Maths Analysis II - Homework #9 20220315

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# Maths Analysis II

## Homework #9 20220315

### 9.2.19

$(3)$

$$
\begin{gather}
\frac{\partial u}{\partial r}=\frac{\partial u}{\partial x}\cdot\frac{\partial x}{\partial r}+\frac{\partial u}{\partial x}\cdot\frac{\partial x}{\partial r}=\frac{2x}{x^2+y^2}\cdot e^{r+s+t};\\
\frac{\partial u}{\partial s}=\frac{\partial u}{\partial x}\cdot\frac{\partial x}{\partial s}+\frac{\partial u}{\partial y}\cdot\frac{\partial y}{\partial s}=\frac{2x}{x^2+y^2}\cdot e^{r+s+t}+\frac{2y}{x^2+y^2}\cdot8s;\\
\frac{\partial u}{\partial t}=\frac{\partial u}{\partial x}\cdot\frac{\partial x}{\partial t}+\frac{\partial u}{\partial y}\cdot\frac{\partial y}{\partial t}=\frac{2x}{x^2+y^2}\cdot e^{r+s+t}+\frac{2y}{x^2+y^2}\cdot8t.
\end{gather}
$$

$(4)$

$$
\frac{\mathrm{d}u}{\mathrm{d}x}=\frac{\partial u}{\partial x}+\frac{\partial u}{\partial y}\cdot\frac{\partial y}{\partial x}+\frac{\partial u}{\partial z}\cdot\frac{\partial z}{\partial x}=\frac{ae^{ax}(y-z)+ae^{ax}\cos x+e^{ax}\sin x}{a^2+1}.
$$

### 9.2.20

$(2)$

$$
\begin{align}
\frac{\mathrm{d}u}{\mathrm{d}t}&=\frac{\partial f}{\partial x}\cdot\frac{\partial x}{\partial t}+\frac{\partial f}{\partial y}\cdot\frac{\partial y}{\partial t}+\frac{\partial f}{\partial z}\cdot\frac{\partial z}{\partial t}\\&=f_x'(x,y,z)\cos t-f_y'(x,y,z)\sin t+f_z'(x,y,z)e^t.
\end{align}
$$

$(4)$

Let $v=x+y+z$ and $w=x^2+y^2+z^2$, then

$$
\begin{align}
\frac{\partial u}{\partial x}&=\frac{\partial f}{\partial v}\cdot\frac{\partial v}{\partial x}+\frac{\partial f}{\partial w}\cdot\frac{\partial w}{\partial x}=\frac{\partial f}{\partial v}+2x\cdot\frac{\partial f}{\partial w}=f_v'(v,w)+2x\cdot f_w'(v,w);\\
\frac{\partial^2u}{\partial x^2}&=\frac{\partial}{\partial x}(\frac{\partial u}{\partial x})=\frac{\partial}{\partial x}(\frac{\partial f}{\partial v})+\frac{\partial}{\partial x}(2x\cdot\frac{\partial f}{\partial w})=\frac{\partial^2 f}{\partial v^2}+4x\cdot\frac{\partial^2f}{\partial w^2}+2\cdot\frac{\partial f}{\partial w}\\
&=f_{vv}''(v,w)+4x\cdot f_{vw}''(v,w)+2x\cdot f_{22}''(v,w)+2f_w'(v,w);\\
\frac{\partial^2u}{\partial x\partial y}&=\frac{\partial}{\partial x}(\frac{\partial u}{\partial y})=\frac{\partial}{\partial x}(\frac{\partial f}{\partial v})+\frac{\partial}{\partial x}(2y\cdot\frac{\partial f}{\partial w})\\&=\frac{\partial^2 f}{\partial v^2}+(2x+2y)\cdot\frac{\partial^2 f}{\partial w\partial v}+4y^2\cdot\frac{\partial^2 f}{\partial w^2}\\&=
f''_{u,u}(u,v)+(2x+2y)\cdot f_{uv}''(u,v)+4y^2\cdot f''_{ww}(u,w).
\end{align}
$$

