Derivatives of Jacobian Matrices: t^2-s^2, ts

[jonroberts74]

x=t^2-s^2, y=ts,u=x,v=-y

a) compute derivative matrices \vec{D}f(x,y) = \left[\begin{array}{cc}2t&-2s\\s&t\end{array}\right]

\vec{D}f(u,v) = \left[\begin{array}{cc}1&0\\0&-1\end{array}\right]

b) express (u,v) in terms of (t,s)

f(u(x,y),v(x,y) = (t^2-s^2,-(ts))

c) Evaluate \vec{D}(u,v)

\vec{D}(u,v) = \left[\begin{array}{cc}1&0\\0&-1\end{array}\right] \left[\begin{array}{cc}2t&-2s\\s&t\end{array}\right]

= \left[\begin{array}{cc}2t&-2s\\-s&-t\end{array}\right]

d) verify if chain rule holdsneed help with this last part, also need to know if I even did the rest correctly

 

[STEMucator]

x=t^2-s^2, y=ts,u=x,v=-y

a) compute derivative matrices \vec{D}f(x,y) = \left[\begin{array}{cc}2t&-2s\\s&t\end{array}\right]

\vec{D}f(u,v) = \left[\begin{array}{cc}1&0\\0&-1\end{array}\right]

b) express (u,v) in terms of (t,s)

f(u(x,y),v(x,y) = (t^2-s2,-(ts))

c) Evaluate \vec{D}(u,v)

\vec{D}(u,v) = \left[\begin{array}{cc}1&0\\0&-1\end{array}\right] \left[\begin{array}{cc}2t&-2s\\s&t\end{array}\right]

= \left[\begin{array}{cc}2t&-2s\\-s&-t\end{array}\right]

d) verify if chain rule holdsneed help with this last part, also need to know if I even did the rest correctly

Part a) looks okay, but the notation is a little rough. Something more like:

<br /> J_{s, t}(x,y) = \begin{pmatrix}<br /> x_s &amp; x_t \\<br /> y_s &amp; y_t<br /> \end{pmatrix}, \quad<br /> <br /> J_{x, y}(u,v) = \begin{pmatrix}<br /> u_x &amp; u_y \\<br /> v_x &amp; v_y<br /> \end{pmatrix}<br />

$J_{s, t}(x,y)$ and $J_{x, y}(u,v)$ are the Jacobian matricies.

I'm sure you meant $u = t^2 - s^2$ for part b).

For part c), I'm sure what is intended is you find the derivative matrix $J_{s, t}(u,v)$ after expressing $u$ and $v$ as functions of $s$ and $t$.

 

[jonroberts74]

Part a) looks okay, but the notation is a little rough. Something more like:

<br /> J_{s, t}(x,y) = \begin{pmatrix}<br /> x_s &amp; x_t \\<br /> y_s &amp; y_t<br /> \end{pmatrix}, \quad<br /> <br /> J_{x, y}(u,v) = \begin{pmatrix}<br /> u_x &amp; u_y \\<br /> v_x &amp; v_y<br /> \end{pmatrix}<br />

$J_{s, t}(x,y)$ and $J_{x, y}(u,v)$ are the Jacobian matricies.

I'm sure you meant $u = t^2 - s^2$ for part b).

For part c), I'm sure what is intended is you find the derivative matrix $J_{s, t}(u,v)$ after expressing $u$ and $v$ as functions of $s$ and $t$.

I fixed the error in (b) and yeah that's what I want for (c), I'm using the notation that is used in my book/what my professor uses

as for part (d) how do I verify this

 

[STEMucator]

I fixed the error in (b) and yeah that's what I want for (c), I'm using the notation that is used in my book/what my professor uses

as for part (d) how do I verify this

For part d), if $f$ is a function of $u$ and $v$, which are functions of $x$ and $y$, which are functions of $s$ and $t$, what is the partial derivative of $f$ with respect to $u$? How about the partial with respect to $v$?

 

[jonroberts74]

(\frac{\partial f}{\partial u})_{t,s} = 2t,-2s \,\,\, (\frac{\partial f}{\partial v})_{t,s} = -s,-t

I know this is kind of an abuse of notation but I've been typing LaTeX all day

 

[STEMucator]

(\frac{\partial f}{\partial u})_{t,s} = 2t,-2s \,\,\, (\frac{\partial f}{\partial v})_{t,s} = -s,-t

I know this is kind of an abuse of notation but I've been typing LaTeX all day

Looks good.