9.6 Tangent Planes and Normal Lines

INTRODUCTION

The notion of the gradient of a function of two or more variables was introduced in the preceding section as an aid in computing directional derivatives. In this section we give a geometric interpretation of the gradient vector.

Geometric Interpretation of the Gradient—Functions of Two Variables

Suppose f(x, y) = c is the level curve of the differentiable function z = f(x, y) that passes through a specified point P(x0, y0); that is, f(x0, y0) = c. If this level curve is parameterized by the differentiable functions

x = g(t), y = h(t) such that x0 = g(t0), y0 = h(t0),

then the derivative of f(g(t), h(t)) = c with respect to t is

(1)

When we introduce the vectors

(1) becomes ∇f · r′ = 0. Specifically, at t = t0, we have

∇f(x0, y0) · r′(t0) = 0. (2)

Thus, if r′(t0) ≠ 0, the vector ∇f(x0, y0) is orthogonal to the tangent vector r′(t0) at P(x0, y0). We interpret this to mean:

∇f is orthogonal to the level curve at P.

See FIGURE 9.6.1.

A curve and 2 vectors. The curve, labeled f(x y) = c, passes through a point P(x subscript 0, y subscript 0). A vector, labeled r prime(t subscript 0), is a tangent vector to the curve at the point P. A vector, labeled nabla f(x subscript 0, y subscript 0), is a gradient vector, begins at the point P, and is perpendicular to the tangent vector.

FIGURE 9.6.1 Gradient is perpendicular to tangent vector at P

EXAMPLE 1 Gradient at a Point

Find the level curve of f(x, y) = –x2 + y2 passing through (2, 3). Graph the gradient at the point.

SOLUTION

Since f(2, 3) = –4 + 9 = 5, the level curve is the hyperbola –x2 + y2 = 5. Now,

∇f(x, y) = –2xi + 2yj and ∇f(2, 3) = –4i + 6j.

FIGURE 9.6.2 shows the level curve and ∇f(2, 3). ≡

A graph. A curve, labeled negative x^2 + y^2 = 5 and consisting of 2 symmetric and disconnected curves, and a vector are graphed on an x y coordinate plane. The first disconnected curve enters the top left of the viewing window in the second quadrant, goes down and to the right, passes through the point (negative 2, 3), reaches a low point at the point (0, sqrt(5)), goes up and to the right symmetrically, passes through the marked point (2, 3), and exits the top right of the viewing window. The second disconnected curve enters the bottom left of the viewing window in the third quadrant, goes up and to the right, passes though a point (negative 2, negative 3), reaches a high point at the point (0, negative sqrt(5)), goes down and to the right symmetrically, passes through the point (2, negative 3), and exits the bottom right of the viewing window. The vector, labeled nabla f(2, 3), begins at the marked point (2, 3), and goes up to the left.

FIGURE 9.6.2 Gradient in Example 1

Geometric Interpretation of the Gradient—Functions of Three Variables

Proceeding as before, let F(x, y, z) = c be the level surface of a differentiable function w = F(x, y, z) that passes through P(x0, y0, z0). If the differentiable functions x = f(t), y = g(t), z = h(t) are the parametric equations of a curve C on the surface for which x0 = f(t0), y0 = g(t0), z0 = h(t0), then the derivative of F(f(t), g(t), h(t)) = 0 implies that

or

(3)

In particular, at t = t0, (3) is

∇F(x0, y0, z0) · r′(t0) = 0. (4)

Thus, when r′(t0) ≠ 0, the vector ∇F(x0, y0, z0) is orthogonal to the tangent vector r′(t0). Since this argument holds for any differentiable curve through P(x0, y0, z0) on the surface, we conclude that:

∇F is normal (perpendicular) to the level surface at P.

See FIGURE 9.6.3.

A curve labeled C passes through a marked point P(x subscript 0, y subscript 0, z subscript 0) on a surface labeled F(x y z) = c. The tangent vector to the curve C at the point P is labeled r prime(t subscript 0). The gradient vector to the curve C at the point P, labeled nabla F(x subscript 0, y subscript 0, z subscript 0), is perpendicular to the tangent vector and normal to the surface F.

FIGURE 9.6.3 Gradient is normal to level surface at P

EXAMPLE 2 Gradient at a Point

Find the level surface of F(x, y, z) = x2 + y2 + z2 passing through (1, 1, 1). Graph the gradient at the point.

SOLUTION

Since F(1, 1, 1) = 3, the level surface passing through (1, 1, 1) is the sphere x2 + y2 + z2 = 3. The gradient of the function is

∇F(x, y, z) = 2xi + 2yj + 2zk,

and so, at the given point, ∇F(1, 1, 1) = 2i + 2j + 2k. The level surface and ∇F(1, 1, 1) are illustrated in FIGURE 9.6.4. ≡

A graph. A sphere labeled x^2 + y^2 + z^2 = 3 and a gradient vector are graphed on a three dimensional x y z coordinate system. The gradient vector labeled nabla F(1, 1, 1), begins on the sphere at the marked point (1, 1, 1), and points outward.

