answer to Problem 1E

The Jacobian of the function is $\left[\begin{array}{cc}v\cos uv& u\cos uv\\ v{e}^{uv}& u{e}^{uv}\end{array}\right]$ .

Explanation of Solution

Given information:

The given function is,

  $F\left(u,v\right)=\left(\sin uv,e^{uv}\right)$

Concept used:

The Jacobian of the function is calculated as,

  $DF\left(u,v\right)=\left[\begin{array}{cc}\frac{\partial f_1}{\partial u}& \frac{\partial f_1}{\partial v}\\ \frac{\partial f_2}{\partial u}& \frac{\partial f_2}{\partial v}\end{array}\right]$

The given function is,

  $\begin{array}{l}F\left(u,v\right)=\left(\sin uv,e^{uv}\right)\\ F\left(u,v\right)=\left(f_1\left(u,v\right),f_2\left(u,v\right)\right)\\ f_1\left(u,v\right)=\sin uv\\ f_2\left(u,v\right)=e^{uv}\end{array}$

The Jacobian of the function is calculated as,

  $DF\left(u,v\right)=\left[\begin{array}{cc}\frac{\partial f_1}{\partial u}& \frac{\partial f_1}{\partial v}\\ \frac{\partial f_2}{\partial u}& \frac{\partial f_2}{\partial v}\end{array}\right]$

Substitute $\sin uv$ for $f_1$ and $e^{uv}$ for $f_2$ .

  $\begin{array}{c}DF\left(u,v\right)=\left[\begin{array}{cc}\frac{\partial \left(\sin uv\right)}{\partial u}& \frac{\partial \left(\sin uv\right)}{\partial v}\\ \frac{\partial \left(e^{uv}\right)}{\partial u}& \frac{\partial \left(e^{uv}\right)}{\partial v}\end{array}\right]\\ =\left[\begin{array}{cc}v\cos uv& u\cos uv\\ v{e}^{uv}& u{e}^{uv}\end{array}\right]\end{array}$

Therefore, the Jacobian of the function is $\left[\begin{array}{cc}v\cos uv& u\cos uv\\ v{e}^{uv}& u{e}^{uv}\end{array}\right]$ .

To determine

To find: The Jacobian of the function.

Answer to Problem 1E

The Jacobian of the function is $\left[\begin{array}{cc}2u& 2v\\ 2\left(u-1\right)& 2v\end{array}\right]$ .

Explanation of Solution

Given information:

The given function is,

  $F\left(u,v\right)=\left(u^2+v^2-1,{\left(u-1\right)}^2+v^2-1\right)$

Concept used:

The Jacobian of the function is calculated as,

  $DF\left(u,v\right)=\left[\begin{array}{cc}\frac{\partial f_1}{\partial u}& \frac{\partial f_1}{\partial v}\\ \frac{\partial f_2}{\partial u}& \frac{\partial f_2}{\partial v}\end{array}\right]$

The given function is,

  $\begin{array}{l}F\left(u,v\right)=\left(u^2+v^2-1,{\left(u-1\right)}^2+v^2-1\right)\\ F\left(u,v\right)=\left(f_1\left(u,v\right),f_2\left(u,v\right)\right)\\ f_1\left(u,v\right)=u^2+v^2-1\\ f_2\left(u,v\right)={\left(u-1\right)}^2+v^2-1\end{array}$

The Jacobian of the function is calculated as,

  $DF\left(u,v\right)=\left[\begin{array}{cc}\frac{\partial f_1}{\partial u}& \frac{\partial f_1}{\partial v}\\ \frac{\partial f_2}{\partial u}& \frac{\partial f_2}{\partial v}\end{array}\right]$

Substitute $u^2+v^2-1$ for $f_1$ and ${\left(u-1\right)}^2+v^2-1$ for $f_2$ .