### 9.2.25

Let $v=xy$ and $w=x+y$, then

$$
\begin{align}
\frac{\partial u}{\partial x}&=\frac{\partial u}{\partial t}\cdot\frac{\partial t}{\partial x}=\frac{\partial u}{\partial t}(\frac{\partial t}{\partial v}\cdot\frac{\partial v}{\partial x}+\frac{\partial t}{\partial w}\cdot\frac{\partial w}{\partial x})\\
&=f'_t(\phi_x'(xy,x+y)y+\phi'_y(xy,x+y));
\\\frac{\partial u}{\partial y}&=\frac{\partial u}{\partial t}\cdot\frac{\partial t}{\partial y}=\frac{\partial u}{\partial t}(\frac{\partial t}{\partial v}\cdot\frac{\partial v}{\partial y}+\frac{\partial t}{\partial w}\cdot\frac{\partial w}{\partial y})\\
&=f'_t(\phi_x'(xy,x+y)x+\phi'_y(xy,x+y));
\\\frac{\partial^2 u}{\partial x\partial y}&=\frac{\partial}{\partial x}(\frac{\partial u}{\partial y})=\frac{\partial^2 u}{\partial t^2}\cdot\frac{\partial t}{\partial x}(\frac{\partial t}{\partial v}x+\frac{\partial t}{\partial w})+\frac{\partial u}{\partial t}(\frac{\partial^2t}{\partial v^2}\frac{\partial v}{\partial x}x+\frac{\partial t}{\partial v})\\
&=f_t''(\phi_v'(v,w)y+\phi_w'(v,w))(\phi_v'(v,w)x+\phi_w'(v,w))
\\&+f_t'(\phi_{vv}''(v,w)v+\phi_v'(v,w)).
\end{align}
$$

### 9.2.26

$$
x\cdot\frac{\partial z}{\partial x}-y\cdot\frac{\partial z}{\partial y}=
xf'(xy)y-yf'(xy)x=0.
$$

**The proof is complete.**

### 9.2.28

$$
\begin{align}
\frac{\partial^2 u}{\partial t^2}&=\frac{\partial}{\partial t}(\phi'(x-at)\cdot(-a)+\psi'(x+at)\cdot a)=a^2(\phi''(x-at)+\psi''(x+at))\\
&=a^2\cdot\frac{\partial^2 u}{\partial x^2}.
\end{align}
$$

**The proof is complete.**

### 9.2.30

It is obvious that $\frac{\partial}{\partial x}$ and $\frac{\partial}{\partial y}$ are linear operators, so that their compounds: $\frac{\partial^2}{\partial x^2},\frac{\partial^2}{\partial y^2}$ and $\frac{\partial^2}{\partial x\partial y}$ are also linear operators. So that

$$
\begin{align}
&\frac{\partial^2 u}{\partial x^2}+2\frac{\partial^2 u}{\partial x\partial y}-3\frac{\partial^2u}{\partial y^2}+2\frac{\partial u}{\partial x}+6\frac{\partial u}{\partial y}
\\&=(\frac{\partial}{\partial x}\circ\frac{\partial}{\partial x}+2\frac{\partial}{\partial x}\circ\frac{\partial}{\partial y}-3\frac{\partial}{\partial y}\circ\frac{\partial}{\partial y}+2\frac{\partial}{\partial x}+6\frac{\partial}{\partial y})u
\\&=(\frac{\partial}{\partial x}+3\frac{\partial}{\partial y})\circ(\frac{\partial}{\partial x}-\frac{\partial}{\partial y}+2I)u
\\&=(4\frac{\partial}{\partial\xi})\circ(4\frac{\partial}{\partial\eta}+2I)u=16(\frac{\partial^2u}{\partial\xi\partial\eta}+\frac{1}{2}\frac{\partial u}{\partial\xi}).
\end{align}
$$

So that we have

$$
\frac{\partial^2 u}{\partial x^2}+2\frac{\partial^2 u}{\partial x\partial y}-3\frac{\partial^2u}{\partial y^2}+2\frac{\partial u}{\partial x}+6\frac{\partial u}{\partial y}=0\Leftrightarrow\frac{\partial^2u}{\partial\xi\partial\eta}+\frac{1}{2}\frac{\partial u}{\partial\xi}=0.
$$