FIGURE 9.6.4 Gradient in Example 2

Tangent Plane

In the study of differential calculus a basic problem was finding an equation of a tangent line to the graph of a function. In 3-space the analogous problem is finding an equation of a tangent plane to a surface. We assume again that w = F(x, y, z) is a differentiable function and that a surface is given by F(x, y, z) = c.

DEFINITION 9.6.1 Tangent Plane

Let P(x0, y0, z0) be a point on the graph of F(x, y, z) = c, where ∇F is not 0. The tangent plane at P is that plane through P that is normal to ∇F evaluated at P.

Thus, if P(x, y, z) and P(x0, y0, z0) are points on the tangent plane and r and r0 are their respective position vectors, then a vector equation of the tangent plane is ∇F(x0, y0, z0) · (r – r0) = 0. See FIGURE 9.6.5. We summarize this last result:

A graph. A surface, a plane and 2 vectors are graphed on a three dimensional x y z coordinate system. The surface is labeled F(x y z) = c. A curve on the surface passes though the point (x subscript 0, y subscript 0, z subscript 0). The plane is tangent to the surface at the point (x subscript 0, y subscript 0, z subscript 0). The first vector begins at the point (x subscript 0, y subscript 0, z subscript 0), and ends at the point (x y z) on the tangent plane. The second vector, labeled nabla F(x subscript 0, y subscript 0, z subscript 0), is perpendicular to the first vector and normal to the tangent plane.

FIGURE 9.6.5 Tangent plane is normal to gradient at P

THEOREM 9.6.1 Equation of Tangent Plane

Let P(x0, y0, z0) be a point on the graph of F(x, y, z) = c, where ∇F is not 0. Then an equation of the tangent plane at P is

(5)

EXAMPLE 3 Equation of Tangent Plane

Find an equation of the tangent plane to the graph of x2 – 4y2 + z2 = 16 at (2, 1, 4).

SOLUTION

By defining F(x, y, z) = x2 – 4y2 + z2, the given surface is the level surface F(x, y, z) = F(2, 1, 4) = 16 passing through (2, 1, 4). Now, Fx(x, y, z) = 2x, Fy(x, y, z) = –8y, and Fz(x, y, z) = 2z, so that

∇F(x, y, z) = 2xi – 8yj + 2zk and ∇F(2, 1, 4) = 4i – 8j + 8k.

It follows from (5) that an equation of the tangent plane is

4(x – 2) – 8(y – 1) + 8(z – 4) = 0 or x – 2y + 2z = 8. ≡

Surfaces Given by z = f(x, y)

For a surface given explicitly by a differentiable function z = f(x, y), we define F(x, y, z) = f(x, y) – z or F(x, y, z) = z – f(x, y). Thus, a point (x0, y0, z0) is on the graph of z = f(x, y) if and only if it is also on the level surface F(x, y, z) = 0. This follows from F(x0, y0, z0) = f(x0, y0) – z0 = 0.

EXAMPLE 4 Equation of Tangent Plane

Find an equation of the tangent plane to the graph of z = x2 + y2 + 4 at (1, –1, 5).

SOLUTION

Define F(x, y, z) = x2 + y2 – z + 4 so that the level surface of F passing through the given point is F(x, y, z) = F(1, –1, 5) or F(x, y, z) = 0. Now, Fx = x, Fy = y, and Fz = –1, so that

∇F(x, y, z) = xi + yj – k and ∇F(1, –1, 5) = i – j – k.

Thus, from (5) the desired equation is

(x – 1) – (y + 1) – (z – 5) = 0 or –x + y + z = 3.

See FIGURE 9.6.6. ≡

A graph. A paraboloid, a plane and a vector are graphed on a three dimensional x y z coordinate system. The plane is tangent to the paraboloid at the point (1, negative 1, 5). The vector, labeled nabla F(1, negative 1, 5), is normal to the tangent plane. A dashed vertical line connects the marked points (1, negative 1, 0) and (1, negative 1, 5).

FIGURE 9.6.6 Tangent plane in Example 4

Normal Line

Let P(x0, y0, z0) be a point on the graph of F(x, y, z) = c, where ∇F is not 0. The line containing P(x0, y0, z0) that is parallel to ∇F(x0, y0, z0) is called the normal line to the surface at P. As indicated by its name, this line is normal to the tangent plane to the surface at P.

EXAMPLE 5 Normal Line to a Surface

Find parametric equations for the normal line to the surface in Example 4 at (1, –1, 5).

