Linearization and Tangent Planes
The Total Derivative
To this point, we have considered only partial derivatives of a function f (x; y) :
In this section, we introduce the concept of a total derivative of a transformation.
Recall that a function f (x) of one variable is said to be di¤erentiable at p if
the following limit exists:
f 0 (p) = lim
h!0
f (p + h)
h
f (p)
(1)
The problem with extending (1) to higher dimensions is that it requires h to
become a vector, and yet we cannot "divide" by a vector. However, we can
rewrite (1) in the form
lim
h!0
f (p + h)
h
f (p)
f 0 (p) = 0
and thus, we obtain also that
lim
h!0
f (p + h)
h
f (p)
f 0 (p) h
=0
h
Consequently, the de…nition of the derivative (1) can be re-interpreted to mean
that f (x) is di¤erentiable at p if there is a number f 0 (p) such that
lim
h!0
jf (p + h)
f (p)
jhj
f 0 (p) hj
=0
In this formulation, division is by jhj ; which naturally generalizes to the norm
of a vector h = hh; ki : Indeed, if we denote f (x; y) by f (x), where x = hx; yi ;
and let p = (p; q) be a point, then we have the following:
De…nition 5.1: A function f (x; y) is di¤ erentiable at a point (p; q)
if there exists a vector rf (p) for which
lim
h!0
jf (p + h)
f (p)
jjhjj
rf (p) hj
=0
When it exists, rf (p) is the total derivative of f (x) at p.
The vector rf (p) is also called the gradient of f (x) at p:
1
(2)
If we write rf (p) = ha; bi, then we can determine the values of a and b
by evaluating the limit in (2) in two di¤erent directions. Along the x-axis,
h = hh; 0i and thus we have
lim
h!0
jf (p + h; q)
f (p; q) ha; bi hh; 0ij
jjhh; 0ijj
f (p + h; q) f (p; q) ah
lim
h!0
h
f (p + h; q) f (p; q)
lim
a
h!0
h
=
0
=
0
=
0
This implies that
a = lim
h!0
f (p + h; q)
h
f (p; q)
= fx (p; q)
Similarly, (see the exercises), it can be shown that b = fy (p; q) ; so that the
gradient is given by
rf (p; q) = hfx (p; q) ; fy (p; q)i
This yields the following:
Theorem 5.2: If f (x; y) is di¤erentiable at a point (p; q) ; then its total
derivative is given by the gradient of f at (p; q) ; which is
rf (p; q) = hfx (p; q) ; fy (p; q)i
The total derivative is also known as the Jacobian of the mapping f (x; y) for
reasons that will become apparent in the next chapter.
EXAMPLE 1
Find the total derivative (i.e., gradient) of
f (x; y) = x2 + 3xy
Solution: Since fx = 2x + 3y and fy = 3x; the total derivative is
rf = h2x + 3y; 3xi
De…nition 5.1 can be applied to a function f of any number of variables, in
which case Theorem 5.2 says that rf is the vector of …rst partial derivatives.
For example, the gradient of a function of 3 variables U (x; y; z) is given by
rU = hUx ; Uy ; Uz i
and rU is the total derivative of U (x; y; z) :
2
Check your Reading: How is the tangent plane to the surface related to the
surface itself?
Linearization and Tangent Planes
If we let x = p + h; then h = x
the following form:
p; so that De…nition 5.1 can be restated in
De…nition 5.1a: A function f (x; y) is di¤ erentiable at a point (p; q)
if there exists a vector rf (p) for which
lim
x!p
jf (x)
f (p) rf (p) (x
jjx pjj
p)j
=0
(3)
When it exists, rf (p) is the total derivative of f (x) at p.
The de…nition of the limit then implies that for any " > 0; there is a neighborhood of p for which
jf (x)
[f (p) + rf (p) (x
p)]j < " jjx
(4)
pjj
for all x in that neighborhood.
