Tangent Planes

Section 3.2 Tangent Planes

If you are confronted with a complicated surface and want to get some idea of what it looks like near a specific point, probably the first thing that you will do is find the plane that best approximates the surface near the point. That is, find the tangent plane to the surface at the point. In general, a good way to specify a plane is to supply

a nonzero vector \(\vn\) (called a normal vector) perpendicular to the plane
¹
Alternatively, you could find two vectors that are in the plane (and not parallel to each other), and then construct a normal vector by taking their cross product.
(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, then 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 entirely in the plane and so is perpendicular to \(\vn\text{.}\) This gives the following very neat the equation for the plane.

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

The following theorem provides formulae for normal vectors \(\vn\) to general surfaces, assuming first that the surface is parametrized, second that the surface is a graph and finally the surface is given by an implicit equation. The formulae are developed in the proof of the theorem.

Theorem 3.2.1. Normal vectors to surfaces.

Let
\begin{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 surface and let \((x_0,y_0,z_0)=\vr(u_0,v_0)\) be a point on the surface. Set
\begin{align*} \vT_u &= \frac{\partial\ }{\partial u}\vr(u,v_0)\Big|_{u=u_0} =\Big(\frac{\partial x}{\partial u}(u_0,v_0)\,,\, \frac{\partial y}{\partial u}(u_0,v_0)\,,\, \frac{\partial z}{\partial u}(u_0,v_0)\Big)\\ \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}{\partial v}(u_0,v_0)\,,\, \frac{\partial z}{\partial v}(u_0,v_0)\Big) \end{align*}
Then
\begin{align*} \vn = \vT_u\times\vT_v =\det\left|\begin{matrix} \hi & \hj & \hk \\ \frac{\partial x}{\partial u}(u_0,v_0) & \frac{\partial y}{\partial u}(u_0,v_0) & \frac{\partial 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}\right| \end{align*}
is normal (i.e. perpendicular) to the surface at \((x_0,y_0,z_0)\text{,}\) if \(\vn \ne \vZero\text{.}\)
Let \((x_0,y_0,z_0)=f(x_0,y_0)\) be a point on the surface \(z=f(x,y)\text{.}\) Then,
\begin{gather*} \vn =-f_x(x_0,y_0)\,\hi - f_y(x_0,y_0)\,\hj + \hk \end{gather*}
is normal to the surface at \((x_0,y_0,z_0)\text{.}\)
Consider the surface given implicitly by the equation \(G(x,y,z) = K\text{,}\) where \(K\) is a constant. Let \((x_0,y_0,z_0)\) be a point on the surface and assume that the gradient \(\vnabla G\big(x_0,y_0,z_0\big)\ne \vZero\text{.}\) Then
\begin{gather*} \vn= \vnabla G\big(x_0,y_0,z_0\big) \end{gather*}
is normal to the surface at \((x_0,y_0,z_0)\text{.}\)

Note that none of the normal vectors \(\vn\) above need be of unit length.

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

Proof.

(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)\big) \end{equation*}

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

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

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

Next fix \(u=u_0\) and let \(v\) vary. Then

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

is a curve on the surface (the blue curve in the figure on the right above) that passes through \((x_0,y_0,z_0)\) when \(v=v_0\text{.}\) 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{equation*} \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}{\partial v}(u_0,v_0)\,,\, \frac{\partial z}{\partial v}(u_0,v_0)\Big) \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 cross product

\begin{align*} \vn = \vT_u\times\vT_v =\det\left|\begin{matrix} \hi & \hj & \hk \\ \frac{\partial x}{\partial u}(u_0,v_0) & \frac{\partial y}{\partial u}(u_0,v_0) & \frac{\partial 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}\right| \end{align*}

is normal (i.e. perpendicular) to the surface at \((x_0,y_0,z_0)\text{.}\) Note however that this vector need not be normalized. That is, it need not be of unit length.

(b) Next assume 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 reuse part (a):

\begin{equation*} \vr(x,y) = \big(x,y, f(x,y)\big) \end{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 &= \frac{\partial\vr }{\partial x}(x_0,y_0) =\big(1\,,\, 0\,,\, f_x(x_0,y_0)\big)\\ \vT_y &= \frac{\partial\vr }{\partial y}(x_0,y_0) =\big(0\,,\, 1\,,\, f_y(x_0,y_0)\big) \end{align*}

and

\begin{align*} \vn &= \vT_x\times\vT_y =\det\left|\begin{matrix} \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)\,\hi - f_y(x_0,y_0)\,\hj + \hk \end{align*}

(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 any curve with \(\vr(0) = (x_0,y_0,z_0)\) that lies on the surface, then

\begin{alignat*}{3} & & G\big(\vr(t)\big)&=K &\qquad &\text{for all } t\\ &\implies & \diff{\ }{t} G\big(x(t),y(t),z(t)\big)&=0 & &\text{for all } t \end{alignat*}

