Chapter 4: Partial Differentiation
Section 4.5: Gradient Vector
Example 4.5.1
Let fx,y=5−2 x2−3 y2 and let P be the point 1,2.
Obtain ∇f at P.
Graph the surface z=fx,y.
On the same set of axes, graph the level curve through P, and ∇f at P.
At P, show that ∇f is orthogonal to a vector tangent to the level curve through P.
At P, obtain ψ=∇f·u, the directional derivative of f in the direction u=cost i+sint j . Show that ψ is a maximum when u is along ∇fP and that this maximum is ∇fP.
Solution
Mathematical Solution
For Part (a), ∇fP=−4 x−6 yx=a|f(x)x=1,y=2 = −4−12.
The graphs for Parts (b) and (c) are given in Figures 4.5.1(a) and 4.5.1(b), respectively.
Figure 4.5.1(a) The surface z=fx,y
use plots, Student:-VectorCalculus in module() local f,p1,p2,p3,V; f:=5-2*x^2-3*y^2; V:=RootedVector(root=[1,2],<-4,-12>); p1:=contourplot(f,x=-5..5,y=-10..3,color=black,contours=[-3,-6,-9,-12,-15,-18,-30,-40,-50]); p2:=PlotVector(V,color=red); p3:=display(p1,p2,scaling=constrained); print(p3); end module: end use:
Figure 4.5.1(b) Level curves & gradient vector
For Part (d), solve f1,2=−9=fx,y for y=67−x2/3, giving an explicit representation of the level curve through P. Then, writing this curve as
R=x67−x2/3
its tangent vector at P is R′1=1−1/3, from which it follows that
∇fP·R′1=−4−12·1−1/3=−4+4=0
In Part (e), the maximum of ψ=−4 cost+3 sint is 4 10≐12.65 at t=π+arctan3≐4.4 radians (about 252°) measured from the positive x-axis.
Figure 4.5.1(c) is a graph of ψ, from which the maximum can be approximated.
Figure 4.5.1(d) is an animation showing ∇fP as the transparent red arrow, and ut as the rotating black arrow. The value of t appears above the y-axis, and the corresponding value of ψ is written in the fourth quadrant. The maximum value of ψ is attained when ut is aligned with ∇fP. This maximum value is ∇fP=410.
Figure 4.5.1(c) Graph of ψ
use Student:-VectorCalculus,plots in module() local p1,p2,V,F; V:=RootedVector(root=[1,2],<-4,-12>); p1:=PlotVector(V,width=.3,color=red,transparency=.5); F:=proc(t) local g1,g2,U; U:=RootedVector(root=[1,2],<cos(t),sin(t)>); g1:=PlotVector(U,color=black,unconstrainedview=false); g2:=textplot([1,-5,convert(evalf(-4*(cos(t)+3*sin(t)),5),string)]); display(g1,g2); end proc; p2:=animate(F,[t],t=0..2*Pi,frames=63,background=p1,paraminfo=true,scaling=constrained); print(p2); end module: end use:
Figure 4.5.1(d) ∇fP and u
Maple Solution - Interactive
Initialize
Tools≻Load Package: Student Multivariate Calculus
Loading Student:-MultivariateCalculus
Context Panel: Assign Name
f=5−2 x2−3 y2→assign
u=cost,sint→assign
Part (a)
Type the name f. Context Panel: Evaluate and Display Inline
Context Panel: Student Multivariate Calculus≻Differentiate≻Gradient See Figure 4.5.1(e).
Context Panel: Select Element≻1
Context Panel: Assign to a Name≻Gf
Figure 4.5.1(e) Gradient dialog
f = −2x2−3y2+5→gradient→select entry 1→assign to a nameGf
Part (b) - Generate Figure 4.5.1(a)
From the Context Panel, launch the Plot Builder on the rule for f.
Plots≻Plot Builder≻3-D plot
Set x∈−2,2 and y∈−2,2 view → 0 to 5
f = −2x2−3y2+5→
Part (c)
To generate Figure 4.5.1(b) interactively, first obtain a 2D contour map of f via the Plot Builder, launched from the Context Panel. Select "2-D contour plot"
Obtain a graph of ∇fP via the Context Panel's Plot option, "Arrow from point" See Figure 4.5.1(f).
Copy/paste the arrow onto the contour map.