Consequently, we de…ne the linearization of f (x) at p to be
Lp (x) = f (p) + rf (p) (x
p)
which in non-vector form is
Lp (x; y) = f (p; q) + fx (p; q) (x
p) + fy (p; q) (y
q)
It follows that
jf (x)
Lp (x)j < " jjx
pjj
which means that the graph of Lp (x; y) is a plane that is practically the same
as the surface z = f (x; y) near the point (p; q; f (p; q)) :
3
We say that z = Lp (x; y) is the tangent plane to z = f (x; y) at (p; q) :
EXAMPLE 2
p = (1; 1) :
Find the tangent plane to f (x; y) = 9 x2 y 2 when
Solution: To begin with, fx (x; y) =
that fx (1; 1) = fy (1; 1) = 2 and
Lp (x; y)
2x and fy (x; y) =
= f (1; 1) + fx (1; 1) (x 1) + fy (1; 1) (y
= 7 2 (x 1) 2 (y 1)
2y; so
1)
which simpli…es to Lp (x; y) = 11 2x 2y: Thus, the equation of
the tangent plane to f (x; y) at the point (1; 1; 7) is
z = 11
2x
2y
which is shown below.
The tangent plane is the plane that “best approximates”a surface in the neighborhood of a point. However, this does not exclude the possibility that a tangent
plane may intersect a surface in an in…nite number of points.
EXAMPLE 3
(1; 2) :
Find the linearization of f (x; y) = x3
4
xy 2 at the point
Solution: Since fx = 3x2
fx (1; 2) = 3
Since f (1; 2) =
is
4=
y 2 and fy =
1;
2xy; we have
fy (1; 2) =
2 1 2=
4
3; the linearization of f (x; y) = sin x2 y at (2; )
Lp (x; y) =
3
1 (x
1)
4 (y
2)
which simpli…es to Lp (x; y) = 6 x 4y: The graph of Lp (x; y) is
shown versus the graph of f (x; y) below:
Check your Reading: Which of the two tangent planes below is tangent to
5
the surface?
Quadratic Approximations
The "second derivative" of a function f (x; y) is de…ned using matrices, where a
matrix A is a 2 dimensional rectangular array of numbers, such as for example
A=
1
1
2
0
5
3
If the length of the rows of A is the same as the length of the columns of B;
then the product AB is de…ned to be the matrix of inner products of rows of A
with columns of B: For example, if
2
3
A=
1
5
and
B=
4
0
3
2
then their product is the matrix of inner products of rows of A with columns of
B:
AB =
2
3
1
5
4
0
3
2
=
2 ( 4) + 1 0 2 3 + 1 2
3 ( 4) + 5 0 3 3 + 5 2
=
8
12
8
19
The transposition of a matrix A is a matrix At whose rows are the columns
6
of A. For example,
2
a1
a2
..
.
6
6
t
[a1 ; a2 ; : : : ; an ] = 6
4
an
3
7
7
7
5
Vectors v = ha; bi will be denoted in matrix form by a column vector,
a
b
v=
The "second derivative" of f (x; y) at a point (p; q) is given by the Hessian
matrix of f (x; y) at a point (p; q) ; which is de…ned to be
fxx (p; q) fxy (p; q)
fxy (p; q) fyy (p; q)
Hf (p) =
Also, the quadratic approximation Qp (x; y) of f at (p; q) is de…ned to be
Qp (x) = Lp (x) +
where (x
t
p) = [x
p; y
1
(x
2
t
p) Hf (p) (x
p)
q] :
EXAMPLE 4 Find the quadratic approximation of f (x; y) = x3
3x + y 2 at (1; 2) :
Solution: First, fx = 3x2 3 implies that fx (1; 2) = 0, and fy = 2y
implies that fy (1; 2) = 4: Since f (1; 2) = 2; the linearization of
f (x; y) at (1; 2) is
Lp (x; y) = 2 + 0 (x
1) + 4 (y
2) = 2 + 4 (y
2)
Since fxx = 6x; fxy = 0; and fyy = 2; the Hessian at (1; 2) is
Hf (1; 2) =
Since (x
p) = [x
1; y
Qp (x; y) = Lp (x; y) +
6
0
0
2
t
2] ; we have
1
[x
2
1; y
2]
6
0
0
2
x
y
1
2
Matrix arithmetic then leads to
Qp (x; y)
1
6 (x 1)
[x 1; y 2]
2 (y 2)
2
1
1
= Lp (x; y) + (x 1) 6 (x 1) + (y
2
2
2
2
= Lp (x; y) + 3 (x 1) + (y 2)
= Lp (x; y) +
7
2) 2 (y
2)
Substitution of Lp (x; y) = 2 + 4 (y
Qp (x; y) = 2 + 4 (y
2) then leads to
2) + 3 (x
2
1) + (y
2
2)
It can be show that for all " > 0; there is a neighborhood of p on which
jf (x)
Qp (x)j < " jjx
2
pjj
Thus, a quadratic approximation is an even better approximation than is a
linearization.
