Chapter 4: Partial Differentiation
Section 4.7: Approximations
Example 4.7.7
Use differentials to estimate the maximum error in determining the surface area of a closed rectangular box measured (in inches) to be 15×6×21, if each measurement is accurate to .2 in.
Solution
Mathematical Solution
If the length, width, and height of the box are x,y, and z respectively, the surface area of the box is given by the function fx,y,z=2y z+x z+x y. The maximal error in measuring the surface area as f15,6,21=1062, will be approximately df, where
df
=fx15,6,21⋅±0.2+fy15,6,21⋅± 0.2+fz15,6,21⋅± 0.2
=54+72+42⋅± 0.2
= ± 33.6
Indeed, the actual value of the maximal error is
f15.2,6.2,21.2−f15,6,21=1095.84−1062=33.84
Maple Solution - Interactive
Define fx,y,z and its first partial derivatives
Context Panel: Assign Function
fx,y,z=2y z+x z+x y→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
Calculate f2.03,3.12,1.48≐f2,3,3/2+df
Context Panel: Evaluate and Display Inline
f__x15,6,21⋅.2+f__y15,6,21⋅.2+f__z15,6,21⋅.2 = 33.6
Compute Δf "exactly"
f15.2,6.2,21.2−f15,6,21 = 33.84
Maple Solution - Coded
Define f.
f≔x,y,z→2y z+x z+x y:
Use the differential operator D to obtain the partial derivatives in df.
df≔D1f15,6,21⋅.2+D2f15,6,21⋅.2+D3f15,6,21⋅.2 = 33.6
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