Question 6 Let F(x,y)=(x2−yxy),G(s,t)=(escostessint).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 Find the Jacobian matrices JFJ_F and JGJ_G. Use matrix chain rule to find JF∘G(0,π/2)J_{F\circ G}(0,\pi/2). Check one entry directly. Show solutionHide solution+Question 6 – Solution Strategy. Evaluate JFJ_F at G(s,t)G(s,t) before multiplying by JGJ_G. Step 1: Jacobians JF=(2x−1yx),JG=(escost−essintessintescost).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,π/2)(0,\pi/2), G=(0,1)G=(0,1), so JF(0,1)=(0−110),JG=(0−110).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 JF∘G(0,π/2)=(−100−1).\displaystyle J_{F\circ G}(0,\pi/2)= \begin{pmatrix}-1&0\\0&-1\end{pmatrix}. The composite Jacobian is −I2.\boxed{\text{The composite Jacobian is }-I_2.} Step 3: Check The first component is e2scos2t−essinte^{2s}\cos^2t-e^s\sin t; its ss-derivative at the point is 0−1=−10-1=-1, matching the upper-left entry.