**The proof is complete.**

### 9.2.31

Similarly to **9.2.30**, we have

$$
\begin{align}
&\frac{\partial^2u}{\partial x^2}+2\frac{\partial^2u}{\partial x\partial y}\cos x-\frac{\partial^2u}{\partial y^2}\sin^2x-\frac{\partial u}{\partial y}\sin x\\
&=(\frac{\partial}{\partial x}\circ\frac{\partial}{\partial x}+(\cos x-1)\frac{\partial}{\partial y}\circ\frac{\partial}{\partial x}+\frac{\partial}{\partial x}\circ(\cos x+1)\frac{\partial}{\partial y}
\\&+(\cos x-1)\frac{\partial}{\partial y}\circ(\cos x+1)\frac{\partial}{\partial y})u\\&=
(\frac{\partial}{\partial x}+(\cos x-1)\frac{\partial}{\partial y})\circ(\frac{\partial}{\partial x}+(\cos x+1)\frac{\partial}{\partial y})u
\\&=4\frac{\partial^2u}{\partial\xi\partial\eta}.
\end{align}
$$

So that we have

$$
\frac{\partial^2u}{\partial x^2}+2\frac{\partial^2u}{\partial x\partial y}\cos x-\frac{\partial^2u}{\partial y^2}\sin^2x-\frac{\partial u}{\partial y}\sin x=0\Leftrightarrow\frac{\partial^2u}{\partial\xi\partial\eta}=0.
$$

**The proof is complete.**

### 9.2.32

$$
\begin{align}
&6\frac{\partial^2z}{\partial x^2}+\frac{\partial^2z}{\partial x\partial y}-\frac{\partial^2z}{\partial y^2}
\\&=(6\frac{\partial}{\partial x}\circ\frac{\partial}{\partial x}+\frac{\partial}{\partial x}\circ\frac{\partial}{\partial y}-\frac{\partial}{\partial y}\circ\frac{\partial}{\partial y})z
\\&=(2\frac{\partial}{\partial x}+\frac{\partial}{\partial y})\circ(3\frac{\partial}{\partial x}-\frac{\partial}{\partial y})z
\\&=\begin{pmatrix}\frac{\partial}{\partial x}&\frac{\partial}{\partial y}\end{pmatrix}\begin{pmatrix}2\\1\end{pmatrix}\circ\begin{pmatrix}\frac{\partial}{\partial x}&\frac{\partial}{\partial y}\end{pmatrix}\begin{pmatrix}3\\-1\end{pmatrix}z
\end{align}
$$

On the other hand,

$$
\begin{align}
\begin{pmatrix}u\\v\end{pmatrix}=\begin{pmatrix}1&-2\\1&a\end{pmatrix}\begin{pmatrix}x\\y\end{pmatrix}&\Rightarrow\begin{pmatrix}\frac{\partial}{\partial x}&\frac{\partial}{\partial y}\end{pmatrix}=\begin{pmatrix}\frac{\partial}{\partial u}&\frac{\partial}{\partial v}\end{pmatrix}\begin{pmatrix}1&-2\\1&a\end{pmatrix}\\
&\Rightarrow\begin{pmatrix}\frac{\partial}{\partial x}&\frac{\partial}{\partial y}\end{pmatrix}\begin{pmatrix}2&3\\1&-1\end{pmatrix}=\begin{pmatrix}\frac{\partial}{\partial u}&\frac{\partial}{\partial v}\end{pmatrix}\begin{pmatrix}0&5\\a+2&3-a\end{pmatrix}

\end{align}
$$

Therefore since the diagonal of the matrix $\begin{pmatrix}0&5\\a+2&3-a\end{pmatrix}$ should all be zero, we have $a=3$.

