매개변수 곡면

A normal vector to the surface $x^2 + 9y^2 + 4z^2 = 17$ at the point $(x,y,z)$ is $( 2x\,,\, 18y\,,\,8z)$. A normal vector to the plane $x - 8z = 0$ is $( 1\,,\,0\,,\,-8)$. So we want $( 2x\,,\, 18y\,,\,8z)$ to be parallel to $( 1\,,\,0\,,\,-8)$, i.e. to be a nonzero constant times $( 1\,,\,0\,,\,-8)$. This is the case whenever $y=0$ and $z=-2x$ with $x\ne 0$. In addition, we want $(x,y,z)$ to lie on the surface $x^2 + 9y^2 + 4z^2 = 17$. So we want $y=0$, $z=-2x$ and

$$\begin{aligned} 17= x^2 + 9y^2 + 4z^2 =x^2 +4(-2x)^2=17x^2 \implies x=\pm 1 \end{aligned}$$

So the allowed points are $\pm(1,0,-2)$.

Let $(x_0,y_0,z_0)$ be a point on the hyperboloid $z^2=4x^2+y^2-1$ where the tangent plane is parallel to the plane $2x-y+z=0$. A normal vector to the plane $2x-y+z=0$ is $( 2,-1,1)$. Because the hyperboloid is $G(x,y,z)=4x^2+y^2-z^2-1$ and ${ \boldsymbol{\nabla} } G(x,y,z) = ( 8x,2y,-2z)$, a normal vector to the hyperboloid at $(x_0,y_0,z_0)$ is ${ \boldsymbol{\nabla} } G(x_0,y_0,z_0)=( 8x_0,2y_0,-2z_0)$. So $(x_0,y_0,z_0)$ satisfies the required conditions if and only if there is a nonzero $t$ obeying

$$\begin{aligned} &( 8x_0,2y_0,-2z_0) =t( 2,-1,1) \text{ and } z_0^2=4x_0^2+y_0^2-1\\ &\iff x_0=\frac{t}{4},\ y_0=z_0=-\frac{t}{2}\text{ and } z_0^2=4x_0^2+y_0^2-1\\ &\iff \frac{t^2}{4}= \frac{t^2}{4}+ \frac{t^2}{4}-1\text{ and } x_0=\frac{t}{4},\ y_0=z_0=-\frac{t}{2}\\ & \iff t=\pm 2\qquad (x_0,y_0,z_0)=\pm \big(\tfrac{1}{2},-1,-1\big) \end{aligned}$$

(a) A vector perpendicular to $x^2+z^2=10$ at $(1,1,3)$ is

$${ \boldsymbol{\nabla} }(x^2+z^2)\big|_{(1,1,3)} =(2x\hat{\boldsymbol{\imath}}+2z\hat{\mathbf{k}})\big|_{(1,1,3)} =2\hat{\boldsymbol{\imath}}+6\hat{\mathbf{k}}\text{ or } \frac{1}{2} ( 2,0,6)=( 1,0,3)$$

(b) A vector perpendicular to $y^2+z^2=10$ at $(1,1,3)$ is

$${ \boldsymbol{\nabla} }(y^2+z^2)\big|_{(1,1,3)} =(2y\hat{\boldsymbol{\jmath}}+2z\hat{\mathbf{k}})\big|_{(1,1,3)} =2\hat{\boldsymbol{\jmath}}+6\hat{\mathbf{k}}\text{ or }\frac{1}{2} ( 0,2,6)=( 0,1,3)$$

A vector is tangent to the specified curve at the specified point if and only if it perpendicular to both $(1,0,3)$ and $(0,1,3)$. One such vector is

$$( 0,1,3)\times(1,0,3) =\det\left[\begin{matrix} \hat{\boldsymbol{\imath}} & \hat{\boldsymbol{\jmath}} & \hat{\mathbf{k}} \\ 0 & 1 & 3 \\ 1 & 0 & 3 \end{matrix}\right] =( 3,3,-1)$$

