Exercise 10.4

Question

\def\where{\,|\,}
Exercise 4: For $(x,y)\in\R2$, define
    $$
        \v f(x,y)=(e^x\cos y-1,e^x\sin y).
    $$
    Prove that $\v F=\v G_2\circ\v G_1$, where
    \begin{align*}
        \v G_1(x,y) &= (e^x\cos y-1,y) \
        \v G_2(u,v) &= \big(u,(1+u)\tan v\big)
    \end{align*}
    are primitive in some neighborhood of $(0,0)$. Compute the Jacobians of $\v G_1$, $\v G_2$, $\v F$ at $(0,0)$. Define
    $$
        \v H_2(x,y)=(x,e^x\sin y)
    $$
    and find
    $$
        \v H_1(u,v)=\big(h(u,v),v\big)
    $$
    so that $\v F=\v H_1\circ\v H_2$ in some neighborhood of $(0,0)$.

analambanomenos · Accepted Answer

We have
    \begin{align*}
        \v G_2\circ\v G_1(x,y) &= \v G_2(e^x\cos y-1,y) \
        &= (e^x\cos y-1,e^x\cos y	an y) \
        &= (e^x\cos y-1,e^x\sin y) \
        &= \v F(x,y)
    \end{align*}
    The derivative matrices are
    \begin{align*}
        \v G_1'(x,y) &= 
            \begin{pmatrix}
                e^x\cos y & -e^x\sin y \
                0 & 1
            \end{pmatrix} \
        \v G_2'(u,v) &= 
            \begin{pmatrix}
                1 & 0 \
                	an v & (1+u)\cos^{-2}v
            \end{pmatrix}
    \end{align*}
    so that $\v G_1'(0,0)=\v G_2'(0,0)=I$, hence $J_{\v G_1}(0,0)=J_{\v G_2}(0,0)=1$. By the chain rule and the properties of determinants, we also have $J_{\v F}(0,0)=1$.

Let $h(u,v)=\sqrt{v^2-e^{2u}}-1$. Then, for $(x,y)$ near the origin,
    \begin{align*}
        \v H_1\circ\v H_2(x,y) &= \v H_1(x,e^x\sin y) \
        &= \bigg(\sqrt{e^{2x}\sin^2y-e^{2x}}-1,e^x\sin y\bigg) \
        &=(e^x\cos y-1,e^x\sin y) \
        &= \v F(x,y)
    \end{align*}

Exercise 10.4

Answers

Comments

Exercise 10.4

Answers

Comments

Add answer