### 9.2.33

$$
\begin{align}
z&=z(x,y)=z(x,x^2)+\int_{x^2}^y\frac{\partial z}{\partial y}\mathrm{d}y
\\&=1+x^2(y-x^2)+y^2-x^4
\\&=-2x^4+y^2+x^2y+1.
\end{align}
$$

#### 9.2.35

$$
\begin{align}
\mathrm{d}u&=\frac{\partial u}{\partial x}\mathrm{d}x+\frac{\partial u}{\partial y}\mathrm{d}y\Rightarrow\frac{\partial u}{\partial y}=\frac{\mathrm{d}x}{\mathrm{d}y}(\frac{\mathrm{d}u}{\mathrm{d}x}-\frac{\partial u}{\partial x})\\&\Rightarrow u'_y(x,2x)=\frac{\mathrm{d}y}{\mathrm{d}x}(\frac{\mathrm{d}u(x,2x)}{\mathrm{d}x}-u'_x(x,2x))
\\&=\frac{1-x^2}{2};
\\\mathrm{d}\frac{\partial u}{\partial x}&=\frac{\partial^2u}{\partial x^2}\mathrm{d}x+\frac{\partial^2u}{\partial y\partial x}\mathrm{d}y\Rightarrow\frac{\mathrm{d}}{\mathrm{d}x}\frac{\partial u}{\partial x}=\frac{\partial^2u}{\partial x^2}+\frac{\partial^2u}{\partial y\partial x}\frac{\mathrm{d}y}{\mathrm{d}x}\\&\Rightarrow2x=u''_{xx}(x,2x)+2u''_{yx}(x,2x);
\\\mathrm{d}\frac{\partial u}{\partial y}&=\frac{\partial^2u}{\partial x\partial y}\mathrm{d}x+\frac{\partial^2u}{\partial y^2}\mathrm{d}y\Rightarrow\frac{\mathrm{d}}{\mathrm{d}x}\frac{\partial u}{\partial y}=\frac{\partial^2u}{\partial x\partial y}+\frac{\partial^2u}{\partial y^2}\frac{\mathrm{d}y}{\mathrm{d}x}\\&\Rightarrow-x=u''_{xy}(x,2x)+2u''_{yy}(x,2x);
\end{align}
$$

From the results above and the fact that $u''_{xx}=u''_{yy}$, we have that

$$
\left\{
\begin{array}{l}
u''_{xx}(x,2x)=u''_{yy}(x,2x)=-\frac{4}{3}x\\
u''_{xy}(x,2x)=\frac{5}{3}x
\end{array}.
\right.
$$

### 9.2.36

$(2)$

$$
\begin{align}
\mathrm{d}u&=\triangledown f\begin{pmatrix}\mathrm{d}\xi\\\mathrm{d}\eta\end{pmatrix}=\triangledown f\begin{pmatrix}\triangledown \xi\\\triangledown \eta\end{pmatrix}\begin{pmatrix}\mathrm{d}x\\\mathrm{d}y\end{pmatrix}
=\begin{pmatrix}f'_x&f'_y\end{pmatrix}\begin{pmatrix}x&y\\\frac{1}{y}&-\frac{x}{y^2}\end{pmatrix}\begin{pmatrix}\mathrm{d}x\\\mathrm{d}y\end{pmatrix}
\\&=(xf'_x+\frac{1}{y}f_y')\mathrm{d}x+(yf_x'-\frac{x}{y^2}f_y')\mathrm{d}y.
\end{align}
$$

$(4)$

$$
\begin{align}
\mathrm{d}u&=\triangledown f\begin{pmatrix}\mathrm{d}x\\\mathrm{d}\xi\\\mathrm{d}\eta\end{pmatrix}
=\triangledown f\begin{pmatrix}\triangledown x\\\triangledown\xi\\\triangledown\eta\end{pmatrix}\begin{pmatrix}\mathrm{d}x\\\mathrm{d}y\end{pmatrix}=\begin{pmatrix}f_x'&f_y'&f_z'\end{pmatrix}\begin{pmatrix}1&0\\2x&2y\\2x&2y\end{pmatrix}\begin{pmatrix}\mathrm{d}x\\\mathrm{d}y\end{pmatrix}\\&=(f_x'+2xf_y'+2xf_z')\mathrm{d}x+(2yf_y'+2yf_z')\mathrm{d}y.
\end{align}
$$

