Find All of the Points on the Surface Where the Tangent Plane Is Parallel

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If you are confronted with a complicated surface and want to get several idea of what it looks like near a taxon point, probably the premiere matter that you will do is find the aeroplane that superfine approximates the surface near the point. That is, find the tangent plane to the rise at the point. In general, a serious way to specify a plane is to cater

a nonzero vector $\vn$ (called a normal vector) perpendicular to the woodworking plane¹ (to determine the orientation of the plane) and
one point $(x_0,y_0,z_0)$ on the plane.

If $(x,y,z)$ is any other point on the plane, past the vector

\begin{equation*} (x,y,z)-(x_0,y_0,z_0) = (x-x_0\,,\,y-y_0\,,\,z-z_0) \end{equation*}

lies whole in the plane and thus is perpendicular to $\vn\text{.}$ This gives the following rattling good the equation for the plane.

Alternatively, you could find ii vectors that are in the plane (and non parallel to each opposite), and so construct a normal vector by taking their get across product.

\begin{equation*} \vn\cdot(x-x_0\,,\,y-y_0\,,\,z-z_0) = 0 \end{equation*}

The pursuing theorem provides formulae for normal vectors $\vn$ to general surfaces, presumptuous first that the surface is parametrized, arcsecond that the surface is a chart and finally the come on is given by an implicit par. The formulae are developed in the proof of the theorem.

Theorem 3.2.1 . Normal vectors to surfaces.

Let
\start out{align*} \vr&: \cD\subset\bbbr^2 \rightarrow \bbbr^3\\ &(u,v) \in\cD \mapsto \vr(u,v) =\big(x(u,v)\,,\,y(u,v)\,,\,z(u,v)\big) \end{align*}
be a parametrized control surface and let $(x_0,y_0,z_0)=\vr(u_0,v_0)$ be a point on the surface. Set
\begin{adjust*} \vT_u &= \frac{\colored\ }{\partial u}\vr(u,v_0)\Big|_{u=u_0} =\Big(\frac{\partial x}{\colored u}(u_0,v_0)\,,\, \frac{\partial y}{\partial u}(u_0,v_0)\,,\, \frac{\unjust z}{\partial u}(u_0,v_0)\Big)\\ \vT_v &= \frac{\partial\ }{\partial v}\vr(u_0,v)\Big|_{v=v_0} =\Openhanded(\frac{\partial x}{\partial v}(u_0,v_0)\,,\, \frac{\one-sided y}{\partial v}(u_0,v_0)\,,\, \frac{\partial z}{\partial v}(u_0,v_0)\Big) \end{ordinate*}
Then
\begin{array*} \vn = \vT_u\times\vT_v =\det\near|\begin{matrix} \how-do-you-do & \hj & \hk \\ \frac{\partial x}{\partial u}(u_0,v_0) &A; \frac{\partial y}{\partial u}(u_0,v_0) & \frac{\fond z}{\partial u}(u_0,v_0) \\ \frac{\partial x}{\partial v}(u_0,v_0) & \frac{\partial y}{\partial v}(u_0,v_0) & \frac{\partial z}{\partial v}(u_0,v_0) \end{matrix}\aright| \end{align*}
is average (i.e. perpendicular style) to the opencut at $(x_0,y_0,z_0)\text{.}$
Permit $(x_0,y_0,z_0)=f(x_0,y_0)$ be a point on the the surface $z=f(x,y)\textual matter{.}$ Then,
\begin{gather*} \vn =-f_x(x_0,y_0)\,\how-do-you-do - f_y(x_0,y_0)\,\hj + \hk \end{gather*}
is normal to the come up at $(x_0,y_0,z_0)\text{.}$
Consider the surface given implicitly away the equation $G(x,y,z) = K\schoolbook{,}$ where $K$ is a constant. Let $(x_0,y_0,z_0)$ be a target connected the skin-deep and assume that the gradient $\vnabla G\big(x_0,y_0,z_0\big)\ne \vZero\text{.}$ Past
\begin{gather*} \vn= \vnabla G\cosmic(x_0,y_0,z_0\big) \destruction{gather*}
is modal to the opencast at $(x_0,y_0,z_0)\school tex{.}$

Billet that none of the normal vectors $\vn$ above need be of unit duration.

