The Tangent Plane to a Surface

The derivative of a function of one variable gives the slope of the tangent line to the graph. The partial derivatives $\pder f x$ and $\pder f y$ of a function of two variables $z = f(x, y)$ determine the tangent plane to the graph.

$$\hbox{\epsfysize=1.75 in \epsffile{tangent-planes-1.eps}}$$

The graph of $z = f(x, y)$ is a surface in 3 dimensions. Suppose we're trying to find the equation of the tangent plane at $(a, b, f(a, b))$ .

To write down the equation of a plane, we need a point on the plane and a vector perpendicular to the plane. We have a point on the plane, namely $(a, b, f(a, b))$ .

To find a vector perpendicular to the plane, we find two vectors in the plane and take their cross product. To do this, look at a small piece of the surface near the point of tangency. A small piece will be nearly flat, and will look like the parallelogram depicted below:

$$\hbox{\epsfysize=1.75 in \epsffile{tangent-planes-2.eps}}$$

The vectors $\vec a$ and $\vec b$ which are the sides of the parallelogram are tangent to the surface at the point of tangency. Consider $\vec a$ . It runs in the x-direction. A small change $dx$ in x produces a change in z --- the amount the vector $\vec a$ "rises".

How much does z change due to a change $dx$ in x? The rate of change of z with respect to x is $\pder z x$ , so the change in z produced by changing x by $dx$ is just $\pder z x dx$ .

Now $\vec a$ is a vector with x-component $dx$ , no y-component, and z-component $\pder z x dx$ . Therefore,

$$\vec a = \left(dx, 0, \pder z x dx\right).$$

A similar argument shows that

$$\vec b = \left(0, dy, \pder z y dy\right).$$

The cross product is

$$\vec a \times \vec b = \left(-\pder z x dx\,dy, -\pder z y dx\,dy, dx\,dy\right) = \left(-\pder z x, -\pder z y, 1\right) dx\,dy.$$

I need any vector perpendicular to the surface. Since vectors which are multiples are parallel, I may use this vector as the perpendicular vector to the surface:

$$\left(-\pder z x, -\pder z y, 1\right).$$

This is often referred to as the normal vector to the surface and denoted by $\vec N$ .

The tangent plane at $(a, b, f(a, b))$ is

$$\left(\left.-\pder f x\right|_{(a, b)}\right) (x - a) + \left(\left.-\pder f y\right|_{(a, b)}\right) (y - b) + \left(z - f(a, b)\right) = 0.$$

The normal line to the surface at $(a, b, f(a, b))$ is the line which passes through $(a, b, f(a, b))$ and is perpendicular to the tangent plane. The normal line is parallel to the normal vector $\left(-\pder z x, -\pder z y, 1\right)$ . Therefore, the parametric equations of the normal line are

$$x - a = \left.-\pder f x\right|_{(a, b)}\cdot t, \quad y - b = \left.-\pder f y\right|_{(a, b)}\cdot t, \quad z - f(a, b) = t.$$

Example. Find the equation of the tangent plane and the parametric equations of the normal line to $z = \dfrac{2 x}{y} - x^2$ at $(1, 1, 1)$ .

The normal vector to the surface is

$$\left(-\pder z x, -\pder z y, 1\right) = \left(-\dfrac{2}{y} + 2 x, \dfrac{2 x}{y^2}, 1\right).$$

Plugging in $x = 1$ and $y = 1$ gives $(0, 2, 1)$ .

The tangent plane is

$$0 \cdot (x - 1) + 2 \cdot (y - 1) + 1 \cdot (z - 1) = 0, \quad\hbox{or}\quad 2 y + z = 3.$$

The normal line is

$$x - 1 = 0, \quad y - 1 = 2 t, \quad z - 1 = t.\quad\halmos$$

Example. Use a tangent plane to approximate $(1.99)^2 - \dfrac{1.99}{1.01}$ .

The idea is to think of this as the result of plugging numbers into a function $z = f(x,y)$ . What is f? Well, the form of the expression suggests that 1.99 corresponds to one of the variables and 1.01 to the other. It's natural to use the function

$$z = f(x, y) = x^2 - \dfrac{x}{y}.$$

I want to approximate $f(1.99, 1.01)$ . The point $(1.99, 1.01)$ is close to $(2, 1)$ , so I'll use the tangent plane at $(2, 1)$ to approximate f.

