Chapter 9: Vector Calculus
Section 9.3: Differential Operators
Example 9.3.1
For fx,y=2 x2+3 y2, obtain ∇f at the point x,y=1,1. Show that at this point, the gradient vector is orthogonal to a vector tangent to the level curve through this point.
Solution
Mathematical Solution
The gradient of f is given by ∇f=fx i+fy j; as a column vector, it becomes 4 x6 y.
Evaluating at 1,1 gives the vector 46.
The level curve through 1,1 is defined implicitly by the equation fx,y=f1,1, that is, by the equation 2 x2+3 y2=5. The branch of this ellipse through 1,1 is yx=5−2 x2/3. Consequently, a position-vector representation of this curve is given by
R=x(5−2 x2)/3
so a vector tangent to this curve is R′x=1−2 x15−6 x2. At x=1, this gives T=1−23.
The orthogonality of the gradient and tangent vectors is verified by the vanishing of their dot product, that is, by
T·∇f=1−2/3·46 = 4−23⋅6=4−4=0
Maple Solution - Interactive
Initialize
Tools≻Load Package: Student Vector Calculus
Loading Student:-VectorCalculus
Tools≻Tasks≻Browse: Calculus - Vector≻ Vector Algebra and Settings≻ Display Format for Vectors
Press the Access Settings button and select "Display as Column Vector"
Display Format for Vectors
Define the function f
Context Panel: Assign Function
fx,y=2 x2+3 y2→assign as functionf
Obtain ∇f at the point 1,1
Common Symbols palette: Del operator
Context Panel: Evaluate and Display Inline
Context Panel: Evaluate at a Point≻x=1,y=1
∇fx,y = →evaluate at point
Alternate approach to obtaining ∇f at the point 1,1
Write fx,y Context Panel: Evaluate and Display Inline
Context Panel: Student Vector Calculus≻Differentiate≻Gradient (See first figure to the right.)
Context Panel: Evaluate at a Point (See the second figure to the right.)
Context Panel: Assign to a Name≻gf
fx,y = 2⁢x2+3⁢y2→gradient4⁢x6⁢y→evaluate at point
Obtain the level curve through 1,1
Write the equation defining the level curve. Press the Enter key.
Context Panel: Solve≻Obtain Solutions for≻y
fx,y=f1,1
2⁢x2+3⁢y2=5
→solutions for y
13⁢−6⁢x2+15,−13⁢−6⁢x2+15
Describe the level curve as the position vector R
Context Panel: Assign to a Name≻R
x,13⁢−6⁢x2+15→assign to a nameR
Obtain a vector tangent to R at 1,1
Write the name R. Context Panel: Student Vector Calculus≻Differentiate≻With Respect To≻x
Context Panel: Evaluate at a Point≻x=1
Context Panel: Simplify≻Simplify
Context Panel: Assign to a Name≻T
R = →differentiate →evaluate at point = simplify →assign to a nameT
Verify orthogonality by the vanishing of the dot product
Common Symbols palette: Dot product operator
T·gf = 0
Maple Solution - Coded
Load the Student VectorCalculus package and execute the BasisFormat command.
withStudent:-VectorCalculus:BasisFormatfalse:
Define the function f.
f≔x,y→2 x2+3 y2:
Obtain the gradient of f
Apply the Gradient command.
gradf≔Gradientfx,y
Evaluate the gradient at x,y=1,1
Apply the evalVF command.
gradf11≔evalVFgradf,1,1
Define the level curve through 1,1 as a position vector R
Solve fx,y=f1,1 for yx by writing the equation 2 x2+3 y2=5 , solving for y, and selecting the branch that goes through 1,1.
Construct R as the free vector x,yx.
R≔x,5−2 x2/3
Obtain Tx, a tangent vector at x along the level curve
Apply the TangentVector command.
T≔TangentVectorR
Evaluate Tx at x=1
Apply the eval and simplify commands.
T1≔simplifyevalT,x=1
Show orthogonality via the vanishing of the dot product
Apply the DotProduct command to the tangent and gradient vectors. Because T1 is derived from a RootedVector, it retains its variable root point; the DotProduct command fails if the root points don't agree. Hence, the conversion of T1 to a free vector via the ConvertVector command.
DotProductConvertVectorT1,free,gradf11 = 0
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