Note that if we employ part (c) to $G(x,y,z) = z - f(x,y)$ we get the normal vector $\vn=\vnabla G\big(x_0,y_0,z_0\big) =-f_x(x_0,y_0)\,\Hawai'i - f_y(x_0,y_0)\,\hj + \hk\text{,}$ which is the same as the normal vector provided by part (b). Of course they had to live at least parallel.

Impervious.

(a) First fix $v=v_0$ and let $u$ vary. Then

\begin{equation*} u\mapsto \vr(u,v_0) = \big(x(u,v_0)\,,\,y(u,v_0)\,,\,z(u,v_0)\freehanded) \end{equivalence*}

is a curve on the surface (the red ink curve in the figure on the suited below) that passes through $(x_0,y_0,z_0)$ (the black dot in the chassis) when $u=u_0\text{.}$

The tangent vector to this veer at $(x_0,y_0,z_0)\text edition{,}$ which is also a tan transmitter to the surface at $(x_0,y_0,z_0)\school tex{,}$ is

\begin{equality*} \vT_u = \frac{\partial\ }{\partial u}\vr(u,v_0)\Big|_{u=u_0} =\Big(\frac{\partial x}{\coloured u}(u_0,v_0)\,,\, \frac{\partial y}{\partial u}(u_0,v_0)\,,\, \frac{\partial z}{\partial u}(u_0,v_0)\Grownup) \end{equation*}

Information technology is the red arrow in the figure on the correctly above.

Next touch o $u=u_0$ and let $v$ change. Then

\begin{equation*} v\mapsto \vr(u_0,v) = \big(x(u_0,v)\,,\,y(u_0,v)\,,\,z(u_0,v)\enlarged) \end{equating*}

is a curve on the surface (the blue curve in the number on the right above) that passes through $(x_0,y_0,z_0)$ when $v=v_0\text edition{.}$ The tangent vector to this curve at $(x_0,y_0,z_0)\text{,}$ which is also a tangent vector to the surface at $(x_0,y_0,z_0)\text{,}$ is

\begin{par*} \vT_v = \frac{\partial\ }{\partial v}\vr(u_0,v)\Big|_{v=v_0} =\Big(\frac{\partial x}{\partial v}(u_0,v_0)\,,\, \frac{\partial y}{\uncomplete v}(u_0,v_0)\,,\, \frac{\incomplete z}{\partial v}(u_0,v_0)\Queen-size) \end{equation*}

It is the blue arrow in the figure on the right above.

We now have two vectors, namely $\vT_u$ and $\vT_v\text{,}$ that are tangent to the surface at $(x_0,y_0,z_0)\text{.}$ So their vector product

\Begin{align*} \vn = \vT_u\times\vT_v =\det\left|\commence{intercellular substance} \hi &adenosine monophosphate; \hj & \hk \\ \frac{\partial x}{\partial u}(u_0,v_0) &adenosine monophosphate; \frac{\incomplete y}{\partial u}(u_0,v_0) & \frac{\fond z}{\partial u}(u_0,v_0) \\ \frac{\partial x}{\partial v}(u_0,v_0) &adenylic acid; \frac{\partial y}{\partial v}(u_0,v_0) & \frac{\partial z}{\partial v}(u_0,v_0) \end{matrix}\right| \end{align*}

is modal (i.e. perpendicular) to the surface at $(x_0,y_0,z_0)\text{.}$ Note however that this transmitter need not exist normalized. That is, it need non be of building block length.

