Chapter 4: Partial Differentiation
Section 4.7: Approximations
Approximate 43.033+56.22 by using the total differential for some appropriate function fx,y.
Define the function fx,y=4 x3+5 y2 and approximate 43.033+56.22 as f3,6+df.
Note that f3.03,6.2=17.42046234.
Maple Solution - Interactive
Define fx,y and its first partial derivatives
Context Panel: Assign Function
fx,y=4 x3+5 y2→assign as functionf
Calculus palette: Partial derivative operator
(Set the symbols fx and fy as Atomic Identifiers)
f__xx,y=∂∂ x fx,y→assign as functionf__x
f__yx,y=∂∂ y fx,y→assign as functionf__y
Context Panel: Evaluate and Display Inline
Context Panel: Approximate≻10 (digits)
f3,6+f__x3,6⋅.03+f__y3,6⋅.2 = 1.026458333⁢288→at 10 digits17.41957555
Compute f3.03,6.2 "exactly"
f3.03,6.2 = 17.42046234
Maple Solution - Coded
Define an appropriate function f
f≔x,y→4 x3+5 y2:
Use the differential operator D to calculate partial derivatives at 3,6.
Use the evalf command to float the result.
df≔evalfD1f3,6⋅.03+D2f3,6⋅.2 = 0.4490128060
Approximate f3.03,6.2 as f3,6+df
Apply the evalf command.
evalff3,6+df = 17.41957555
Obtain f3.03,6.2 "exactly"
Evaluate f at 3.03,6.2.
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