Chain Rule — Question 6

Question 6

Let $$F(x,y)=\begin{pmatrix}x^2-y\\xy\end{pmatrix},\qquad G(s,t)=\begin{pmatrix}e^s\cos t\\e^s\sin t\end{pmatrix}.$$ Tasks

1. Find the Jacobian matrices $J_F$ and $J_G$.

2. Use matrix chain rule to find $J_{F\circ G}(0,\pi/2)$.

3. Check one entry directly.

Question 6 – Solution

Strategy. Evaluate $J_F$ at $G(s,t)$ before multiplying by $J_G$.

Step 1: Jacobians $$J_F=\begin{pmatrix}2x&-1\\y&x\end{pmatrix},\qquad J_G=\begin{pmatrix}e^s\cos t&-e^s\sin t\\e^s\sin t&e^s\cos t\end{pmatrix}.$$ At $(0,\pi/2)$, $G=(0,1)$, so $$J_F(0,1)=\begin{pmatrix}0&-1\\1&0\end{pmatrix},\quad J_G=\begin{pmatrix}0&-1\\1&0\end{pmatrix}.$$

Step 2: Multiply $$\displaystyle J_{F\circ G}(0,\pi/2)= \begin{pmatrix}-1&0\\0&-1\end{pmatrix}.$$ $$\boxed{\text{The composite Jacobian is }-I_2.}$$

Step 3: Check The first component is $e^{2s}\cos^2t-e^s\sin t$; its $s$-derivative at the point is $0-1=-1$, matching the upper-left entry.