The normal vector is

$$\left(-2 x + \dfrac{1}{y}, -\dfrac{x}{y^2}, 1\right).$$

Plug in $x = 2$ , $y = 1$ . This gives $(-3, -2, 1)$ .

When $x = 2$ and $y = 1$ , $z = 2$ . The point of tangency is $(2, 1, 2)$ .

The tangent plane is

$$-3(x - 2) - 2(y - 1) + (z - 2) = 0, \quad\hbox{or}\quad z = 3 x + 2 y - 6.$$

Now set $x = 1.99$ , $y = 1.01$ . This gives $z \approx 1.99$ . (The actual value is 1.989803.)

Here is an equivalent way to think of things that is similar to the "approximation by differentials" technique you may have seen in first-year calculus. The change $\Delta f$ in f produced by small changes in $dx$ in x and $dy$ in y is approximated by

$$df = \pder f x\,dx + \pder f y\,dy.$$

Thus,

$$f(x + dx, y + dy) \approx f(x, y) + df.$$

Here $(x, y)$ denotes the "nice" point ($(2, 1)$ in the last example) and $(x + dx, y + dy)$ denotes the "ugly" point ($(1.99, 1.01)$ in the last example).

If you redo the example using this differential approach, you'd have

$$f(1.99, 1.01) \approx f(2, 1) + \left(\pder f x\right)(dx) + \left(\pder f y\right)(dy) = 2 + (3)(-0.01) + (2)(0.01) = 1.99.\quad\halmos$$

Suppose a surface is given parametrically:

$$x = f(u, v), \quad y = g(u, v), \quad z = h(u, v).$$

Consider the vectors

$$\vec{T}_u = \left(\pder x u, \pder y u, \pder z u\right) \quad\hbox{and}\quad \vec{T}_v = \left(\pder x v, \pder y v, \pder z v\right).$$

These vectors are tangent to the curves in the surface determined by letting one of u or v vary and holding the other constant. For example, if u varies and $v = c$ is constant, I get the curve

$$x = f(u, c), \quad y = g(u, c), \quad z = h(u, c).$$

The velocity vector for this curve is $\vec{T}_u$ .

Likewise, consider the curve obtained by setting u to a constant:

$$x = f(c, v), \quad y = g(c, v), \quad z = h(c, v).$$

The velocity vector for this curve is $\vec{T}_v$ .

$$\hbox{\epsfysize=1.75 in \epsffile{tangent-planes-3.eps}}$$

The cross product of $\vec{T}_u$ and $\vec{T}_v$ is a normal vector to the surface:

$$N = \vec{T}_u \times \vec{T}_v.$$

$$\hbox{\epsfysize=1.75 in \epsffile{tangent-planes-4.eps}}$$

Example. Find the equation of the tangent plane and the parametric equations for the normal line to

$$x = u^2 - v^2, \quad y = uv, \quad z = u^2 + v^2 \quad\hbox{at}\quad (u, v) = (2, 1).$$

First, the point of tangency is obtained by plugging $u = 2$ and $v = 1$ into x, y, and z. I get $x = 3$ , $y = 2$ , and $z = 5$ . The point is $(3, 2, 5)$ .

Next,

$$\vec{T}_u = (2 u, v, 2 u) \quad\hbox{and}\quad \vec{T}_v = (-2 v, u, 2 v).$$

When $u = 2$ and $v = 1$ ,

$$\vec{T}_u = (4, 1, 4) \quad\hbox{and}\quad \vec{T}_v = (-2, 2, 2).$$

The normal vector is

$$\vec{T}_u \times \vec{T}_v = \left|\matrix{\ihat & \jhat & \khat \cr 4 & 1 & 4 \cr -2 & 2 & 2 \cr}\right| = (-6, -16, 10).$$

The tangent plane is

$$-6(x - 3) - 16(y - 2) + 10(z - 5) = 0.$$

The normal line is

$$x - 3 = -6 t, \quad y - 2 = -16 t, \quad z - 5 = 10 t.\quad\halmos$$

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