Find the quadratic approximation of f (x; y) = sin x2 + xy + y 2
EXAMPLE 5
at (0; 0)
Solution: First, fx = cos x2 + xy + y 2 (2x + y) and fy = cos x2 + xy + y 2 (x + 2y) :
Thus, f (0; 0) = fx (0; 0) = fy (0; 0) = 0 and similarly, L (x; y) = 0:
However,
fxx
=
=
@
cos x2 + xy + y 2 (2x + y)
@x
2
2 cos x2 + xy + y 2
(2x + y) sin x2 + xy + y 2
from which it follows that fxx (0; 0) = 2: Likewise,
fxy
=
cos x2 + xy + y 2
fyy
=
2 cos x2 + xy + y 2
sin x2 + xy + y 2 (x + 2y) (2x + y)
2
sin x2 + xy + y 2 (x + 2y)
Thus, fxy (0; 0) = 1, fyy (0; 0) = 2 and the Hessian is
2
1
Hf (0; 0) =
1
2
Consequently, the quadratic approximation is
Qp (x; y) = L (x; y) +
1
[x
2
0; y
0]
2
1
1
2
x
y
0
0
Matrix arithmetic and L (x; y) = 0 then leads to
Qp (x; y) =
1
[x; y]
2
2x + y
x + 2y
=
1
1
x (2x + y) + y (x + 2y)
2
2
However, this in turn simpli…es to Qp (x; y) = x2 + xy + y 2 :
8
Below we have plotted Qp (x; y) = x2 +xy+y 2 versus f (x; y) = sin x2 + xy + y 2
.
They are practically the same at (0; 0) ; and indeed, notice that both surfaces
have a minimum at (0; 0) :
Check your Reading: Is the product of a row of length n and a column of
length n the same as an inner product?
Di¤erentiability and Di¤erentials
If in De…nition 5.1 we instead let h =
lim
x!0
j f
x and
rf (p)
jj xjj
f = f (p +
xj
x)
f (p) ; then
=0
Thus, for any " > 0; there is a neighborhood of 0 such that
j f
for all nonzero
rf (p)
k xk
xj
<"
x in that neighborhood. Equivalently,
j f
rf (p)
xj < " k xk
and if we further allow that " ( x) is a continuous function for which " (0) = 0;
then we obtain the following:
9
De…nition 5.1b: A function f (x) of n-variables is di¤ erentiable
at a point p with total derivative rf (p) if there is a continuous
function " ( x) with " (0) = 0 such that
j f
rf (p)
xj < " ( x) k xk
on some neighborhood of 0.
If we let dz = rf (p)
that
x and let
x = hdx; dyi ; then Theorem 5.2 implies
dz = fx (p; q) dx + fy (p; q) dy
De…nition 5.1b then implies that
z
dz
whenever dx and dy are su¢ ciently close to 0.
EXAMPLE 6 Compute z and dz for f (x; y) = x2 + xy when
(p; q) = (1; 2) ; dx = x = 0:01; and dy = y = 0:03:
Solution: To begin with, f (1; 2) = 12 +1 2 = 3 and f (1:01; 2:03) =
2
(1:01) + 1:01 2:03 = 3: 0704, so that
z = f (1:01; 2:03)
f (1; 2) = 3: 0704
3 = 0:0704
Moreover, fx (x; y) = 2x + y and fy (x; y) = x; so that
fx (1; 2) = 2 1 + 2 = 4
and
fy (1; 2) = 1
As a result, we have
dz = fx (1; 2) dx + fy (1; 2) dy = 4 0:01 + 1 0:03 = 0:07
which is within 0:0004 of
z:
Let’s look at an application of the di¤erential, one in which dx and dy are both
interpreted to be tolerances due to the inaccuracy of measuring devices.
