Chapter 4: Partial Differentiation
Section 4.7: Approximations
Approximate 12.03+13.02+14.97 by using the total differential for some appropriate function fx,y,z.
Define the function fx,y,z=1x+1y+1z and approximate 12.03+13.02+14.97 as f2,3,5+df.
Note that f2.03,3.02,4.97=1.024943909.
Maple Solution - Interactive
Define fx,y and its first partial derivatives
Context Panel: Assign Function
fx,y,z=1x+1y+1z→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
f2,3,5+f__x2,3,5⋅.03+f__y2,3,5⋅.02+f__z2,3,5⋅−.03 = 1.024811111
Compute f2.03,3.02,4.97 "exactly"
f2.03,3.02,4.97 = 1.024943909
Maple Solution - Coded
Define an appropriate function f
Use the differential operator D to calculate partial derivatives at 2,3,5.
Approximate f2.03,3.02,4.97 as f2,3,5+df
Apply the evalf command.
evalff2,3,5+df = 1.024811111
Obtain f2.03,3.02,4.97 "exactly"
Evaluate f at 2.03,3.02,4.97.
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