### 9.2.37

Since that $x=r\cos\phi\sin\theta,y=r\sin\phi\sin\theta,z=r\cos\theta$, we have

$$
\begin{align}
\begin{pmatrix}\frac{\partial}{\partial r}\\\frac{\partial}{\partial \phi}\\\frac{\partial}{\partial\theta}\end{pmatrix}&=
\begin{bmatrix}\frac{\partial x}{\partial r}&\frac{\partial y}{\partial r}&\frac{\partial z}{\partial r}\\\frac{\partial x}{\partial \phi}&\frac{\partial y}{\partial\phi}&\frac{\partial z}{\partial\phi}\\\frac{\partial x}{\partial\theta}&\frac{\partial y}{\partial\theta}&\frac{\partial z}{\partial\theta}\end{bmatrix}
\begin{pmatrix}\frac{\partial}{\partial x}\\\frac{\partial}{\partial y}\\\frac{\partial}{\partial z}\end{pmatrix}\\&=\begin{bmatrix}\cos\phi\sin\theta&\sin\phi\sin\theta&\cos\theta\\-r\sin\phi\sin\theta&r\cos\phi\cos\theta&0\\r\cos\phi\cos\theta&r\sin\phi\cos\theta&-r\sin\theta\end{bmatrix}\begin{pmatrix}\frac{\partial}{\partial x}\\\frac{\partial}{\partial y}\\\frac{\partial}{\partial z}\end{pmatrix}

\end{align}
$$

denote which by $A$ and

$$
\begin{pmatrix}\hat{r}&\hat{\phi}&\hat{\theta}\end{pmatrix}=\begin{pmatrix}\hat{x}&\hat{y}&\hat{z}\end{pmatrix}\begin{bmatrix}\sin\theta\cos\phi&-\sin\phi&\cos\theta\cos\phi\\\sin\theta\sin\phi&\cos\phi&\cos\theta\sin\phi\\\cos\theta&0&-\sin\theta\end{bmatrix}
$$

denote which by $B$. Then,

$$
\begin{align}
\triangledown&=\begin{pmatrix}\hat{x}&\hat{y}&\hat{z}\end{pmatrix}\begin{pmatrix}\frac{\partial}{\partial x}\\\frac{\partial}{\partial y}\\\frac{\partial}{\partial z}\end{pmatrix}=\begin{pmatrix}\hat{r}&\hat{\phi}&\hat{\theta}\end{pmatrix}B^{-1}A^{-1}\begin{pmatrix}\frac{\partial}{\partial r}\\\frac{\partial}{\partial \phi}\\\frac{\partial}{\partial\theta}\end{pmatrix}
\\&=\begin{pmatrix}\hat{r}&\hat{\phi}&\hat{\theta}\end{pmatrix}\cdot
\begin{bmatrix}\sin\theta\cos\phi&\sin\theta\sin\phi&\cos\theta\\-\sin\phi&\cos\theta&0\\\cos\theta\sin\phi&\cos\theta\sin\phi&-\sin\theta\end{bmatrix}\\&\cdot
\begin{bmatrix}\sin\theta\cos\phi&-\frac{\sin\phi}{r\sin\theta}&\frac{1}{r}\cos\theta\cos\phi\\\sin\theta\sin\phi&\frac{\cos\phi}{r\sin\theta}&\frac{1}{r}\cos\theta\sin\phi\\\cos\theta&0&-\frac{1}{r}\sin\theta\end{bmatrix}
\cdot\begin{pmatrix}\frac{\partial}{\partial r}\\\frac{\partial}{\partial \phi}\\\frac{\partial}{\partial\theta}\end{pmatrix}
\\&=\begin{pmatrix}\hat{r}&\hat{\phi}&\hat{\theta}\end{pmatrix}
\begin{bmatrix}1&0&0\\0&\frac{1}{r\sin\theta}&0\\0&0&\frac{1}{r}\end{bmatrix}
\begin{pmatrix}\frac{\partial}{\partial r}\\\frac{\partial}{\partial \phi}\\\frac{\partial}{\partial\theta}\end{pmatrix}
\\&=\frac{\partial}{\partial r}\hat{r}+\frac{1}{r\sin\theta}\frac{\partial}{\partial\phi}\hat{\phi}+\frac{1}{r}\frac{\partial}{\partial\theta}\hat{\theta}.
\end{align}
$$