Figure 4.5.1(f) Dialog for graphing an arrow
f
−2x2−3y2+5
→
Gf
→plot arrow
Part (d)
Expression palette: Evaluation template Form the equation fx,y=f1,2
Context Panel: Solve≻Obtain Solutions for≻y
Control-drag to form the equation Yx=…
Context Panel: Assign Function
f=fx=a|f(x)x=1,y=2
−2x2−3y2+5=−9
→solutions for y
−6x2+423,−−6x2+423
Yx=13−6x2+42→assign as functionY
T=1,Y′1→assign
Common Symbols palette: Dot product operator
Context Panel: Evaluate and Display Inline
Context Panel: Simplify≻Simplify
T·Gf = −4+2363= simplify 0
Part (e)
Common Symbols palette: Dot Product operator
ψ=Gf·u→assign
Context Panel: Optimization≻Maximize (local)
ψ = −4cost−12sint→maximize12.6491106406735181,t=−1.89254688119156
Write ψ and press the Enter key.
Context Panel: Differentiate≻With Respect To≻t
Context Panel: Conversions≻Equate to 0
Context Panel: Solve≻Solve (general solution)
Expression palette: Evaluation template Context Panel: Evaluate and Display Inline
ψ
−4cost−12sint
→differentiate w.r.t. t
4sint−12cost
→equate to 0
4sint−12cost=0
→solve
t=arctan3+π_Z1
ψx=a|f(x)t=π+arctan3 = 410
Context Panel: Norm≻Euclidean
Gf = →Euclidean-norm410
Form a sequence of the vectors ∇fP and i.
Context Panel: Student Multivariate Calculus≻Lines & Planes≻Angle
Context Panel: Assign to a Name≻lambda
Gf,1,0 = ?,?→angleπ−arccos1010→assign to a nameλ
Context Panel: Approximate≻5 (digits)
π+arctan3−2 π−λ→at 5 digits0.
Maple Solution - Coded
Install the Student MultivariateCalculus package.
withStudent:-MultivariateCalculus:
Define f.
f≔5−2 x2−3 y2:
Use the eval and Gradient commands to obtain ∇fP.
Gf≔Gradientf,x,y=1,2
Gf≔
Part (b)
Figure 4.5.1(a), a graph of the surface z=fx,y, can be drawn with the plot3d command, below.
plot3df,x=−3..3,y=−3..3view=−8..5,scaling=constrained,axes=frame,tickmarks=4,4,8,labels=x,y,z:
The following conjunction of the RootedVector, contourplot, PlotVector, and display commands will generate Figure 4.5.1(b).
V≔VectorCalculus:-RootedVectorroot=1,2,−4,−12;p1≔plots:-contourplotf,x=−5..5,y=−10..3,color=black,contours=−3,−6,−9,−12,−15,−18,−30,−40,−50;p2≔VectorCalculus:-PlotVectorV,color=red;plots:-displayp1,p2,scaling=constrained;
Use the solve and eval commands to obtain an explicit representation of yx, the level curve passing through 1,2.
Y≔solveevalf,x=1,y=2=f,y1
Y≔−6x2+423
Use the simplify, eval, and diff commands to obtain at 1,2, a vector T tangent to the level curve yx.
T≔simplifyevaldiffx,Y,x,x=1
T≔
Use the DotProduct command to obtain ∇fP·T.
DotProductGf,T = 0
The vanishing of the dot product between the tangent vector and the gradient vector verifies that the gradient vector is orthogonal to the level curve.
Define the unit vector u.
u≔cost,sint:
Use the DirectionalDiff command in the Student VectorCalculus package to obtain ψ. Although there is a DirectionalDerivative command in the Student MultivariateCalculus package, it accepts only numeric direction vectors, so would fail with the generic unit vector u.
ψ≔Student:-VectorCalculus:-DirectionalDifff,u,point=1,2
ψ≔−4cost−12sint
Obtain Figure 4.5.1(c) with the plot command.
plotψ,t=0..2 π:
Maximize ψ by using the solve and diff commands to obtain all solutions of ψ′=0.
solvediffψ,t=0,AllSolutions
arctan3+π_Z2
Use the eval command to obtain the maximum value of ψ, namely, ψπ+arctan3.
evalψ,t=π+arctan3 = 410
Apply the Norm command to the gradient vector and evaluate at the point 1,2.
NormGf1,2 = 410
Use the eval command to obtain uπ+arctan3, the direction giving the maximum value of the directional derivative, ψ. This vector is a positive multiple of ∇fP.
evalu,t=π+arctan3 =
Obtain the (smaller) angle between ∇fP and the horizontal vector i by using the simplify and Angle commands. Note that the angle, clockwise from i to ∇fP, is returned as an arccosine.
λ≔simplifyAngleGf,1,0 = π−arccos11010
Define Λ as the counterclockwise angle from i to ∇fP.
Λ≔2 π−λ:
Use the evalf command to show that the difference between Λ and π+arctan3 is zero.
evalfπ+arctan3−Λ = 0.
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