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cfsv:chapter-4 [2012/06/25 12:44] (current)
dcs created
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 +====The Calculus of Functions of Several Variables====
  
 +===Answers for selected problems===
 +
 +**Chapter 4**
 +
 +----
 +
 +**Section 4.1**
 +
 +2. The surface is the graph of $ f $.
 +
 +3. $ F(s, t) = (s, t, s^2 + t^2) $
 +
 +6. $ \{\mathbf{x} \ : \ \mathbf{x} \in \mathbb{R}^n,​ \mathbf{x} \ne \mathbf{0}\} $
 +
 +----
 +
 +**Section 4.2**
 +
 +1. (a) $ \displaystyle{A(x,​y) = \begin{bmatrix}2 & 4\\ 6 & 3\end{bmatrix} \begin{bmatrix}x - 1\\ y - 2\end{bmatrix} + \begin{bmatrix}5\\ 6\end{bmatrix} = \begin{bmatrix}2x + 4y - 5\\ 6x + 3y - 6\end{bmatrix}}$
 +
 +%%(c)%% ​ $ \displaystyle{A(s,​t) = \begin{bmatrix}-6 & \phantom{-}1\\ \phantom{-}1 & -1\\ 36 & -24\\ -1 & \phantom{-}4\end{bmatrix} \begin{bmatrix}s + 1\\ t - 3\end{bmatrix} + \begin{bmatrix}\phantom{-}6\\ -4\\ -36\\ \phantom{-}13\end{bmatrix} = \begin{bmatrix}-6s + t - 3\\ s - t\\ 36s - 24t + 72\\ -2 + 4t\end{bmatrix}}$
 +
 +2. (a)  $ x = -2s + \pi$, $ y = t$, $ z = t$
 +
 +%%(c)%% $ \displaystyle{x = -\frac{1}{\sqrt{2}}\left(s - \frac{\pi}{2}\right)}$
 +
 +$ \displaystyle{y = \frac{1}{\sqrt{2}}\left(t - \frac{\pi}{4}\right) + \frac{1}{\sqrt{2}}}$
 +
 +$ \displaystyle{z = -\frac{1}{\sqrt{2}}\left(t - \frac{\pi}{2}\right) + \frac{1}{\sqrt{2}}}$
 +
 +(e) $ \displaystyle{x = -(2\sqrt{2} + 1)\left(s - \frac{3\pi}{4}\right) + \left(t - \frac{\pi}{4}\right) - 2\sqrt{2} - 1}$
 +
 +$ \displaystyle{y = -(2\sqrt{2} + 1)\left(s - \frac{3\pi}{4}\right) - \left(t - \frac{\pi}{4}\right) + 2\sqrt{2} + 1}$
 +
 +$ \displaystyle{y = \sqrt{2}\left(t - \frac{\pi}{4}\right) + \sqrt{2}}$
 +
 +4. (a) $ \displaystyle{D(f \circ g)(1, -2) =  \begin{bmatrix}720 & -468\\ -8 & 15\end{bmatrix}}$
 +
 +%%(c)%% $ \displaystyle{D(f \circ g)(1, -2, 3) =  \begin{bmatrix}92 & -89 & 3\\ 10920 & -9880 & 1040\\ -16 & 17 & 1\end{bmatrix}}$
 +
 +6. $ \displaystyle{\frac{\partial x}{\partial s} = (2uv)(-2s) + (u^2)\left(-\frac{4t}{s^2}\right)}$
 +
 +$ \displaystyle{\frac{\partial x}{\partial t} = (2uv)(8t) + (u^2)\left(\frac{4}{s}\right)}$
 +
 +$ \displaystyle{\frac{\partial y}{\partial s} = (3)(-2s) + (-1)\left(-\frac{4t}{s^2}\right)}$
 +
 +$ \displaystyle{\frac{\partial y}{\partial t} = (3)(8t) + (-1)\left(\frac{4}{s}\right)}$
 +
 +8. (a)  $ \displaystyle{\left. \frac{\partial T}{\partial r}\right|_{r=4,​ \theta=\frac{\pi}{6}} = \frac{80}{17\sqrt{17}}}$,​ $ \displaystyle{\left. \frac{\partial T}{\partial \theta}\right|_{r=4,​ \theta=\frac{\pi}{6}} = 0}$
 +
 +(b) The level curves of $ T$ are circles.
 +
 +----
 +
 +**Section 4.3**
 +
 +1. (a) $ \displaystyle{\int_C F \cdot ds = \frac{104}{5}}$
 +
 +%%(c)%% $ \displaystyle{\int_C F \cdot ds = -\frac{384}{5}}$
 +
 +2. (a) $ \displaystyle{\int_C 3xdx + 4ydy = 0}$
 +
 +3. (a) $ \displaystyle{\int_C 3xdx + 4ydy + zdz = 2\pi^2}$
 +
 +4. (a) $ \displaystyle{\int_C x^2ydx + (3y+x)dy = 0}$
 +
 +5. $ \displaystyle{\int_C F \cdot ds = -6\pi}$
 +
 +6. (a) $ \displaystyle{\int_C 3ydx = -\frac{27\pi}{2}}$
 +
 +7. $ \displaystyle{\int_C \frac{x}{x^2+y^2}dx + \frac{y}{x^2+y^2}dy = \frac{1}{2}\log(13)}$
 +
 +----
 +
 +**Section 4.4**
 +
 +1. (a) $ \displaystyle{\int_{\partial D} 2xydx + 3x^2dy = 80}$
 +
 +2. (a) $ \displaystyle{\int_{\partial D} 2xy^2dx + 4xdy = \frac{16}{3}}$
 +%%(c)%% $ \displaystyle{\int_{\partial D} ydx - xdy = -8}$
 +
 +4. $ \displaystyle{\frac{3}{8}\pi a^2}$
 +
 +5. $ \displaystyle{\frac{\pi}{8}}$
 +
 +6. $ \displaystyle{\frac{9\pi}{2}}$
 +
 +----