So that

$$
\triangledown\cdot f=\text{div}f=\frac{1}{r^2}\frac{\partial}{\partial r}(r^2f_r)+\frac{1}{r\sin\theta}\frac{\partial}{\partial\theta}\sin\theta f_{\theta}+\frac{1}{r\sin\theta}\frac{\partial}{\partial\phi}f_{\phi}
$$

from which we can get

$$
\triangledown^2f=\triangledown\cdot\triangledown\cdot f=\frac{1}{r^2}\cdot\frac{\partial}{\partial r}(r^2\frac{\partial f}{\partial r})+\frac{1}{r^2\sin\theta}\frac{\partial}{\partial\theta}(\sin\theta\frac{\partial f}{\partial\theta})+\frac{1}{r^2\sin^2\theta}\frac{\partial^2f}{\partial^2\phi}.
$$

### 9.3.2

$(1)$ Since that

$$
F(x,y)=\sin xy-e^{xy}-x^2y,
$$

we have

$$
\frac{\mathrm{d}y}{\mathrm{d}x}=-\frac{F_x'}{F_y'}=-\frac{y\cos xy-ye^{ey}-2xy}{x\cos xy-xe^{xy}=x^2}.
$$

$(3)$ Since that

$$
F(x,y)=\frac{\ln x}{x}-\frac{\ln y}{y},
$$

we have

$$
\begin{align}
\frac{\mathrm{d}y}{\mathrm{d}x}&=-\frac{F'_x}{F'_y}=\frac{y^2(\ln x-1)}{x^2(\ln y-1)}\\\Rightarrow\frac{\mathrm{d}^2y}{\mathrm{d}x^2}&=\frac{\mathrm{d}}{\mathrm{d}x}(\frac{y^2(\ln x-1)}{x^2(\ln y-1)})\\
&=\frac{y^2}{x^4}\cdot\frac{(\ln x-1)^2(2\ln y-3)y-(\ln y-1)^2(2\ln x-3)x}{(\ln y-1)^3}.
\end{align}
$$

$(5)$ Since that

$$
F(x,y,z)=\frac{x}{z}-\ln\frac{z}{y},
$$

we have

$$
\begin{align}
\frac{\partial z}{\partial x}&=-\frac{F'_x}{F'_z}=-\frac{\frac{1}{z}}{-\frac{1}{z}-\frac{x}{z^2}}=\frac{z}{x+z};
\\\frac{\partial z}{\partial y}&=-\frac{\frac{\partial F(xz,yz)}{\partial y}}{\frac{\partial F(xz,yz)}{\partial z}}=\frac{zF'_{yz}(xz,yz)}{xF'_{xz}(xz,yz)+yF'_{yz}(xz,yz)}.
\end{align}
$$

### 9.3.5

$$
\begin{align}
1+xy=k(x-y)&\Rightarrow x\mathrm{d}x+y\mathrm{d}x=k(\mathrm{d}x-\mathrm{d}y)\\&\Rightarrow(x-y)(x\mathrm{d}x+y\mathrm{d}x)=k(x-y)(\mathrm{d}x-\mathrm{d}y)
\\&\Rightarrow\frac{\mathrm{d}x}{1+x^2}=\frac{\mathrm{d}y}{1+y^2}.
\end{align}
$$

**The proof is complete.**

### 9.3.7

Let $u=cx-az,v=cy-bz$, then

$$
\begin{align}
a\frac{\partial z}{\partial x}+b\frac{\partial z}{\partial y}&=-a\cdot\frac{c\phi_u'}{-a\phi_u'-b\phi_v'}-b\cdot\frac{c\phi_v'}{-a\phi_u'-b\phi_v'}\\
&=\frac{ac\phi_u'+bc\phi_v'}{a\phi_u'+b\phi_v'}
\\&=c.
\end{align}
$$