(c) The specified tangent line passes through $(1,1,3)$ and has direction vector $( 1,1,3)$ and so has vector parametric equation $$\mathbf{r}(t)=( 1,1,3)+t( 3,3,-1)$$

$\mathbf{r}(t)=( x(t)\,,\,y(t)\,,\,z(t))$ intersects $z^3 + xyz -2 = 0$ when

$$z(t)^3+x(t)\,y(t)\,z(t)-2=0\iff \big(t^2\big)^3 + \big(t^3)(t)\big(t^2\big)-2=0 \iff 2t^6=2\iff t=1$$

since $t$ is required to be positive. The direction vector for the curve at $t=1$ is

$$\mathbf{r}'(1)=3\,\hat{\boldsymbol{\imath}}+\hat{\boldsymbol{\jmath}}+2\,\hat{\mathbf{k}}$$

A normal vector for the surface at $\mathbf{r}(1)=( 1,1,1)$ is

$${ \boldsymbol{\nabla} }(z^3+xyz)\big|_{(1,1,1)}=[yz\hat{\boldsymbol{\imath}}+xz\hat{\boldsymbol{\jmath}}+(3z^2+xy)\hat{\mathbf{k}}]_{(1,1,1)} =\hat{\boldsymbol{\imath}}+\hat{\boldsymbol{\jmath}}+4\hat{\mathbf{k}}$$

The angle $\theta$ between the curve and the normal vector to the surface is determined by

$$\begin{aligned} \big|( 3,1,2)\big|\,\big|( 1,1,4)\big|\cos\theta =( 3,1,2) \cdot( 1,1,4) &\iff \sqrt{14}\sqrt{18}\cos\theta=12 \\ &\iff \sqrt{7\times 36}\cos\theta=12 \\ &\iff \cos\theta=\frac{2}{\sqrt{7}} \\ &\iff \theta=40.89^\circ \end{aligned}$$

The angle between the curve and the surface is $90-40.89=49.11^\circ$ (to two decimal places).

Let $(x_0,y_0,z_0)$ be any point on the surface. A vector normal to the surface at $(x_0,y_0,z_0)$ is

$$\begin{aligned} &{ \boldsymbol{\nabla} }\Big(xy e^{-(x^2+y^2)/2}-z\Big)\bigg|_{(x_0,y_0,z_0)} \\& =\left( y_0 e^{-(x_0^2+y_0^2)/2}-x_0^2y_0 e^{-(x_0^2+y_0^2)/2}, x_0 e^{-(x_0^2+y_0^2)/2}-x_0y_0^2 e^{-(x_0^2+y_0^2)/2},-1\right) \end{aligned}$$

The tangent plane to the surface at $(x_0,y_0,z_0)$ is horizontal if and only if this vector is vertical, which is the case if and only if its $x$- and $y$-components are zero, which in turn is the case if and only if

$$\begin{aligned} &y_0(1-x_0^2)=0\text{ and }x_0(1-y_0^2)=0\\ &\iff\big\{y_0=0\text{ or }x_0=1\text{ or }x_0=-1\big\} \text{ and }\big\{x_0=0\text{ or }y_0=1\text{ or }y_0=-1\big\}\\ &\iff (x_0,y_0)=(0,0)\text{ or }(1,1)\text{ or }(1,-1) \text{ or }(-1,1)\text{ or }(-1,-1) \end{aligned}$$

The values of $z_0$ at these points are $0$, $e^{-1}$, $-e^{-1}$, $-e^{-1}$ and $e^{-1}$, respectively. So the horizontal tangent planes are $z=0$, $z=e^{-1}$ and $z=-e^{-1}$. At the highest and lowest points of the surface, the tangent plane is horizontal. So the largest and smallest values of $z$ are $e^{-1}$ and $-e^{-1}$, respectively.