  $\begin{array}{c}DF\left(u,v\right)=\left[\begin{array}{cc}\frac{\partial \left(u^2+v^2-1\right)}{\partial u}& \frac{\partial \left(u^2+v^2-1\right)}{\partial v}\\ \frac{\partial \left({\left(u-1\right)}^2+v^2-1\right)}{\partial u}& \frac{\partial \left({\left(u-1\right)}^2+v^2-1\right)}{\partial v}\end{array}\right]\\ =\left[\begin{array}{cc}2u& 2v\\ 2\left(u-1\right)& 2v\end{array}\right]\end{array}$

Therefore, the Jacobian of the function is $\left[\begin{array}{cc}2u& 2v\\ 2\left(u-1\right)& 2v\end{array}\right]$ .

To determine

To find: The Jacobian of the function.

Answer to Problem 1E

The Jacobian of the function is $\left[\begin{array}{ccc}2u& 1& -2w\\ vw\cos uvw& uw\cos uvw& uv\cos uvw\\ v{w}^4& u{w}^4& 4uv{w}^3\end{array}\right]$ .

Explanation of Solution

Given information:

The given function is,

  $F\left(u,v,w\right)=\left(u^2+v-w^2,\sin uvw,uv{w}^4\right)$

Concept used:

The Jacobian of the function is calculated as,

  $DF\left(u,v,w\right)=\left[\begin{array}{ccc}\frac{\partial f_1}{\partial u}& \frac{\partial f_1}{\partial v}& \frac{\partial f_1}{\partial w}\\ \frac{\partial f_2}{\partial u}& \frac{\partial f_2}{\partial v}& \frac{\partial f_2}{\partial w}\\ \frac{\partial f_3}{\partial u}& \frac{\partial f_3}{\partial v}& \frac{\partial f_3}{\partial w}\end{array}\right]$

The given function is,

  $\begin{array}{l}F\left(u,v,w\right)=\left(u^2+v-w^2,\sin uvw,uv{w}^4\right)\\ F\left(u,v\right)=\left(f_1\left(u,v,w\right),f_2\left(u,v,w\right),f_3\left(u,v,w\right)\right)\end{array}$

The functions are,

  $\begin{array}{l}{f}_1\left(u,v,w\right)=u^2+v-w^2\\ f_2\left(u,v,w\right)=\sin uvw\\ f_3\left(u,v,w\right)=uv{w}^4\end{array}$

The Jacobian of the function is calculated as,

  $DF\left(u,v,w\right)=\left[\begin{array}{ccc}\frac{\partial f_1}{\partial u}& \frac{\partial f_1}{\partial v}& \frac{\partial f_1}{\partial w}\\ \frac{\partial f_2}{\partial u}& \frac{\partial f_2}{\partial v}& \frac{\partial f_2}{\partial w}\\ \frac{\partial f_3}{\partial u}& \frac{\partial f_3}{\partial v}& \frac{\partial f_3}{\partial w}\end{array}\right]$

Substitute $u^2+v-w^2$ for $f_1$ , $\sin uvw$ for $f_2$ , and $uv{w}^4$ for $f_3$ .

  $\begin{array}{c}DF\left(u,v,w\right)=\left[\begin{array}{ccc}\frac{\partial \left(u^2+v-w^2\right)}{\partial u}& \frac{\partial \left(u^2+v-w^2\right)}{\partial v}& \frac{\partial \left(u^2+v-w^2\right)}{\partial w}\\ \frac{\partial \left(\sin uvw\right)}{\partial u}& \frac{\partial \left(\sin uvw\right)}{\partial v}& \frac{\partial \left(\sin uvw\right)}{\partial w}\\ \frac{\partial \left(uv{w}^4\right)}{\partial u}& \frac{\partial \left(uv{w}^4\right)}{\partial v}& \frac{\partial \left(uv{w}^4\right)}{\partial w}\end{array}\right]\\ =\left[\begin{array}{ccc}2u& 1& -2w\\ vw\cos uvw& uw\cos uvw& uv\cos uvw\\ v{w}^4& u{w}^4& 4uv{w}^3\end{array}\right]\end{array}$

Therefore, the Jacobian of the function is $\left[\begin{array}{ccc}2u& 1& -2w\\ vw\cos uvw& uw\cos uvw& uv\cos uvw\\ v{w}^4& u{w}^4& 4uv{w}^3\end{array}\right]$ .