Applying the chain rule gives

\begin{align*} \frac{\partial G}{\partial x}\big(x(t),y(t),z(t)\big)\diff{x}{t}(t) &+\frac{\partial G}{\partial y}\big(x(t),y(t),z(t)\big)\diff{y}{t}(t)\\ &+\frac{\partial G}{\partial z}\big(x(t),y(t),z(t)\big)\diff{z}{t}(t) =0 \end{align*}

The left hand side is exactly the dot product of \(\big(\frac{\partial G}{\partial x} \,,\,\frac{\partial G}{\partial y} \,,\,\frac{\partial G}{\partial z}\big)=\vnabla G\) with \(\big(\diff{x}{t} \,,\,\diff{y}{t} \,,\,\diff{z}{t}\big) =\diff{\vr}{t}\text{,}\) so that

\begin{align*} \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 & \end{align*}

This tell us that \(\vnabla G\big(x_0,y_0,z_0\big)\) is perpendicular to \(\vr'(0)\text{,}\) 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 normal vector 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 &=y(u,v) = u\sin v\\ z &=z(u,v) = u \end{align*}

Observe that

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

So our surface is also

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

We shall sketch it shortly. But first, let’s find it’s tangent plane at \((x_0,y_0,z_0)=\vr(u_0,v_0)\text{.}\) In fact, let’s do it twice. Once using the parametrization and once using its implicit equation. First, using the parametrization \(\vr(u,v) = u\cos v\,\hi + u\sin v\,\hj + u\,\hk\text{,}\) we have

\begin{align*} \vT_u &= \frac{\partial\vr}{\partial u}(u_0,v_0) = \cos v_0\,\hi + \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 \end{align*}

so that

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

Next using the implicit equation \(G(x,y,z) = x^2+y^2-z^2=0\text{,}\) we have the normal 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 other. Either can be 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{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*}

provided \((x_0,y_0,z_0)\ne \vZero\text{.}\) In the event that \((x_0,y_0,z_0) = \vZero\) the “tangent plane equation” reduces to \(0=0\) and there is clearly a problem.

More generally, if \(\vT_u\times\vT_v=\vZero\) (or \(\vnabla G(x_0,y_0,z_0)=\vZero\)), then either

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

the surface fails to have a tangent plane at \((x_0,y_0,z_0)\text{,}\) or
our parametrization is screwy
³
Of course “screwy” is not a mathematically precise word. One way a parametrization \(\vr(u,v)\) could be “screwy” is if it failed to give a one-to-one correspondence between parameter values \((u,v)\) and points on (part of) the surface. For example, polar coordinates \(\vr(u,v) = u\cos v\,\hi + u\sin v\,\hj\) give \(\vr(0,v)=(0,0)\) for all values of \(v\text{.}\)
there. For example, we can parametrize the \(xy\)-plane, \(z=0\text{,}\) by \(\vr(u,v) = u\cos v\,\hi + u\sin v\,\hj\text{.}\) (This is just polar coordinates.) Then \(\vT_u = \cos v_0\,\hi + \sin v_0\,\hj \) and \(\vT_v = -u_0\sin v_0\,\hi + u_0\cos v_0\,\hj\text{,}\) so that \(\vT_u\times\vT_v = u_0\hk\) is \(\vZero\) when \(u_0=0\text{.}\) But the 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 surface with the horizontal plane \(z=z_0\) is \(x^2+y^2=z_0^2\text{,}\) 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 radius increases linearly as we move away from the \(xy\)-plane. Our surface is a cone. It does not have a tangent plane at \((0,0,0)\text{.}\)

Example 3.2.3.

This time we shall find the tangent planes to the surface

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

As for the cone of the last example, the intersection of this surface with the horizontal plane \(z=z_0\) is a circle — the circle of radius \(\sqrt{1+z_0^2}\) centred 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 move away from the \(xy\)-plane. Here is a sketch of the surface.

It is called a hyperboloid

⁴

There are also hyperboloids of two sheets. See Appendix A.8.

of one sheet.