**The proof is complete.**

### 9.3.9

First we have that

$$
\left\{\begin{array}{l}
y=f(x+t)\\y+g(x,t)=0
\end{array}\right.\Rightarrow f(x+t)+g(x,t)=0\Rightarrow\frac{\mathrm{d}t}{\mathrm{d}x}=-\frac{F_x'}{F_t'}=-\frac{f'(x+t)+g'_x(x,t)}{f'(x+t)+g'_t(x,t)}
$$

So that

$$
\begin{align}
\frac{\mathrm{d}y}{\mathrm{d}x}&=\frac{\mathrm{d}f(x+t)}{\mathrm{d}x}
=f'(x+t)(1+\frac{\mathrm{d}t}{\mathrm{d}x})
\\&=f'(x+t)(1-\frac{f'(x+t)+g'_x(x,t)}{f'(x+t)+g'_t(x,t)}).
\end{align}
$$

### 9.3.11

$(1)$

$$
\frac{\partial(u,v)}{\partial(x,y)}=(\frac{\partial(F,G)}{\partial(u,v)})^{-1}\cdot\frac{\partial(F,G)}{\partial(x,y)}=\begin{vmatrix}2u&2v\\1&1\end{vmatrix}^{-1}\cdot\begin{vmatrix}2x&2y\\1&1\end{vmatrix}=\frac{x-y}{u-v}.
$$

$(2)$

$$
\frac{\partial(u,v)}{\partial(x,y)}=(\frac{\partial(F,G)}{\partial(u,v)})^{-1}\cdot\frac{\partial(F,G)}{\partial(x,y)}=\begin{vmatrix}x&-y\\y&x\end{vmatrix}^{-1}\cdot\begin{vmatrix}u&-v\\v&u\end{vmatrix}=\frac{u^2+v^2}{x^2+y^2}.
$$

$(3)$

$$
\begin{align}
\frac{\partial(u,v)}{\partial(x,y)}&=(\frac{\partial(F,G)}{\partial(u,v)})^{-1}\cdot\frac{\partial(F,G)}{\partial(x,y)}=\begin{vmatrix}xf_x'-1&f_y'\\g_x'&2yvg_y'-1\end{vmatrix}^{-1}\cdot\begin{vmatrix}uf_x'&f_y'\\-g_x'&v^2g_y'\end{vmatrix}
\\&=\frac{uv^2f_x'g_y'+f'_yg'_x}{(xf'_x-1)(2yvg_y'-1)-f_y'g_x'}.
\end{align}
$$

### 9.3.14

Let $G(x,y,z)=xf(x+y)-z$, then

$$
\left\{
\begin{array}{l}F(x,y,z)=0\\G(x,y,z)=0\end{array}
\right.\Rightarrow\frac{\mathrm{d}z}{\mathrm{d}x}=\frac{\frac{\partial(F,G)}{\partial(x,y)}}{\frac{\partial(F,G)}{\partial(y,z)}}=\frac{\begin{vmatrix}F'_x&F'_y\\f+xf'&xf'\end{vmatrix}}{\begin{vmatrix}F'_y&F'_z\\xf'&-1\end{vmatrix}}=\frac{(f+xf')F_y-xf'F_x}{F_y'+xf'F'_z}.
$$

### 9.3.16

Let $F(x,y,z,t,u)=f(x,y,z,t)-u$, then we have

$$
\left\{
\begin{array}{l}F(x,y,z,t,u)=0\\g(y,z,t)=0\\h(z,t)=0\end{array}
\right.
$$