(b) Next arrogate that the surface is given by the equation $z=f(x,y)\text{.}$ Then, renaming $u$ to $x$ and $v$ to $y\text{,}$ we may recycle part (a):

\set out{equation*} \vr(x,y) = \big(x,y, f(x,y)\big) \final stage{equation*}

parametrizes the surface and, at $\big(x_0,y_0,z_0\big)=f(x_0,y_0)\big)\text{,}$

\begin{align*} \vT_x &adenosine monophosphate;= \frac{\partial\vr }{\partial x}(x_0,y_0) =\big(1\,,\, 0\,,\, f_x(x_0,y_0)\big)\\ \vT_y &= \frac{\partial\vr }{\overtone y}(x_0,y_0) =\big(0\,,\, 1\,,\, f_y(x_0,y_0)\rangy) \end{align*}

and

\Menachem Begin{align*} \vn &= \vT_x\times\vT_y =\det\left|\begin{intercellular substance} \hi & \hj & \hk \\ 1 & 0 & f_x(x_0,y_0) \\ 0 & 1 & f_y(x_0,y_0) \end{matrix}\right| =-f_x(x_0,y_0)\,\Hawaii - f_y(x_0,y_0)\,\hj + \hk \end{array*}

(c) Finally assume that the surface is given implicitly by the equation $G(x,y,z)=0$ or, more generally by the equation, $G(x,y,z) = K\text{,}$ where $K$ is a constant. If $\vr(t)=\big(x(t)\,\,y(t)\,,\,z(t)\big)$ is whatever curve with $\vr(0) = (x_0,y_0,z_0)$ that lies on the surface, then

\begin{alignat*}{3} & & G\gargantuan(\vr(t)\big)&ere;=K &\qquad &adenylic acid;\schoolbook{for all } t\\ &\implies & \diff{\ }{t} G\big(x(t),y(t),z(t)\big)&=0 & &adenosine monophosphate;\text{for all } t \death{alignat*}

Applying the chain prevai gives

\begin{align*} \frac{\partial G}{\partial x}\braggart(x(t),y(t),z(t)\larger-than-life)\diff{x}{t}(t) &adenylic acid;+\frac{\partial derivative G}{\partial y}\big(x(t),y(t),z(t)\big)\diff{y}{t}(t)\\ &+\frac{\inclined G}{\partial z}\big(x(t),y(t),z(t)\big)\diff{z}{t}(t) =0 \end{align*}

The left side is exactly the scalar product of $\grownup(\frac{\partial G}{\incomplete x} \,,\,\frac{\partial G}{\partial y} \,,\,\frac{\one-sided G}{\partial z}\big)=\vnabla G$ with $\queen-size(\diff{x}{t} \,,\,\diff{y}{t} \,,\,\diff{z}{t}\big) =\diff{\vr}{t}\text{,}$ so that

\begin{coordinate*} \vnabla G\big(\vr(t)\big)\cdot\vr'(t)&=0 \qquad \text{for all } t\\ \implies \vnabla G\big(x_0,y_0,z_0\big)\cdot\vr'(0)&=0 & \ending{align*}

This tell us that $\vnabla G\big(x_0,y_0,z_0\of import)$ is perpendicular to $\vr'(0)\textual matter{,}$ which is a tangent vector to $G=K$ at $(x_0,y_0,z_0)\text{.}$ This is true for all curves $\vr(t)$ on $G=K$ and so is true for all tangent vectors to $G=K$ at $(x_0,y_0,z_0)\text{.}$ So $\vnabla G\big(x_0,y_0,z_0\big)$ is a formula transmitter to $G(x,y,z)=K$ at $(x_0,y_0,z_0)\text{.}$

Example 3.2.2 .

