Math 2224 Multivariable Calc–Sec. 13.6: Tangent Planes and Differentials
I.
Tangent Planes and Normal Line
A. Definitions
1. The tangent plane at the point P0 (x0 , y0 , z0 ) on the level surface
differentiable function f is the plane through P0 normal to
2. The normal line of the surface at
B. Equations
1. Tangent Plane to
!f
P0
f (x, y, z) = c of a
.
P0 is the line through P0 parallel to !f
P0
.
f (x, y, z) = c at P0 (x0 , y0 , z0 )
fx (P0 )(x ! x0 ) + fy (P0 )(y ! y0 ) + fz (P0 )(z ! z0 ) = 0
OR from math 1224:
where
!
! !
Ax + By + Cz = ! ! P0 " f x (P0 ) x + f y (P0 ) y + fz (P0 ) z = !f (P0 )" P0 ,
!
! = A, B, C = f x (P0 ) , f y (P0 ) , fz (P0 ) = !f (P0 )
2. Normal Line to
f (x, y, z) = c at P0 (x0 , y0 , z0 )
x = x0 + fx (P0 )t
3. Tangent Plane to a Surface
y = y0 + fy (P0 )t
z = z0 + fz (P0 )t
z = f (x, y) at (x0 , y0 , f (x0 , y0 ))
The tangent plane to the surface
z = f (x, y) of a differentiable function f at the point
P0 (x0 , y0 , z0 ) = (x0 , y0 , f (x0 , y0 )) is f x (x0 , y0 )(x ! x0 ) + f y (x0 , y0 )(y ! y0 ) ! (z ! z0 ) = 0
C. Examples
1. Find the equation of the tangent plane to the level surface
P(1, 2, !2) .
x + y + z 2 = 7 at the point
2. Find the equation of the tangent plane to the level surface
point P(1, 0, !2) .
3. Find the equation of the tangent plane to the curve
xe y z 2 + ln x = 4 at the
z = x 2 + y 2 at the point P(3, 4) .
II.
Differentials
A. Estimating the change in
f in a Direction u
To estimate the change in the value of a differentiable function f when we move a small
!
distance ds from a point P0 in a particular direction u , use the formula
(
df = !f
P0
)
!
" u ds
directional derivative
distance increment
B. Alternate Differential Formula
1. If
z = f (x, y) is a differentiable function and dx = !x and dy = !y ,
" !f %
" !f %
dz = ( fx )dx + ( fy )dy = $ ' dx + $ ' dy
# !x &
# !y &
2. If we move from
(x0 , y0 ) to a point nearby, the resulting differential in f is
df = ( fx (x0 , y0 ))dx + ( fy (x0 , y0 ))dy .
C. Examples
1. Estimate how much the value of
from
f (x, y) = x 2 + 3xy ! y 2 will change if (x,y) changes
(2,3) to (2.05, 2.96).
f (x, y, z) = x + xz ! yz 2 + y will change if (x,y,z)
moves from (1,-1,0) a distance of ds=0.1 unit toward the point (2,1,-2).
2. Estimate how much the value of
f (x, y, z) = x + x cos z ! ysin z + y will change if (x,y,z)
moves from (2,-1,0) a distance of ds=0.2 unit toward the point (0,1,2).
3. Estimate how much the value of
III.
Linearization
A. Definition
The linearization of a function
f (x, y) at a point (x0 , y0 ) where f is differentiable is the
function L(x, y) = f (x0 , y0 ) + f x (x0 , y0 )(x ! x0 ) + f y (x0 , y0 )(y ! y0 ) . The approximation
f (x, y) ! L(x, y) is the standard linear approximation of f at (x0 , y0 ) .
B. The Error in the Standard Linear Approximation
If f has continuous first and second partial derivatives throughout an open set containing
a rectangle R centered at (x0 , y0 ) and if M is any upper bound for the values of
fxx , fyy , and fxy on R, then the error E(x,y) incurred in replacing f (x, y) on R by its
linearization
inequality
L(x, y) = f (x0 , y0 ) + fx (x0 , y0 )(x ! x0 ) + fy (x0 , y0 )(y ! y0 ) satisfies the
E(x, y) ! 12 M ( x " x0 + y " y0 )2 .
C. Example
1. Find the linearization of
f (x, y) = x y at the point (1,4).
2. Find the linearization of
f (x, y, z) = xz ! 3yz + 2 at the point (1,1,2).
3. Find the linearization of
f (x, y) = ln(x ! 3y) at the point (7,2) and use it to
approximate
f(6.9,2.06).