I.f Tangent Planes and Normal Lines

Again we begin by:

Recall: (1) Given two vectors A = a1i + a2j + a3k, B = b1i + b2j + b3k then A×B

is a vector perpendicular to both A and B. Then length of A × B is |A| |B| sin θ,

while its direction is given by the right hand rule: point fingers of the right hand in

the direction of A, curl them in the direction of B. Then the thumb points in the

direction of A × B. To compute A × B, it is convenient to use the formula:

A × B = det

i j k

a1 a2 a3

b1 b2 b3

.

A BB

A

Observe that A× B = −(B × A), while if A and B are parallel, A × B = 0.

(2) A plane is determined by a vector N = αi + βj + γk (called the normal

vector), and a point P0 = (x0, y0, z0) on the plane. Any other point P = (x, y, z) of

space is then on the plane iff the vector−−→P0P is perpendicular to N, i.e.,

−−→P0P ·N = 0,

or

(x − x0)α + (y − y0)β + (z − z0)γ = 0.

50

N

P

P

0

(3) Instead of a vector N and a point P0, we may specify three points of the

plane P0, P1, P2 (not collinear).

P 0 P 1

P 2

N

We then construct the vectors−−→P0P1,

−−→P0P2 and set N =

−−→P0P1 ×

−−→P0P2. For planes it

does matter whether we choose N or −N as the normal!

(4) Curves in space are given in parametric form as:

x = x(t)

y = y(t)

z = z(t)

a ≤ t ≤ b.

This is a very convenient way for us! Remember, once again, that z = f(x, y)

gives a surface, not a curve. Suppose the curve C is given in the above form. Then51

r(t0) = x(t0)i+ y(t0)j+ z(t0)k is the vector from the origin (0, 0, 0) to the point on

C at t = t0. We recall that dr/dt = (dx/dt)i + (dy/dt)j + (dz/dt)k gives a tangent

vector to the curve C.

d r

d t

( t )r

C

x

y

z

Suppose now we are given a surface in (x, y, z) space. It is convenient for us to write

the surface S in the form F (x, y, z) = 0, so if we had the equation z = f(x, y), then

we would get F (x, y, z) = z − f(x, y). Let P0 = (x0, y0, z0) be a point of S. Then

the Tangent Plane T of S at P0 is the plane which: (1) contains P0 and (2) if C is

any curve on S through P0, then the tangent vector to C at P0 is on the plane.

C1

C 2C3

P 0

S

The tangent plane plays the same role for surfaces that the tangent line played for

y = f(x) in the (x, y) plane. We obtain the equation of the tangent plane T at P0

as follows: We know one point of T (namely P0), so we need only find the normal

vector N. To do this, let C be any given curve through P0 such that C lies on S.52

Then C is given by

x = x(t)

y = y(t)

z = z(t)

a ≤ t ≤ b

for some functions x(t), y(t), z(t). On the other hand, since C is on the surface S,

we must have F (x(t), y(t), z(t)) = 0 for all t! I.e., differentiating both sides with

respect to t, we obtain

∂F

∂x

dx

dt+

∂F

∂y

dy

dt+

∂F

∂z

dz

dt= 0.

F

P 0

rdd t

C

We know that dr/dt = (dx/dt)i + (dy/dt)j + (dz/dt)k is tangent to C at P0,

and thus ∇F = (∂F/∂x)i + (∂F/∂y)j + (∂F/∂z)k is perpendicular to dr/dt! But

C was not specified! So ∇F is perpendicular to the tangent of any curve, and we

conclude that we may take ∇F as the normal to the tangent plane!

In summary the equation of the tangent plane to F (x, y, z) = 0 at P0 =

(x0, y0, z0) is

∂F

∂x(x0, y0, z0)(x − x0) +

∂F

∂y(x0, y0, z0)(y − y0) +

∂F

∂z(x0, y0, z0)(z − z0) = 0.

By definition, the line through P0 in the direction of ∇F (x0, y0, z0) is called53

the Normal Line. Its equation is

x = x0 + t∂F

∂x(x0, y0, z0)

y = y0 + t∂F

∂y(x0, y0, z0)

z = z0 + t∂F

∂z(x0, y0, z0)

−∞ < t < ∞.

Finally, two surfaces are perpendicular (or orthogonal) at a point of intersection

iff the normal vectors N1, N2 of their respective tangent planes are perpendicular:

N1 · N2 = 0.

Remark. Observe that ∇F is also perpendicular to the equipotential surfaces

F (x, y, z) = c, with c = constant, by exactly the same calculations. Earlier we

showed that ∇F gave the direction of maximal rate of increase for the function

F (x, y, z). We now also know that ∇F is perpendicular to the surface F (x, y, z) = 0.

F

F = c

We thus have that the maximal rate of increase of F (x, y, z) can be obtained by

going in a perpendicular direction to the equipotential surfaces: F (x, y, z) = c.

We illustrate the above results with the following examples.

Example 1. Find the directional derivative of

f = x sin(ez) + yex at the point (0, 1, 1)54

in the direction of the vector v = i − j + k.

Answer. ∇f = (sin(ez) + yex)i + exj + x cos(ez)ezk, so

∇f∣

∣

∣

(0,1,1)= (sin e + 1)i + j + 0k

while a unit vector u in the direction of v is

u =i − j + k√

12 + 12 + 12.

Finally,

df

dx

∣

∣

∣

∣

v

= ∇f · u =(sin e + 1) − 1√

3=

sin(e)√3

.

Example 2. Let the electrical potential V be given by

V = 5x2 + 3y3z + xz4.

You are at the point (1, 1, 0).

(a) In which direction should you head for the maximal rate of change of V with

respect to distance? What is the maximal rate?