EXAMPLE 7 A large co¤ee can has a height of 6:500 to within an
accuracy of 1=1600 and a base with a radius of 3:2500 to within an
10
accuracy of 1=1600 : Find the volume V and the approximate error
dV in the volume of the can.
Solution: If h denotes the height and r denotes the radius of the
base of the can, then
V = r2 h
When h = 6:5 and r = 3:25; then V =
inches. Moreover,
2
(3:25)
6:5 = 215: 69 cubic
dV = Vh (6:5; 3:25) dh + Vr (6:5; 3:25) dr
where Vh = r2 and Vr = 2 rh: Since dh = dr = 1=1600 , we thus
have
dV =
2
(3:25)
1
1
+ 2 (3:25) (6:25)
= 10:05 in3
16
16
Thus, the volume of the co¤ee can is 215:7 in3 ; give or take approximately 10 in3 :
Exercises
Find the linearization L (x; y) of f (x; y) at the given point in the xy-plane.
Then graph L (x; y) and f (x; y) to illustrate that z = L (x; y) is tangent to
z = f (x; y).
1.
3.
5.
7.
9.
11.
f (x; y) = x2 + y 3 at (1; 2)
f (x; y) = x2 + xy + 3x at (0; 0)
f (x; y) = ln x2 + y 2 at (1; 1)
f (x; y) = x2 sin (y) at (0; )
f (x; y) = 3x + 7y + 1 at (1; 2)
f (x; y) = x tan (xy) at (1; )
13.
f (x; y) =
1
(x2 +y 2 )3=2
at (0; 1)
11
2.
4.
6.
8.
10.
12.
f (x; y) = x2 y + y 2 at (1; 2)
2
f (x; y) = (x + 2y) at (1; 3)
f (x; y) = x sin (xy) at (1; )
f (x; y) = ex ln y 2 + 1 at (1; 0)
f (x; y) = 3x + 7y + 1 at (1; 2)
f (x; y) = csc (x + y) at 2 ; 3
14.
f (x; y) =
1
1 xy
at (0; 0)
Find the Hessian matrix and the quadratic approximation of f at the given
point.
15.
17.
: 19.
21.
23.
f (x; y) = x2 + y 3 at (1; 2)
f (x; y) = x2 + xy + 3x at (0; 0)
f (x; y) = 3x 2y + 1
f (x; y) = ln x2 + y 2 at (1; 1)
f (x; y) = x2 sin (y) at (0; )
Find
z and dz at the given point in the xy-plane.
25.
27.
29.
z = x2 + 2y at (1; 2)
dx = x = 0:1; dy = y = 0:01
z = x tan (xy) at (1; )
dx = x = 0:01; dy = y = 0:01
z = tan 1 xy at (1; 1)
dx = x = 0:01; dy = y = 0:01
16.
18.
20.
22.
24.
26.
28.
30.
f (x; y) = x2 + 2xy + y 3 at (2; 2)
2
f (x; y) = (x + 2y) at (1; 3)
f (x; y) = 5
f (x; y) = x sin (xy) at (1; )
f (x; y) = ex ln y 2 + 1 at (1; 0)
z = x2 y + y 2 at (1; 1)
dx = x = 0:1; dy = y = 0:01
2
z = (x + 2y) at (1; 1)
dx = x = 0:01; dy = y = 0:1
2
2
z = e x y at (0; 0)
dx = x = 0:1; dy = y = 0:2
31. A rectangular box has a height of 4 feet to within an accuracy of 1 in and
a square base with width 2 feet to within an accuracy of 1 in. Find the volume
V and the approximate error dV in the volume of the can.
