Chapter 4: Partial Differentiation
Section 4.7: Approximations
Use the total differential to approximate the value of fx,y,z=x lny+ zx2+y z at x,y,z=2.03,3.12,1.48.
Define the function fx,y,z=x lny+ zx2+y z and approximate f2.03,3.12,1.48 as f2,3,3/2+df.
Note that f2.03,3.12,1.48 = −1.302614961.
Maple Solution - Interactive
Define fx,y,z and its first partial derivatives
Context Panel: Assign Function
fx,y,z=x lny+ zx2+y z→assign as functionf
Calculus palette: Partial derivative operator
(Set the symbols fx, fy and fz as Atomic Identifiers)
f__xx,y,z=∂∂ x fx,y,z→assign as functionf__x
f__yx,y,z=∂∂ y fx,y,z→assign as functionf__y
f__zx,y,z=∂∂ z fx,y,z→assign as functionf__z
Context Panel: Evaluate and Display Inline
Context Panel: Approximate≻10 (digits)
f2,3,3/2+f__x2,3,3/2⋅.03+f__y2,3,3/2⋅.12+f__z2,3,3/2⋅−.02 = 2.03⁢ln⁡917−0.01202614379→at 10 digits−1.303083340
Compute f2.03,3.12,1.48 "exactly"
f2.03,3.12,1.48 = −1.302614961
Maple Solution - Coded
Define an appropriate function f
f≔x,y,z→x lny+ zx2+y z:
Use the differential operator D to calculate partial derivatives at 2,3,3/2.
Use the evalf command to float the result.
Approximate f2.03,3.12,1.48 as f2,3,3/2+df
Apply the evalf command.
evalff2,3,3/2+df = −1.303083340
Obtain f2.03,3.02,4.97 "exactly"
Evaluate f at 2.03,3.12,1.48.
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