1.3. TANGENT PLANES 25
1.3 TANGENT PLANES
One of the first things we built after developing skill with the derivative was tangent lines to
a curve. With a function z D f .x; y/ the analogous object is a tangent plane. ere are several
ways to specify a plane.
Knowledge Box 1.7
Formulas for planes
If a , b, c, and d are constants, all of the following formulas specify planes:
z D ax C by C c
ax C by C cz D d:
.a; b; c/ .x; y; z/ D d
Note that the third formula, while phrased in terms of a dot product, is actually
the same as the second.
Let’s get some practice with converting between the different possible forms of a plane.
Example 1.48 If
.3; 2; 5/ .x; y; z/ D 2
find the other two forms of the plane.
Solution:
.3; 2; 5/ .x; y; z/ D 2
3x 2y C 5z D 2 Second form
5z D 3x C 2y C 2
z D 0:6x C 0:4y C 0:4 First form
˙
e most common way to find a tangent line for a function is to take the point of tangency –
which must be on the line – together with the slope of the line found by computing the derivative
and use the point slope form to find the formula for the line. It turns out that there is a way of
specifying planes that is similar to the point slope form of a line. Remember that if Ev and Ew are
vectors that are at right angles to one another, then Ev Ew D 0.
26 1. ADVANCED DERIVATIVES
Knowledge Box 1.8
Formula for a plane at right angles to a vector
Suppose that Ev is a vector at right angles to a plane in three dimensions and that
.a; b; c/ is a point on that plane. en a formula for the plane is
Ev .x a; y b; z c/ D 0:
Notice that .a; b; c/ is constructively on the plane and that our knowledge of the
dot product tells us it is at right angles to Ev.
Example 1.49 Find the plane at right angles to Ev D .1; 2; 1/ through the point .2; 1; 5/.
Solution:
Simply substitute into the formula given in Knowledge Box 1.8.
.1; 2; 1/ .x 2; y C 1; z 5/ D 0
1.x 2/ C 2.y C 1/ C1.z 5/ D 0
x C 2y C z 2 C2 5 D 0
x C 2y C z D 5
˙
Knowledge Box 1.8 seems very special purpose, but it turns out to be very useful in light of
another fact. Suppose that
f .x; y; z/ D c
specifies a surface. We need to expand our definition of the gradient just a bit to
rf .x; y; z/ D
f
x
.x; y; z/; f
y
.x; y; z/; f
z
.x; y; z/
:
In this case the vector rf .a; b; c/ points directly outward from the surface f .x; y; z/ D c at
.a; b; c/ and it is at right angles to the tangent plane.
1.3. TANGENT PLANES 27
Knowledge Box 1.9
Formula for the tangent plane to a surface
If f .x; y; z/ D c defines a surface, and .a; b; c/ is a point on the surface, then a
formula for the tangent plane to that surface at .a; b; c/ is:
f
x
.a; b; c/; f
y
.a; b; c/; f
z
.a; b; c/
.x a; y b; z c/ D 0:
Defining a surface in the form f .x; y; z/ D c is a little bit new – but in fact this is another
version of level curves, just one dimension higher. Let’s practice.
Example 1.50 Find the tangent plane to the surface x
2
C y
2
C z
2
D 3 at the point .1; 1; 1/.
Solution:
Check that the point is on the surface: .1/
2
C .1/
2
C .1/
2
D 3 – so it is. Next find the
gradient
rx
2
C y
2
C z
2
D .2x; 2y; 2z/:
is means that the gradient at .1; 1; 1/ is Ev D .2; 2; 2/. is makes the plane
.2; 2; 2/ .x 1; y C 1; z 1/ D 0
2x 2y C 2z 2 2 2 D 0
2x 2y C 2z D 6
x y C z D 3
Notice that we simplified the form of the plane; this is not required but it does make for neater
answers.
˙
e problem with finding the tangent plane to a surface f .x; y; z/ D c is that it does not solve the
original problem – finding tangent planes to z D f .x; y/. A modest amount of algebra solves
this problem. If z D f .x; y/ then g.x; y; z/ D z f .x; y/ D 0 is in the correct form for our
surface techniques. is gives us a new way of finding tangent planes to a function that defines
a surface in 3-space.
28 1. ADVANCED DERIVATIVES
Knowledge Box 1.10
Formula for the tangent plane to a functional surface
If z D f .x; y/ defines a surface, then the tangent plane to the surface at .a; b/
may be obtained as
f
x
.a; b/; f
y
.a; b/; 1
.x a; y b; z f .a; b// D 0:
is is the result of applying the gradient-of-a-surface formula to the surface z
f .x; y/ D 0 at the point .a; b; f .a; b//.
Example 1.51 Find the tangent plane to f .x; y/ D x
2
y
3
at the point .2; 1/.
Solution:
Assemble the pieces and plug into Knowledge Box 1.10.
f
x
.x; y/ D 2x
f
x
.2; 1/ D 4
f
y
.x; y/ D 3y
2
f
y
.2; 1/ D 3
f .2; 1/ D 4 .1/ D 5
Put the plane together
.4; 3; 1/ .x 2; y C 1; z 5/ D 0
4x C 8 C 3y C 3 C z 5 D 0
4x C 3y C z D 6
Which is the tangent plane desired.
˙
1.3. TANGENT PLANES 29
Example 1.52 Find the tangent plane to
g.x; y/ D xy
2
at (3,1).
Solution:
Assemble the pieces and plug into Knowledge Box 1.10.
f
x
.x; y/ D y
2
f
x
.3; 1/ D 1
f
y
.x; y/ D 2xy
f
y
.3; 1/ D 6
f .3; 1/ D 3
Put the plane together
.1; 6; 1/ .x 3; y 1; z 3/ D 0
x C 3 6y C 6 C z 3 D 0
x 6y C z D 6
˙
PROBLEMS
Problem 1.53 Find the plane through (1,-1,1) at right angles to Ev D .2; 2; 1/.
Problem 1.54 Find the plane through (2,0,5) at right angles to Ev D .1; 1; 1/.
Problem 1.55 Find the plane through (3,2,1) at right angles to Ev D .1; 1; 2/.
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