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

\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*}

and we may take \((x_0,y_0,-z_0)\) 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 surface. In particular, 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 under rotation about the \(z\)-axis. So it is natural to parametrize the surface using cylindrical coordinates.

\begin{align*} x &= r\cos\theta\\ y &= r\sin\theta\\ z &= z \end{align*}

In cylindrical coordinates the surface \(x^2+y^2-z^2=1\) is \(r^2-z^2=1\text{,}\) and we could parametrize it by \(\vr(\theta,z) = \sqrt{1+z^2}\,\cos\theta\,\hi +\sqrt{1+z^2}\,\sin\theta\,\hj +z\,\hk\text{.}\) Alternatively, we can eliminate the square roots in the parametrization by exploiting the hyperbolic trig functions

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

The functions have properties

⁵

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

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{gather*}

We can set \(r=\cosh u\text{,}\) \(z=\sinh u\) to yield the parametrization

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

As an exercise in working with hyperbolic trig functions, we’ll use this parametrization to find \(\hn\text{.}\)

\begin{align*} x&= \cosh u\,\cos\theta & x_u&= \sinh u\,\cos\theta & x_\theta&= -\cosh u\,\sin\theta\\ y&= \cosh u\,\sin\theta & y_u&= \sinh u\,\sin\theta & y_\theta&= \phantom{-}\cosh u\,\cos\theta\\ z&=\sinh u & z_u&=\cosh u & z_\theta&=0 \end{align*}

\begin{align*} \vn = \vT_u\times\vT_\theta &=\det\left|\begin{matrix} \hi & \hj & \hk \\ \sinh u\,\cos\theta & \sinh u\,\sin\theta & \cosh u \\ -\cosh u\,\sin\theta & \cosh u\,\cos\theta & 0 \end{matrix}\right|\\ &=\big(-\cosh^2u\,\cos\theta\,,\, -\cosh^2u\,\sin\theta\,,\, \sinh u\cosh u\big) \end{align*}

Exercises Exercises

Exercise Group.

Exercises — Stage 1

1.

Is it reasonable to say 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.

Let the point \(\vr_0= (x_0,y_0,z_0)\) lie on the surface \(G(x,y,z)=0\text{.}\) Assume that \(\vnabla G(x_0,y_0,z_0)\ne\vZero\text{.}\) Suppose 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 curve at \(\vr_0\) lies in the tangent plane to \(G=0\) at \(\vr_0\text{.}\)

3.

Find the parametric equations of the normal line to the surface \(z=f(x,y)\) at the point \(\big(x_0\,,\,y_0\,,\,z_0\!=\!f(x_0,y_0)\big)\text{.}\) By definition, the normal line in question is the line through \((x_0,y_0,z_0)\) whose direction vector is perpendicular to the surface 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 parallel to each other. Find the equation of the normal plane to the curve of intersection of the surfaces \(F(x,y,z)=0\) and \(G(x,y,z)=0\) at \((x_0,y_0,z_0)\text{.}\) By definition, that normal plane is the plane through \((x_0,y_0,z_0)\) whose normal vector is the tangent vector to the curve of intersection at \((x_0,y_0,z_0)\text{.}\)

5.

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

Exercise Group.

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 point \(\left( -1\,,\,1\,,\,\frac{1}{3}\right)\text{.}\)

7. (✳).

Find the tangent 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 surface \(z = f(x,y)\) defined implicitly by the equation \(xyz^2 + y^2 z^3 = 3 + x^2\text{.}\) Use a 3--dimensional gradient vector to find the equation of the tangent plane to this surface at the point \((-1, 1, 2)\text{.}\) Write your answer in the form \(z = ax + by + c\text{,}\) where \(a\text{,}\) \(b\) and \(c\) are constants.

9. (✳).

A surface is given by

\begin{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 value of \(a\) is the tangent plane parallel to the plane \(x - y + z = 1\text{?}\)

10. (✳).

A surface S is given by the parametric equations

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

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

11. (✳).

Let \(S\) be the surface given by

\begin{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 \end{equation*}

Find the tangent plane to the surface 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 on the surface \(x^2 + 9y^2 + 4z^2 = 17\) where the tangent plane is parallel to the plane \(x - 8z = 0\text{.}\)

14. (✳).

Let \(S\) be the surface \(z = x^2 + 2y^2 + 2y - 1\text{.}\) Find all points \(P (x_0,y_0,z_0)\) on \(S\) with \(x_0 \ne 0\) such that the normal line at \(P\) contains the origin \((0,0,0)\text{.}\)

15. (✳).

Find all points on the hyperboloid \(z^2=4x^2+y^2-1\) where the tangent plane is parallel to the plane \(2x-y+z=0\text{.}\)

Exercise Group.

Exercises — Stage 3

16. (✳).

Find a vector perpendicular at the point \((1,1,3)\) to the surface with equation \(x^2+z^2=10\text{.}\)
Find a vector tangent at the same point to the curve of intersection of the surface in part (a) with surface \(y^2+z^2=10\text{.}\)
Find parametric equations for the line tangent to that curve at that point.

17. (✳).

Let \(P\) be the point where the curve

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

intersects the surface

\begin{equation*} z^3 + xyz -2 = 0 \end{equation*}

Find the (acute) angle between the curve and the surface at \(P\text{.}\)

18.

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

\begin{equation*} z=xy e^{-(x^2+y^2)/2} \end{equation*}

What are the largest and smallest values of \(z\) on this surface?

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