So that

$$
\frac{\partial u}{\partial x}=-\frac{\frac{\partial(F,g,h)}{\partial(x,z,t)}}{\frac{\partial(F,g,h)}{\partial(u,z,t)}}=-\frac{\begin{vmatrix}f'_x&f'_z&f'_t\\0&g'_z&g'_t\\0&h'_z&h'_t\end{vmatrix}}{\begin{vmatrix}-1&f'_z&f'_t\\0&g'_z&g'_t\\0&h'_z&h'_t\end{vmatrix}}=f'_x;
$$

Similarly we can get

$$
\frac{\partial u}{\partial y}=f'_y-g'_y\cdot\frac{f'_zh'_t-f'_th'_z}{g'_zh'_t-g'_th'_t}.
$$

Last modification：March 28, 2022

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Maths Analysis II - Homework #9 20220315

Bi_Gu • 2022 年 03 月 21 日

# Maths Analysis II

## Homework #9 20220315

### 9.2.19

$(3)$

$(4)$

### 9.2.20

$(2)$

$(4)$

Let $v=x+y+z$ and $w=x^2+y^2+z^2$, then

### 9.2.25

Let $v=xy$ and $w=x+y$, then

### 9.2.26

$$
x\cdot\frac{\partial z}{\partial x}-y\cdot\frac{\partial z}{\partial y}=
xf'(xy)y-yf'(xy)x=0.
$$

**The proof is complete.**

### 9.2.28

**The proof is complete.**

### 9.2.30

So that we have

**The proof is complete.**

### 9.2.31

Similarly to **9.2.30**, we have

So that we have

**The proof is complete.**

### 9.2.32

On the other hand,

\end{align}
$$

Therefore since the diagonal of the matrix $\begin{pmatrix}0&5\\a+2&3-a\end{pmatrix}$ should all be zero, we have $a=3$.

### 9.2.33

$$
\begin{align}
z&=z(x,y)=z(x,x^2)+\int_{x^2}^y\frac{\partial z}{\partial y}\mathrm{d}y
\\&=1+x^2(y-x^2)+y^2-x^4
\\&=-2x^4+y^2+x^2y+1.
\end{align}
$$

#### 9.2.35

From the results above and the fact that $u''_{xx}=u''_{yy}$, we have that

$$
\left\{
\begin{array}{l}
u''_{xx}(x,2x)=u''_{yy}(x,2x)=-\frac{4}{3}x\\
u''_{xy}(x,2x)=\frac{5}{3}x
\end{array}.
\right.
$$

### 9.2.36

$(2)$

$(4)$

### 9.2.37

Since that $x=r\cos\phi\sin\theta,y=r\sin\phi\sin\theta,z=r\cos\theta$, we have

\end{align}
$$

denote which by $A$ and

denote which by $B$. Then,

So that

from which we can get

### 9.3.2

$(1)$ Since that

$$
F(x,y)=\sin xy-e^{xy}-x^2y,
$$

we have

$$
\frac{\mathrm{d}y}{\mathrm{d}x}=-\frac{F_x'}{F_y'}=-\frac{y\cos xy-ye^{ey}-2xy}{x\cos xy-xe^{xy}=x^2}.
$$

$(3)$ Since that

$$
F(x,y)=\frac{\ln x}{x}-\frac{\ln y}{y},
$$

we have

$(5)$ Since that

$$
F(x,y,z)=\frac{x}{z}-\ln\frac{z}{y},
$$

we have

### 9.3.5

**The proof is complete.**

### 9.3.7

Let $u=cx-az,v=cy-bz$, then

**The proof is complete.**

### 9.3.9

First we have that

So that

### 9.3.11

$(1)$

$(2)$

$(3)$

### 9.3.14

Let $G(x,y,z)=xf(x+y)-z$, then

### 9.3.16

Let $F(x,y,z,t,u)=f(x,y,z,t)-u$, then we have

$$
\left\{
\begin{array}{l}F(x,y,z,t,u)=0\\g(y,z,t)=0\\h(z,t)=0\end{array}
\right.
$$

So that

Similarly we can get

$$
\frac{\partial u}{\partial y}=f'_y-g'_y\cdot\frac{f'_zh'_t-f'_th'_z}{g'_zh'_t-g'_th'_t}.
$$

Maths Analysis II - Homework #9 20220315

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