习题1.7

第 1 题

求下列曲面在给定点的切平面方程和法线方程.

(1) \(z = x^2 + y^2\), 点 \(P(1, 2, 5)\);

(3) \((2a^2 - z^2)x^2 = a^2y^2\), 点 \(P(a, a, a)(a \neq 0)\);

(5) \(\begin{cases}x = u\cos v, \\ y = u\sin v, \\ z = av,\end{cases}\) 点 \((u, v) = (u_0, v_0)\);

解

(1) 因为 \(\dfrac{\partial z}{\partial x} = 2x, \dfrac{\partial z}{\partial y} = 2y\), 所以切平面的表达式为 \(z = z_0 + 2x_0(x - x_0) + 2y_0(y - y_0)\). 将 \(P(1, 2, 5)\) 代入得 \(z = 2x + 4y - 5\). 法线的表达式为 \(\dfrac{x - 1}{2} = \dfrac{y - 2}{4} = \dfrac{z - 5}{-1}\).

(3) 记 \(F(x, y, z) = (2a^2 - z^2)x^2 - a^2y^2\). 则 \(\dfrac{\partial F}{\partial x} = 2x(2a^2 - z^2), \dfrac{\partial F}{\partial y} = -2a^2y, \dfrac{\partial F}{\partial z} = -2x^2z\). 代入 \(P(a, a, a)\) 得切平面表达式为 \(x - y - z + a = 0\), 法线表达式为 \(x - a = a - y = a - z\).

(5)

\[ \begin{aligned} A &= \begin{vmatrix} \dfrac{\partial y}{\partial u} & \dfrac{\partial y}{\partial v} \\ \dfrac{\partial z}{\partial u} & \dfrac{\partial z}{\partial v} \end{vmatrix} = \begin{vmatrix} \sin v & u\cos v \\ 0 & a \end{vmatrix} = a\sin v\\ B &= \begin{vmatrix} \dfrac{\partial z}{\partial u} & \dfrac{\partial z}{\partial v} \\ \dfrac{\partial x}{\partial u} & \dfrac{\partial x}{\partial v} \end{vmatrix} = \begin{vmatrix} 0 & a \\ \cos v & -u\sin v \end{vmatrix} = -a\cos v\\ C &= \begin{vmatrix} \dfrac{\partial x}{\partial u} & \dfrac{\partial x}{\partial v} \\ \dfrac{\partial y}{\partial u} & \dfrac{\partial y}{\partial v} \end{vmatrix} = \begin{vmatrix} \cos v & -u\sin v \\ \sin v & u\cos v \end{vmatrix} = u \end{aligned} \]

代入 \((u_0, v_0)\) 可得切平面为 \(a\sin v_0(x - u_0\cos v_0) - a\cos v_0(y - u_0\sin v_0) + u_0(z - av_0) = 0\), 法线为 \(\dfrac{x - u_0\cos v_0}{a\sin v_0} = \dfrac{y - u_0\sin v_0}{-a\cos v_0} = \dfrac{z - av_0}{u_0}\).

第 2 题

在椭球面 \(\dfrac{x^2}{a^2} + \dfrac{y^2}{b^2} + \dfrac{z^2}{c^2} = 1\) 上求一点 \(P\), 使得过 \(P\) 点的法线与坐标轴正方向成等角.

解

记 \(F(x, y, z) = \dfrac{x^2}{a^2} + \dfrac{y^2}{b^2} + \dfrac{z^2}{c^2} - 1\). 则 \(\dfrac{\partial F}{\partial x} = \dfrac{2x}{a^2}, \dfrac{\partial F}{\partial y} = \dfrac{2y}{b^2}, \dfrac{\partial F}{\partial z} = \dfrac{2z}{c^2}\). 因此在 \((x_0, y_0, z_0)\) 处的法向量为 \(\left(\dfrac{2x_0}{a^2}, \dfrac{2y_0}{b^2}, \dfrac{2z_0}{c^2}\right)\). 由于与三个坐标轴的正方向成等角, 因此有 \(\dfrac{2x_0}{a^2} = \dfrac{2y_0}{b^2} = \dfrac{2z_0}{c^2}\). 结合 \(F(x_0, y_0, z_0) = 0\) 解得满足条件的点 \(P\) 有两个, \(P_1 = \left(\dfrac{a^2}{\sqrt{a^2 + b^2 + c^2}}, \dfrac{b^2}{\sqrt{a^2 + b^2 + c^2}}, \dfrac{c^2}{\sqrt{a^2 + b^2 + c^2}}\right), P_2 = \left(-\dfrac{a^2}{\sqrt{a^2 + b^2 + c^2}}, -\dfrac{b^2}{\sqrt{a^2 + b^2 + c^2}}, -\dfrac{c^2}{\sqrt{a^2 + b^2 + c^2}}\right)\).

第 3 题

求曲面 \(x^2 + 2y^2 + 3z^2 = 21\) 上平行于 \(x + 4y + 6z = 0\) 的切平面.

解

设切点为 \(P(x_0, y_0, z_0)\). 记 \(F(x, y, z) = x^2 + 2y^2 + 3z^2 = 21\). 因为 \(\dfrac{\partial F}{\partial x} = 2x, \dfrac{\partial F}{\partial y} = 4y, \dfrac{\partial F}{\partial z} = 6z\), 所以切平面的法向量为 \((2x_0, 4y_0, 6z_0)\). 因为改平面平行于 \(x + 4y + 6z = 0\), 所以 \(\dfrac{2x_0}{1} = \dfrac{4y_0}{4} = \dfrac{6z_0}{6}\). 解得法平面为 \(x + 4y + 6z = 21\) 或 \(x + 4y + 6z = -21\).

