Chapter 4: Partial Differentiation
Section 4.4: Directional Derivative
Example 4.4.2
At the point P:1,1,0, and in the direction of the point Q:2,−3,6, obtain the directional derivative of fx,y,z=x3−x y2−z.
Solution
Mathematical Solution
Let v=Q−P = 2−36−110 = 1−46 be the vector from point P to point Q.
Let u=v/v = v/53 be the unit vector from point P in the direction of point Q,
Let R=P+t u be the line through P in the direction defined by u. The parametric equations of this line are
x=1+t53,y=1−4 t53,z=6 t53
Along this line the function values of f are given by
wt=fxt,yt,zt=4 t53−553t2−152809t353
The requisite directional derivative is then w′0=4/53.
Alternatively, obtain the vector ∇fx=a|f(x)P = fx i+fy j+fz kx=a|f(x)P = 2−2−1 and compute its dot product with u so that
DufP=2−2−1·1531−46=453
Maple Solution - Interactive
Initialize
Tools≻Load Package: Student Multivariate Calculus
Loading Student:-MultivariateCalculus
Context Panel: Assign Name
f=x3−x y2−z→assign
v=1,−4,6→assign
u=v/v2→assign
Instant answer via the Context Panel
Type the name f. Context Panel: Evaluate and Display Inline
Context Panel: Student Multivariate Calculus≻Differentiate≻Directional Derivative Fill in the "Variables, Point, and Vector" dialog as shown in Figure 4.4.2(a) Click the OK button.
Figure 4.4.2(a) Variables, Point, and Vector dialog
f = x3−xy2−z→directional derivative45353
Obtain ∇fx=a|f(x)P
Context Panel: Student Multivariate Calculus≻Differentiate≻Gradient (See Figure 4.4.2(b).)
Context Panel: Select Element≻1
Context Panel: Assign to a Name≻Gf
Figure 4.4.2(b) Gradient dialog
x3−x y2−z→gradient2−2−1→select entry 12−2−1→assign to a nameGf
Obtain DufP=(∇fx=a|f(x)P)·u
Common Symbols palette: Dot product operator
Context Panel: Evaluate and Display Inline
Gf·u = 45353
Solution from first principles:
Obtain R=P+t u, the line through P with direction u
Write a sequence of point P and unit vector u.
Context Panel: Student Multivariate Calculus≻ Lines & Planes≻Line
Context Panel: Student Multivariate Calculus≻ Lines & Planes≻Representation≻parametric (See Figure 4.4.2(c).)
Context Panel: Assign to a Name≻L
Figure 4.4.2(c) Line representation dialog
1,1,0,u→make line<< Line 1 >>→representationx=1+t5353,y=1−4t5353,z=6t5353→assign to a nameL
Obtain wt=fxt,yt,zt and DufP=w′0
Expression palette: Evaluation template Press the Enter key.
Context Panel: Simplify≻Simplify
Context Panel: Differentiate≻With Respect To≻t
Context Panel: Evaluate at a Point≻t=0
fx=a|f(x)L
1+153t533−1+153t531−453t532−653t53
= simplify
453t53−553t2−152809t353
→differentiate w.r.t. t
45353−1053t−452809t253
→evaluate at point
45353
Maple Solution - Coded
Install the Student MultivariateCalculus package.
withStudent:-MultivariateCalculus:
Define f.
f≔x3−x y2−z:
Define v as a list.
v≔1,−4,6:
Obtain the directional derivative
Apply the DirectionalDerivative command.
DirectionalDerivativef,x,y,z=1,1,0,v
There is a DirectionalDerivative command in the Student VectorCalculus package, and a DirectionalDiff command in the Physics:-Vectors package. These alternatives will not be explored further; instead, the following two computations are provided.
Compute DufP=(∇fx=a|f(x)P)·u
Apply the Gradient command, evaluating the resulting vector at point P.
Gf≔Gradientf,x,y,z=1,1,0
Apply to the list v the convert/Vector command, then apply the Normalize command to normalize the resulting vector, thereby obtaining the unit vector u.
V≔convertv,Vector:u≔NormalizeV
Invoke the Dotproduct command.
DotProductGf,u = 45353
Obtain the directional derivative from first principles
Use the Line and GetRepresentation commands to obtain the parametric form of the line through P in the direction of u.
L≔GetRepresentationLine1,1,0,u,form=parametric
x=1+153t53,y=1−453t53,z=653t53
Use the eval command to obtain the value of f along line L, then apply the simplify command.
w≔simplifyevalf,L
Apply the diff command to wt, then the eval command to obtain w′0.
evaldiffw,t,t=0
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