LINEAR APPROXIMATION

Math21a, O. Knill

HOMEWORK: 13.7: 12, 16, 30, 34, 38

ESTIMATION. We continue the example and compare the value of f with the value of the linear approximation.
0.00943407 = f (1 + 0.01, 1 + 0.01) L(1 + 0.01, 1 + 0.01) = 0.01 20.01 + 3 = 0.00942478.
EXAMPLE (3D) Find the linear approximation to f (x, y, z) = xy + yz + zx at the point (1, 1, 1).

LINEAR APPROXIMATION.
1D: The linear approximation of a function f (x) at a point
x0 is the linear function
L(x) = f (x0 ) + f (x0 )(x x0 ) .
The graph of L is tangent to the graph of f at x0 .
2D: The linear approximation of a function f (x, y) at (x0 , y0 )
is
L(x, y) = f (x0 , y0 ) + fx (x0 , y0 )(x x0 ) + fy (x0 , y0 )(y y0 )
The level curve of g is tangent to the level curve of f at (x0 , y0 ).
The graph of L is tangent to the graph of f .
3D: The linear approximation of a function f (x, y, z) at
(x0 , y0 , z0 ) by
L(x, y, z) = f (x0 , y0 , z0 ) + fx (x0 , y0 , z0 )(x x0 ) +
.
fy (x0 , y0 , z0 )(y y0 ) + fz (x0 , y0 , z0 )(z z0 )
The level surface of L is tangent to the level surface of f at
(x0 , y0 , z0 ).
Using f = hfx , fy i, the linearization can be written as
L(~x) = f (~x0 ) + f (~x0 ) (~x ~x0 )

HOW CAN IT BE USED? Linearization is important because linear functions are easier to deal with. Using
linearization, one can estimate function values near known points.

We have f (1, 1, 1) = 3, f (x, y, z) = (y + z, x + z, y + x), f (1, 1, 1) = (2, 2, 2). Therefore L(x, y, z) =

EXAMPLE (3D). Use the best linear approximation to f (x, y, z) = ex yz to estimate the value of f at the
point (0.01, 24.8, 1.02).
Solution.
Take (x0 , y0 , z0 ) = (0, 25, 1), where f (x0 , y0 , z0 ) = 5.
The gradient is f (x, y, z) =

(ex yz, exz/(2 y), ex y). At the point (x0 , y0 , z0 ) = (0, 25, 1) the gradient is the vector (5, 1/10, 5). The linear
approximation is L(x, y, z) = f (x0 , y0 , z0 ) + f (x0 , y0 , z0 )(x x0 , y y0 , z z0 ) = 5 + (5, 1/10, 5)(x 0, y
25, z 1) = 5x + y/10 + 5z 2.5. We can approximate f (0.01, 24.8, 1.02) by 5 + (5, 1/10, 5) (0.01, 0.2, 0.02) =
5 + 0.05 0.02 + 0.10 = 5.13. The actual value is f (0.01, 24.8, 1.02) = 5.1306, very close to the estimate.
ABOUT DIMENSIONS. Do not mix up dimensions! For functions f (x, y) of two variables, the linear approximation is a function L(x, y) of two variables. We have tangency in two different dimensions: the level curves of
f are tangent to the level curves of L at (x0 , y0 ). But we also know that the graph of L is tangent to the graph
of f .
TANGENT LINES REVIEW. Because ~n = f (x0 , y0 ) = ha, bi is perpendicular to the level curve f (x, y) = c
through (x0 , y0 ), the equation for the tangent line is
ax + by = d, a = fx (x0 , y0 ), b = fy (x0 , y0 ), d = ax0 + by0
The tangent line is a level curve of L(x, y).
Example: Find the tangent to the graph of the function g(x) = x2
at the point (2, 4). Solution: the level curve f (x, y) = y x2 = 0
is the graph of a function g(x) = x2 and the tangent at a point
(2, g(2)) = (2, 4) is obtained by computing the gradient ha, bi =
f (2, 4) = hg (2), 1i = h4, 1i and forming 4x + y = d, where
d = 4 2 + 1 4 = 4. The answer is 4x + y = 4 which is the
line y = 4x 4 of slope 4. Graphs of 1D functions are curves in
the plane, you have computed tangents in single variable calculus.

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JUSTIFYING THE LINEAR APPROXIMATION.

TANGENT PLANES REVIEW. The tangent plane to the surface g(x, y, z) = z f (x, y) = 0 at (x0 , y0 , z0 =
f (x0 , z0 ) is fx x fy y + z = fx x0 fy y0 + z0 . This can be read as z = z0 + fx (x x0 ) + fy (y y0 ). Calling
the right hand side L(x, y) shows that the graph of L is tangent to the graph of f at (x0 , y0 ).

If the second variable y = y0 is fixed, then we have a one-dimensional situation where the only variable is x. Now
f (x, y0 ) = f (x0 , y0 ) + fx (x0 , y0 )(x x0 ) is the linear approximation. Similarly, if x = x0 is fixed y is the single
variable, then f (x0 , y) = f (x0 , y0 ) + fy (x0 , y0 )(y y0 ). Knowing the linear approximations in both the x and y
varibles, we can get the general linear approximation by f (x, y) = f (x0 , y0 )+fx (x0 , y0 )(xx0 )+fy (x0 , y0 )(yy0 ).

TOTAL DIFFERENTIAL. Aiming to estimate the change f = f (x, y) f (x0 , y0 ) of f for points (x, y) =
(x0 , y0 ) + (x, y) near (x0 , y0 ), we can estimate it with the linear approximation which is L(x, y) =
fx (x0 , y0 )x + fy y. In an old-fashioned notation, one writes also df = fx dx + fy dy and calls df the total
differential. One can totally avoid the notation of the total differential.

An other justification uses the chain rule: the vector ha, bi = hfx , fy i is perpendicular to the level curve at
(x0 , y0 ). Because the line ax + by = ax0 + by0 has also the vector (a, b) perpendicular to the curve and the curve
and line pass through the same point (x0 , y0 ), they are tangent. The line is the best among all lines passing
through (x0 , y0 ).

IN CLASS PROBLEM: Find the linear approximation L(x, y) to

EXAMPLE (2D) Find the linear approximation of the function f (x, y) = sin(xy 2 ) at the point (1, 1).
We have
(fx (x, y), yf (x, y) = (y 2 cos(xy 2 ), 2y cos(xy 2 )) which is at
the point (1, 1) equal to f (1, 1) = ( cos(), 2 cos()) =
(, 2). The linear function approximating f is L(x, y) =
f (1, 1)+(fx(1, 1), fy (1, 1))(x1, y1) = 0(x1)2(y1) =
x2y +3. The level curves of G are the lines x+2y = const.
The line which passes through (1, 1) satisfies x + 2y = 3.

at the point (0, /2) and Estimate f (1.01, /2 0.03).