SOLUTION

A direction vector for the normal line at (1, –1, 5) is ∇F(1, –1, 5) = i – j – k. It follows that parametric equations for the normal line are

x = 1 + t, y = –1 – t, z = 5 – t. ≡

Expressed as symmetric equations the normal line to a surface F(x, y, z) = c at P(x0, y0, z0) is given by

.

In Example 5, you should verify that symmetric equations of the normal line at (1, –1, 5) are

.

REMARKS

Water flowing down a hill chooses a path in the direction of the greatest change in altitude. FIGURE 9.6.7 shows the contours, or level curves, of a hill. As shown in the figure, a stream starting at point P will take a path that is perpendicular to the contours. After reading Sections 9.5 and 9.6, the student should be able to explain why.

A visual representation of contours of a hill and a stream that is perpendicular to them. The contours are labeled in the following way, beginning from the outermost contour: 30, 40, 60, 80, and 100. A stream begins at a point labeled P on the innermost contour, and takes a path that is perpendicular to the contours.

FIGURE 9.6.7 Stream is perpendicular to contours

9.6 Exercises Answers to selected odd-numbered problems begin on page ANS-24.

In Problems 1–12, sketch the level curve or surface passing through the indicated point. Sketch the gradient at the point.

  1. f(x, y) = x – 2y; (6, 1)
  2. f(x, y) = ; (1, 3)
  3. f(x, y) = y – x2; (2, 5)
  4. f(x, y) = x2 + y2; (–1, 3)
  5. f(x, y) = ; (–2, –3)
  6. f(x, y) = ; (2, 2)
  7. f(x, y) = (x – 1)2 – y2; (1, 1)
  8. f(x, y) = ; (π/6,
  9. F(x, y, z) = y + z; (3, 1, 1)
  10. F(x, y, z) = x2 + y2 – z; (1, 1, 3)
  11. F(x, y, z) = ; (3, 4, 0)
  12. F(x, y, z) = x2 – y2 + z; (0, –1, 1)

In Problems 13 and 14, find the points on the given surface at which the gradient is parallel to the indicated vector.

  1. z = x2 + y2; 4i + j + k
  2. x3 + y2 + z = 15; 27i + 8j +k

In Problems 15–24, find an equation of the tangent plane to the graph of the given equation at the indicated point.

  1. x2 + y2 + z2 = 9; (–2, 2, 1)
  2. 5x2 – y2 + 4z2 = 8; (2, 4, 1)
  3. x2 – y2 – 3z2 = 5; (6, 2, 3)
  4. xy + yz + zx = 7; (1, –3, –5)
  5. z = 25 – x2 – y2; (3, –4, 0)
  6. xz = 6; (2, 0, 3)
  7. z = cos(2x + y); (π/2, π/4, –1/)
  8. x2y3 + 6z = 10; (2, 1, 1)
  9. z = ln(x2 + y2); (1/, 1/, 0)
  10. z = 8e−2y sin 4x; (π/24, 0, 4)

In Problems 25 and 26, find the points on the given surface at which the tangent plane is parallel to the indicated plane.

  1. x2 + y2 + z2 = 7; 2x + 4y + 6z = 1
  2. x2 – 2y2 – 3z2 = 33; 8x + 4y + 6z = 5
  3. Find points on the surface x2 + 4x + y2 + z2 – 2z = 11 at which the tangent plane is horizontal.
  4. Find points on the surface x2 + 3y2 + 4z2 – 2xy = 16 at which the tangent plane is parallel to (a) the xz-plane, (b) the yz-plane, and (c) the xy-plane.

In Problems 29 and 30, show that the second equation is an equation of the tangent plane to the graph of the first equation at (x0, y0, z0).

  3. Show that every tangent plane to the graph of z2 = x2 + y2 passes through the origin.
  4. Show that the sum of the x-, y-, and z-intercepts of every tangent plane to the graph of a > 0, is the number a.

In Problems 33 and 34, find parametric equations for the normal line at the indicated point. In Problems 35 and 36, find symmetric equations for the normal line.

  1. x2 + 2y2 + z2 = 4; (1, –1, 1)
  2. z = 2x2 – 4y2; (3, –2, 2)
  3. z = 4x2 + 9y2 + 1; (, , 3)
  4. x2 + y2 – z2 = 0; (3, 4, 5)
  5. Show that every normal line to the graph x2 + y2 + z2 = a2 passes through the origin.
  6. Two surfaces are said to be orthogonal at a point P of intersection if their normal lines at P are orthogonal. Prove that the surfaces given by F(x, y, z) = 0 and G(x, y, z) = 0 are orthogonal at P if and only if FxGx + FyGy + FzGz = 0.

In Problems 39 and 40, use the result of Problem 38 to show that the given surfaces are orthogonal at a point of intersection.

  1. x2 + y2 + z2 = 25; –x2 + y2 + z2 = 0
  2. x2 – y2 + z2 = 4; z = 1/xy2