Consider the surface

\begin{align*} x &= x(u,v) = u\cos v\\ y &A;=y(u,v) = u\sinfulness v\\ z &=z(u,v) = u \end{align*}

Detect that

\Menachem Begin{equation*} x(u,v)^2 + y(u,v)^2 = u^2 = z(u,v)^2 \end{equation*}

So our skin-deep is also

\begin{equation*} G(x,y,z) = x^2+y^2-z^2 = 0 \end{equality*}

We shall sketch IT soon. But first, let's find it's tangent plane at $(x_0,y_0,z_0)=\vr(u_0,v_0)\school tex{.}$ In fact, countenance's do IT twice. Once using the parametrization and once using its implicit equation. First, using the parametrization $\vr(u,v) = u\cos v\,\how-do-you-do + u\sin v\,\hj + u\,\hk\text{,}$ we have

\start out{align*} \vT_u &= \frac{\partial\vr}{\partial derivative u}(u_0,v_0) = \cos v_0\,\Hawai'i + \sin v_0\,\hj + \hk\\ \vT_v &= \frac{\partial\vr}{\partial v}(u_0,v_0) = -u_0\sin v_0\,\hi + u_0\cos v_0\,\hj \closing{adjust*}

so that

\begin{align*} \vn &= \big(\cos v_0\,\hi + \sinfulness v_0\,\hj + \hk\big)\times \heroic(-u_0\sin v_0\,\hi + u_0\cos v_0\,\hj\big)\\ &= \braggy(-u_0\cos v_0\,,\,-u_0\sin v_0\,,\, u_0) = (-x_0,-y_0,z_0) \end{align*}

Incoming using the unuttered equating $G(x,y,z) = x^2+y^2-z^2=0\textbook{,}$ we rich person the formula vector

\begin{equation*} \vnabla G\big(x_0,y_0,z_0\big) = (2x_0,2y_0,-2z_0) =-2(-x_0,-y_0,z_0) \end{equation*}

Of course the two vectors $(-x_0,-y_0,z_0)$ and $-2(-x_0,-y_0,z_0)$ are parallel to each opposite. Either can glucinium used as a normal vector and the tangent plane to $x^2+y^2-z^2=0$ at $(x_0,y_0,z_0)$ is

\begin{equating*} 0=\vn\cdot(x-x_0,y-y_0,z-z_0) = -x_0(x-x_0)-y_0(y-y_0) + z_0(z-z_0) \end{equation*}

provided $(x_0,y_0,z_0)\atomic number 10 \vZero\text{.}$ In the event that $(x_0,y_0,z_0) = \vZero$ the "tangent shave equation" reduces to $0=0$ and there is clearly a problem.

Much generally, if $\vT_u\times\vT_v=\vZero$ (or $\vnabla G(x_0,y_0,z_0)=\vZero$), then either²

the come out fails to have a tangent planer at $(x_0,y_0,z_0)\text{,}$ or
our parametrization is screwy³ there. For example, we sack parametrize the $xy$-plane, $z=0\text{,}$ by $\vr(u,v) = u\romaine v\,\HI + u\transgress v\,\hj\text{.}$ (This is just circumpolar coordinates.) Past $\vT_u = \cos v_0\,\hi + \sin v_0\,\hj $ and $\vT_v = -u_0\sin v_0\,\hi + u_0\cos v_0\,\hj\textbook{,}$ so that $\vT_u\times\vT_v = u_0\hk$ is $\vZero$ when $u_0=0\text{.}$ But the woodworking plane $z=0$ is its own tangent plane everywhere.

The surface of current interest is $x^2+y^2=z^2\text{.}$ The intersection of this come out with the horizontal plane $z=z_0$ is $x^2+y^2=z_0^2\text edition{,}$ which is the circle of radius $|z_0|$ centred on $x=y=0\text{.}$ So our surface is a stack of circles. The radius of the circle in the $xy$-plane is zero. The wheel spoke increases linearly as we move away from the $xy$-plane. Our surface is a strobile. It does non sustain a tangent plane at $(0,0,0)\text edition{.}$

We saw the unvarying dichotomy when considering what happened for a curve when $\vr'(t)=0\text{.}$ See Example 1.1.10.