第 4 题

(1) 曲面 \(xyz = a^3\) 上的任一点的切平面与坐标平面围成的四面体的体积为定值;

(3) 曲面 \(z = x^2 + y^2\) 与直线 \(l:\begin{cases}x + 2z = 1, \\ y + 2z = 2\end{cases}\) 垂直的切平面;

(5) 设 \(f\) 可微, 曲面 \(z = yf\left(\dfrac{x}{y}\right)\) 的所有切平面相交于一个定点.

证明

(1) 设 \(F(x, y, z) = xyz - a^3\). 则 \(\dfrac{\partial F}{\partial x} = yz, \dfrac{\partial F}{\partial y} = xz, \dfrac{\partial F}{\partial z} = xy\). 假设切点为 \((x_0, y_0, z_0)\), 则切平面的表达式为 \(y_0z_0(x - x_0) + x_0z_0(y - y_0) + x_0y_0(z - z_0) = 0\) 与坐标轴交点分别为 \((x_0, 0, 0), (0, y_0, 0), (0, 0, z_0)\). 体积即为 \(\dfrac{1}{6}x_0y_0z_0 = \dfrac{1}{12}a^3\) 为定值.

(3) 由于 \(\dfrac{\partial z}{\partial x} = 2x, \dfrac{\partial z}{\partial y} = 2y\), 所以设切点为 \((x_0, y_0, z_0)\), 则法向量为 \((2x_0, 2y_0, -1)\). 而该直线能被表示为 \(\dfrac{x - 1}{2} = \dfrac{y - 2}{2} = \dfrac{z}{-1}\), 因此有 \(2x_0 = 2y_0 = 2\). 解得切点为 \((1, 1, 2)\). 所以切平面为 \(2(x - 1) + 2(y - 1) - (z - 2) = 0\).

(5) \(\dfrac{\partial z}{\partial x} = yf'\left(\dfrac{x}{y}\right)\cdot \dfrac{1}{y} = f'\left(\dfrac{x}{y}\right), \dfrac{\partial z}{\partial y} = f\left(\dfrac{x}{y}\right) + y\cdot f'\left(\dfrac{x}{y}\right)\cdot\left(-\dfrac{x}{y^2}\right) = f\left(\dfrac{x}{y}\right) - \dfrac{xf'\left(\dfrac{x}{y}\right)}{y}\). 因此在 \((x_0, y_0)\) 处的切平面的表达式为 \(f'\left(\dfrac{x_0}{y_0}\right)(x - x_0) + [f\left(\dfrac{x_0}{y_0}\right) - \dfrac{x_0f'\left(\dfrac{x_0}{y_0}\right)}{y_0}](y - y_0) + y_0f'\left(\dfrac{x_0}{y_0}\right) - z = 0\), 过定点 \((0, 0, 0)\).

第 5 题

求曲线 \(l:\begin{cases}x^2 + y^2 + z^2 = 6, \\ x + y + z = 0\end{cases}\) 在点 \(P(1, -2, 1)\) 处的切线方程与法平面方程.

解

记 \(F(x, y, z) = x^2 + y^2 + z^2 - 6, G(x, y, z) = x + y + z\). 则 \(\dfrac{\partial F}{\partial x} = 2x, \dfrac{\partial F}{\partial y} = 2y, \dfrac{\partial F}{\partial z} = 2z\), \(\dfrac{\partial G}{\partial x} = \dfrac{\partial G}{\partial y} = \dfrac{\partial G}{\partial z} = 1\). 因此代入 \(P(1, -2, 1)\), 有切线方程满足

\[ \begin{cases} 2(x - 1) - 4(y + 2) + 2(z - 1) = 0, \\ (x - 1) + (y + 2) + (z - 1) = 0. \end{cases} \]

法平面的方程则为 \(-(x - 1) + (z - 1) = 0\), 即为 \(x - z = 0\).

第 6 题

证明: 螺旋线 \(\begin{cases}x = a\cos t, \\ y = a\sin t, \\ z = bt\end{cases}\) 的切线与 \(z\) 轴形成定角.

证明

设切点在 \(t = t_0\) 处. 则切向量为 \(\mathbf{n} = (-a\sin t_0, a\cos t_0, b)\), 记沿着 \(z\) 轴正方向的单位向量为 \(\mathbf{k} = (0, 0, 1)\). 该切线与 \(z\) 轴正方向所夹角度 \(\theta\) 满足 \(\cos\theta = \dfrac{\mathbf{n}\cdot \mathbf{k}}{\left|\mathbf{n}\right|} = \dfrac{b}{\sqrt{a^2 + b^2}}\) 为定值.

第 7 题

已知函数 \(f\) 可微, 若 \(T\) 为曲面 \(S:f(x, y, z) = 0\) 在点 \(P(x_0, y_0, z_0)\) 处的切平面, \(l\) 为 \(T\) 上任意一条过 \(P\) 的曲线, 求证: 在 \(S\) 上存在一条曲线, 该曲线在 \(P\) 处的切线恰好为 \(l\).

证明

不妨设要求的曲线为 \(L\). 现在构造 \(L\) 如下: 过 \(P(x_0, y_0, z_0)\) 有法向量 \(\mathbf{n}\), 由于 \(\mathbf{n}\) 和 \(l\) 相互垂直, 因此张成了一个平面 \(Q\). 取 \(L\) 为 \(Q\) 和 \(S\) 的交线, 则 \(L\) 与平面 \(Q\) 相切, 且它在 \(P\) 处的切线恰好为 \(l\). 因此命题得证.