32. The length of a pendulum is measured to be l = 5 feet to within 1 in of
accuracy. The period of the pendulum as it swings is measured to be T = 2:5
seconds to within 0:1 seconds. The acceleration due to gravity is related to the
swinging of the pendulum by
4 2l
g= 2
T
What is the acceleration due to gravity with respect to these measurements and
about how accurate is the computed value of the acceleration?
33. The diameter of the top of a soup can is measured to be 3 inches to within
1/16 of an inch. The height of the can is measured to be 4 inches to within 1/16
of an inch.
a. Write the surface area S as a function of the height and diameter of the
can.
b. Find S when the height is 400 and the diameter is 300 :
c. Find dS when the height is 400 ; the diameter is 300 ; and the di¤erentials
of height and diameter are both 1=1600 : About how much variation
in the area computation is possible given that height and diameter
are accurate only to a sixteenth of an inch?
34. Find the volume V and the di¤erential dV for the can in exercise 33. About
how much variation in the volume computation is possible given that height and
diameter are accurate only to a sixteenth of an inch?
35. Show that the tangent plane to any plane of the form z = ax + by + c is
the plane itself.
12
36. Show that the quadratic approximation at (0; 0) of
Q (x; y) = ax2 + bxy + cy 2
is Q (x; y) itself.
37. The mirror for a telescope should be in the shape of a paraboloid with
focus (0; 0; p), which is the graph of a function of the form
Q (x; y) =
x2 + y 2
4p
(5)
For example, a telescope mirror with focus at (0; 0; 1) is shown below:
However, if the telescope mirror is small enough, it more economical to manufacture a spherical cross-section that approximates (5). The graph of a spherical
cross-section of radius R is given by
p
R2 x2 y 2
f (x; y) = R
Find the Quadratic approximation of f at (0; 0) and show that it is of the form
(5). What is R in terms of p? How does this relate to the manufacture of
spherical mirrors as approximations to parabolic mirrors?
38. In problem 37, what value of R leads to its quadratic approximation having
a focus of (0; 0; 2)?
39. Use the de…nition of the total derivative to show that if rf (p; q) = ha; bi ;
then b = fy (p; q) :
40. Show that if f (x) is di¤erentiable at p; then there exists a neighborhood
of p on which
rf (p) (x
p)
" jjx
pjj < f (x)
f (p) < rf (p) (x
p) + " jjx
pjj
for all x in that neighborhood. Use this to explain why f (x) is continuous at
p . (that is, why
lim f (x) = f (p) )
x!p
41. Show that the tangent plane to the graph of f (x; y) = x2
form
z = 2px 2qy
p2 q 2
13
y 2 is of the
(6)
Then show that the intersection of z = x2 y 2 with its tangent plane (6) results
in a pair of straight lines. How many lines does the surface z = x2 y 2 contain?
42. Write to Learn: Write a short essay which explains why if f (x; y) is a
polynomial such as f (x; y) = x2 y + y 2 ; then fx and fy are the coe¢ cients of h
and k in the expansion of f (x + h; y + k) : How does this relate to the concept
of di¤erentiability?
43.
p Write to Learn: In a short essay, explain why the function f (x; y) =
x2 + y 2 is not di¤erentiable at (0; 0) : (Hint: consider the graph of f (x; y)
and also consider the de…nition of di¤erentiability).
44. Show that fx (0; 0) exists but fy (0; 0) does not exist when
p
f (x; y) = x4 + y 2
Does the total derivative of f (x; y) exist at (0; 0)? Explain.
45. Explain how De…nition 5.1 applies to functions of 3 variables U (x; y; z)
and then mimic the derivation which follows to show that
rU (p) = hUx (p) ; Uy (p) ; Uz (p)i
46. For f (x) di¤erentiable at a point p;let us de…ne a function A ( x) by
A ( x) = rf (p)
x
Show that A ( x) is linear (i.e., that A ( x + y) = A ( x) + A ( y) and
A (k x) = kA ( x) for any scalar k ). Also, explain the signi…cance of vectors
x for which A ( x) = 0:
14