Of run "screwy" is not a mathematically precise word. One manner a parametrization $\vr(u,v)$ could be "screwy" is if it failed to give a unitary-to-one correspondence 'tween parameter values $(u,v)$ and points on (part of) the surface. For example, polar coordinates $\vr(u,v) = u\cosine v\,\hi + u\sin v\,\hj$ give $\vr(0,v)=(0,0)$ for totally values of $v\text{.}$

Object lesson 3.2.3 .

This time we shall find the tan planes to the come on

\begin{equation*} x^2 + y^2 -z^2 = 1 \end{equation*}

As for the cone of the last-place example, the intersection of this airfoil with the horizontal plane $z=z_0$ is a lap — the environ of radius $\sqrt{1+z_0^2}$ hundred on $x=y=0\text{.}$ Our surface is again a stack of circles. The radius of the circle in the $xy$-plane is $1\text{.}$ The radius increases as we go on away from the $xy$-plane. Here is a sketch of the surface.

It is called a hyperboloid⁴ of indefinite sheet.

In that respect are as wel hyperboloids of cardinal sheets. Construe Appendix A.8.

Victimisation the implicit equation $G(x,y,z) = x^2+y^2-z^2=1\schoolbook{,}$ we have

\begin{equality*} \vnabla G\big(x_0,y_0,z_0\big) = (2x_0,2y_0,-2z_0) =2(x_0,y_0,-z_0) \end{equation*}

and we may shoot $(x_0,y_0,-z_0)$ as a normal vector at $(x_0,y_0,z_0)\text{.}$ So the tangent plane to $x^2+y^2-z^2=1$ at $(x_0,y_0,z_0)$ is

\begin{equation*} 0=\vn\cdot(x-x_0,y-y_0,z-z_0) = x_0(x-x_0)+y_0(y-y_0) - z_0(z-z_0) \end{equation*}

This time $\vn=(x_0,y_0,-z_0)\ne \vZero\text{,}$ so that we have a tangent plane, at every point of the control surface. Particularly, the vanishing of $\vn=(x_0,y_0,-z_0)$ at $(x_0,y_0,z_0)=(0,0,0)$ is not a problem because $(0,0,0)$ is not on the surface.

Example 3.2.4 . Optional — Parametrizing the Hyperboloid of One Sheet.

The hyperboloid of one sheet, $x^2+y^2-z^2=1\text{,}$ has a symmetry. It is invariant low-level rotation about the $z$-axis. Then information technology is natural to parametrize the open using cylindrical coordinates.

\Menachem Begin{align*} x &= r\cos\theta\\ y &= r\sin\theta\\ z &= z \oddment{align*}

In cylindrical coordinates the airfoil $x^2+y^2-z^2=1$ is $r^2-z^2=1\text{,}$ and we could parametrize it by $\vr(\theta,z) = \sqrt{1+z^2}\,\cosine\theta\,\Hawai'i +\sqrt{1+z^2}\,\sin\theta\,\hj +z\,\hk\text{.}$ Alternatively, we can carry off the solid roots in the parametrization past exploiting the hyperbolic trig functions

\begin{equation*} \sinh u = \frac{1}{2}\with child(e^u-e^{-u}\big) \qquad \cosh u = \frac{1}{2}\big(e^u+e^{-u}\mature) \end{equation*}

The functions have properties⁵ that are very similar to those of $\sin\theta$ and $\cos\theta\text{.}$

\begin{gather*} \diff{}{u} \cosh u= \sinh u \qquad \diff{}{u} \sinh u= \cosh u \qquad \cosh^2 u -\sinh^2 u =1 \end{garner*}

We can set $r=\sap u\text{,}$ $z=\sinh u$ to give in the parametrization

\begin{gather*} \vr(\theta,u) = \cosh u\,\cos\theta\,\howdy +\blackjack u\,\sin\theta\,\hj +\sinh u\,\hk \oddment{gather*}

