Chapter 4: Partial Differentiation
Section 4.7: Approximations
Example 4.7.1
If x changes from 3 to 3.04 and y changes from −2 to −2.2, compare Δf and df for the function fx,y=3 x2+4 x y−5 y2+7.
Solution
Mathematical Solution
Δf=f3.04,−2.2−f3,−2=6.2272
df=fx3,−2⋅0.04+fy3,−2⋅−0.2=−6.00
Maple Solution - Interactive
Obtain Δf
Context Panel: Assign Function
fx,y=3 x2+4 x y−5 y2+7→assign as functionf
Context Panel: Evaluate and Display Inline
f3.04,−2.2−f3,−2 = −6.2272
Obtain df
Context Panel: Assign Function (Set fx as an Atomic Identifier.)
f__xx,y=∂∂ x fx,y→assign as functionf__x
Context Panel: Assign Function (Set fy as an Atomic Identifier.)
f__yx,y=∂∂ y fx,y→assign as functionf__y
f__x3,−2⋅0.04+f__y3,−2⋅−0.2 = −6.00
Maple Solution - Coded
Define the function f.
f≔x,y→3 x2+4 x y−5 y2+7:
Calculate Δf.
Use the differential operator D to calculate df=fx3,−2⋅0.04+fy3,−2⋅−0.2.
D1f3,−2⋅.04+D2f3,−2⋅−.2 = −6.00
So, Δf= −6.2272, but df= −6.00.
<< Chapter Overview Section 4.7 Next Example >>
© Maplesoft, a division of Waterloo Maple Inc., 2024. All rights reserved. This product is protected by copyright and distributed under licenses restricting its use, copying, distribution, and decompilation.
For more information on Maplesoft products and services, visit www.maplesoft.com
Download Help Document