(b) Do the same for the minimal (i.e., most negative) rate.

Answer. ∇V = (10x + z4)i + 9y2zj + (3y3 + 4z3x)k. Therefore at (1, 1, 0),

∇V

∣

∣

∣

∣

(1,1,0)

= 10i + 0j + 3k.

(a) We have the direction of maximal rate of change = direction of ∇V = 10i+3k.

The rate = |∇V | =√

109.

(b) The direction of the most negative rate of change = direction of −∇V = −10i−

3k. This rate= −|∇V | = −√

109.55

Example 3. Let (u, v) be differentiable functions. Show:

∇(uv) = v∇u + u∇v.

Answer.

∇(uv) =∂

∂x(uv)i +

∂

∂y(uv)j +

∂

∂z(uv)k

=

[

∂u

∂xv + u

∂v

∂x

]

i +

[

∂u

∂yv + u

∂v

∂y

]

j +

[

∂u

∂zv + u

∂v

∂z

]

i

= v

[

∂u

∂xi +

∂u

∂yj +

∂u

∂zk

]

+ u

[

∂v

∂xi +

∂v

∂yj +

∂v

∂zk

]

= v∇u + u∇v.

Example 4. Find the tangent plane and normal line to the surface z = x2 + y2

at the point x = 1, y = −1.

Answer. Write the surface as F (x, y, z) = 0. Here F (x, y, z) = x2 + y2 − z. Then

the normal vector to the tangent plane is

∇F = 2xi + 2yj − k

∣

∣

∣

∣

(1,−1,2)

= 2i− 2j − k.

So the equation of the tangent plane is

(2i − 2j − k) · ((x − 1)i + (y − (−1))j + (z − 2)k) = 0

or 2(x − 1) − 2(y + 1) − (z − 2) = 0, while the normal line is:

x = 1 + 2t

y = −1 − 2t

z = 2 − t

−∞ < t < ∞.

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Example 5. Find the point(s) on the sphere z2 + x2 + y2 = 1 where the tangent

plane is parallel to the plane x + y + z = 17.

Answer. The normal vector N1, to the given plane is

N1 = i + j + k.

We now calculate the normal N2 to the sphere x2 + y2 + z2 = 1. Here F (x, y, z) =

x2 + y2 + z2 − 1 and so

N2 = ∇F = 2xi + 2yj + 2zk.

We wish N2 and N1 to be parallel, i.e., N1 × N2 = 0 or

det

i j k

1 1 12x 2y 2z

= 0.

Equivalently,

i[2z − 2y] − j[2z − 2x] + k[2y − 2x] = 0.

We have

z = y, z = x, y = x.

Note that only two of these three equations are useful, since they imply the third.

We choose: y = z and x = z. Then, we keep in mind that (x, y, z) is on the sphere!

I.e., x2 + y2 + z2 = 1, and so we must have

z2 + z2 + z2 = 1 or z2 =1

3.

We thus find z = 1/√

3, z = −1/√

3 and the two points on the sphere are (1/√

3, 1/√

3, 1/√

3)

and (−1/√

3,−1/√

3,−1/√

3).57

Example 6. Find parametric equations for the tangent line to the curve of inter-

section of the paraboloid z = 4(x2 + y2) and the ellipsoid x2 + 4y2 + z2 = 21 at the

point (1, 1, 4).

Answer. Note that the intersection curve(s) satisfy

x2 + (z − 4x2) + z2 = 21

or

−3x2 + z2 + z = 21

and

4y2 = 21 − z2 − x2 = 21 − z2 +21 − z2 − z

3.

These last two equations give x and y as functions of z by taking square roots.

Clearly these equations imply that the intersection is complicated! Proceeding

directly does not seem too wise. Instead, note that the curve of intersection must

lie on both surfaces simultaneously. It follows that its tangent line must have a

direction vector V which is perpendicular to the normals N1 and N2 of the two

surfaces. We may therefore take V = N1 ×N2. Now z = 4(x2 + y2) can be written

as 4(x2 + y2) − z = 0. Its normal N1 is thus

N1 = 8xi + 8yj− 1k

and

N1

∣

∣

∣

∣

(1,1,4)

= 8i + 8j − k.

58

In the same way, the ellipsoid x2 + 4y2 + z2 = 21 has normal N2 given by

N2 = 2xi + 8yj + 2zk

and

N2

∣

∣

∣

∣

(1,1,4)

= 2i + 8j + 8k.

We then have

V = det

i j k

8 8 −12 8 8

= 72i− 66j + 48k

and the equation of the tangent line becomes:

x = 1 + 72t

y = 1 − 66t

z = 4 + 48t

−∞ < t < ∞.

59

Further Exercises:

1) Find the tangent plane and normal line to z2 = x2 + y2 at (1, 0, 1).

2) Find the tangent plane and normal line to x2 +y2

4+

z2

9= 1 at (1, 0, 0).

3) Find the tangent plane and normal line to the cylinder x2+y2 = 8 at (2, 2,−1).

4) Find all points of the surface x = y2 + z2 where the normal line intersects the

plane x + y + z = 1 at right angles. Find the points of the plane where the

normal lines intersect the plane.

5) Find all points of the surface z = 4x2 +9y2 where the tangent plane is parallel

to the plane 16x + 18y + 2z = 1.

6) Find a tangent vector at the point (1, 0, 1) to the curve of intersection of the

surfaces z2 = x2 + y2 and z = x + 4y.

7) Show that the normal line at any point on the surface (x − 1)2 + y2 + z2 = 1

passes through the point (1, 0, 0).

8) Show that the normal line to the surface z = x2 + y2 at the point (x0, y0, z0)

must contain the point (0, 0, 1 + z0).

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