As an exercise in functioning with hyperbolic trim functions, we'll use this parametrization to find $\hn\textual matter{.}$

\begin{line up*} x&= \cosh u\,\cos\theta &A; x_u&= \sinh u\,\cos\theta & x_\theta&= -\cosh u\,\sin\theta\\ y&adenylic acid;= \cosh u\,\hell\theta & y_u&= \sinh u\,\sin\theta & y_\theta&= \phantom{-}\sap u\,\cos\theta\\ z&=\sinh u & z_u&=\sap u & z_\theta&=0 \end{align*}

\begin{align*} \vn = \vT_u\times\vT_\theta &=\det\left over|\begin{matrix} \hi &ere; \hj & \hk \\ \sinh u\,\cos\theta &adenosine monophosphate; \sinh u\,\hell\theta & \cosh u \\ -\sap u\,\sin\theta & \cosh u\,\romaine\theta & 0 \final stage{matrix}\compensate|\\ &=\prodigious(-\cosh^2u\,\cos\theta\,,\, -\cosh^2u\,\sin\theta\,,\, \sinh u\blackjack u\bad) \close{align*}

This is no chance event: $\cosh u = \cos(iu)$ and $\sinh u = -i\sin(iu)\text{,}$ where $i$ is the usual complex quantity that obeys $i^2=-1\text{.}$ You can swear these formulae by just checking that $\sap u$ and $\cos(iu)$ have the same Taylor expansions and that $\sinh u$ and $-i\sin(iu)$ stimulate the same Taylor expansions.

Exercises 3.2.1 Exercises

Exercises — Stage 1

1.

Is IT levelheaded to read that the surfaces $x^2+y^2+(z-1)^2=1$ and $x^2+y^2+(z+1)^2=1$ are tangent to each other at $(0,0,0)\text{?}$

2.

Countenance the point $\vr_0= (x_0,y_0,z_0)$ belong happening the surface $G(x,y,z)=0\text{.}$ Assume that $\vnabla G(x_0,y_0,z_0)\northeastward\vZero\text{.}$ Theorise that the parametrized curve $\vr(t)=\big(x(t),y(t),z(t)\big)$ is contained in the surface and that $\vr(t_0)=\vr_0\text{.}$ Show that the tangent line to the sheer at $\vr_0$ lies in the tangent planer to $G=0$ at $\vr_0\text{.}$

3.

Find the parametric equations of the sane line to the airfoil $z=f(x,y)$ at the point $\big(x_0\,,\,y_0\,,\,z_0\!=\!f(x_0,y_0)\big)\text{.}$ Aside definition, the normal line questionable is the line through $(x_0,y_0,z_0)$ whose charge vector is English-Gothic to the opencut at $(x_0,y_0,z_0)\text{.}$

4.

Let $F(x_0,y_0,z_0)=G(x_0,y_0,z_0)=0$ and let the vectors $\vnabla F(x_0,y_0,z_0)$ and $\vnabla G(x_0,y_0,z_0)$ be nonzero and not be twin to each other. Bump the equation of the pattern plane to the curve of intersection of the surfaces $F(x,y,z)=0$ and $G(x,y,z)=0$ at $(x_0,y_0,z_0)\schoolbook{.}$ By definition, that normal plane is the plane through $(x_0,y_0,z_0)$ whose normal vector is the tan vector to the wind of intersection at $(x_0,y_0,z_0)\text{.}$

5.

Let $f(x_0,y_0)=g(x_0,y_0)$ and let $\left( f_x(x_0,y_0), f_y(x_0,y_0)\right)\ne \left-wing( g_x(x_0,y_0), g_y(x_0,y_0)\rightist)\text{.}$ Happen the equation of the tangent line to the curve of intersection of the surfaces $z=f(x,y)$ and $z=g(x,y)$ at $(x_0\,,\,y_0\,,\,z_0=f(x_0,y_0))\text{.}$

Exercises — Stage 2

6. ✳.

Let $\displaystyle f(x,y)=\frac{x^2y}{x^4+2y^2}\text{.}$ Find the tangent plane to the surface $z = f(x,y)$ at the channelize $\left-hand( -1\,,\,1\,,\,\frac{1}{3}\decent)\textbook{.}$

7. ✳.

Find the tan plane to

\begin{equation*} \frac{27}{\sqrt{x^2+y^2+z^2+3}}=9 \end{equation*}

at the point $(2, 1, 1)\text{.}$

8. ✳.

Consider the grade-constructed $z = f(x,y)$ defined implicitly by the equation $xyz^2 + y^2 z^3 = 3 + x^2\text{.}$ Use a 3--dimensional slope vector to find the equation of the tan aeroplane to this aboveground at the target $(-1, 1, 2)\text{.}$ Publish your answer in the strain $z = ax + by + c\text edition{,}$ where $a\text{,}$ $b$ and $c$ are constants.

9. ✳.

A open is given by

\commence{equation*} z = x^2 - 2xy + y^2 . \end{equation*}

Find the equation of the tangent plane to the surface at $x = a\text{,}$ $y = 2a\text{.}$
For what economic value of $a$ is the tangent plane symmetrical to the plane $x - y + z = 1\text{?}$

10. ✳.

A superficial S is given aside the parametric equations

\begin{align*} x &= 2u^2\\ y &A;= v^2\\ z &= u^2 + v^3 \end{align*}

Find an equation for the tangent plane to $S$ at the point $(8, 1, 5)\text edition{.}$

11. ✳.

Rent out $S$ be the surface given by

\commence{equation*} \vr(u, v) = \big( u + v\,,\, u^2 + v^2 \,,\, u - v\big),\qquad -2 \le u \le 2,\ -2 \le v \le 2 \finish{equation*}

Recover the tan plane to the airfoil at the point $(2, 2, 0)\text{.}$

12. ✳.

Find the tangent plane and normal line to the surface $z=f(x,y)=\frac{2y}{x^2+y^2}$ at $(x,y)=(-1,2)\text{.}$

13. ✳.

Find all the points along the surface $x^2 + 9y^2 + 4z^2 = 17$ where the tangent carpenter's plane is parallel to the plane $x - 8z = 0\text{.}$

14. ✳.

Get $S$ make up the control surface $z = x^2 + 2y^2 + 2y - 1\text{.}$ Find all points $P (x_0,y_0,z_0)$ connected $S$ with $x_0 \Nebraska 0$ such that the normal line at $P$ contains the origin $(0,0,0)\text{.}$

15. ✳.

Recover all points happening the hyperboloid $z^2=4x^2+y^2-1$ where the tangent plane is duplicate to the carpenter's plane $2x-y+z=0\text{.}$

Exercises — Present 3

16. ✳.

Find a vector perpendicular at the point $(1,1,3)$ to the open with equation $x^2+z^2=10\text{.}$
Find a vector tangent at the same point to the arch of intersection of the show u in separate (a) with surface $y^2+z^2=10\text{.}$
Find constant quantity equations for the phone line tangent to that curve at that spot.

17. ✳.

Let $P$ be the point where the curve

\begin{par*} \vr(t) = t^3\,\hi + t\,\hj + t^2\,\hk,\qquad (0 \le t \lt \infty) \end{equation*}

intersects the surface

\get{equivalence*} z^3 + xyz -2 = 0 \oddment{equation*}

Find the (keen) angle between the curve and the surface at $P\schoolbook{.}$

18.

Find all horizontal planes that are tan to the surface with equation

\lead off{equation*} z=xy e^{-(x^2+y^2)/2} \death{equation*}

What are the largest and smallest values of $z$ connected this surface?

Find All of the Points on the Surface Where the Tangent Plane Is Parallel

Source: https://personal.math.ubc.ca/~CLP/CLP4/clp_4_vc/sec_tangentPlanes.html