Chapter 4: Partial Differentiation
Section 4.3: Chain Rule
Example 4.3.30
The composition of fx,y with x=r cost, y=r sint produces the function Fr,t=fxr,t,yr,t. Express fxx+fyy in terms of Fr,Frr, and Ftt.
Solution
Mathematical Solution
Begin with the composition statement
fx,y=fxr,t,yr,t=Fr,t=Frx,y,tx,y
Obtain the first derivatives
fx=Fr rx+Ft tx
and
fy=Fr ry+Ft ty
and the second derivatives
fxx=Frr rx+Frt txrx+Fr rxx+Ftr rx+Ftt txtx+Ft txx
fyy=Frr ry+Frt tyry+Fr ryy+Ftr ry+Ftt tyty+Ft tyy
Assuming the equality of mixed partial derivatives, obtain the sum
fxx+fyy=rx2+ry2Frr+2tx rx+ty ryFrt +tx2+ty2Ftt+rxx+ryyFr+txx+tyyFt
Obtain the derivatives listed in Table 4.3.30(a).
rx=xx2+y2=r costr=cost
rxx=y2x2+y232=r2sin2tr3=sin2tr
ry=yx2+y2=r sintr=sint
ryy=x2x2+y232=r2cos2tr3=cos2tr
tx=−y/x21+y/x2=−yx2+y2=−sintr
txx=2⁢y⁢xx2+y22=2 r2sintcostr4=2 sintcostr2
ty=1/x1+y/x2=xx2+y2=costr
tyy=−2⁢y⁢xx2+y22=−2 sintcostr2
Table 4.3.30(a) First and second derivatives of rx,y and tx,y
Using the expressions in Table 4.3.30(a), obtain the coefficients listed in Table 4.3.30(b).
rx2+ry2=cos2t+sin2t=1
tx rx+ty ry=−sintrcost+costrsint=0r=0
tx2+ty2=−sintr2+costr2=1r2
rxx+ryy=sin2tr+cos2tr=1r
txx+tyy=2 sintcostr2+−2 sintcostr2=0
Table 4.3.30(b) Coefficients of fxx+fyy
Consequently, fxx+fyy=Frr+Fttr2+Frr.
Maple Solution - Interactive
Assign the starting composition statement the name a
fx,y=Fx2+y2,arctany,x→assign to a namea
Assign the sum of second partial derivatives the name b
Calculus palette: Second-partial operator
Context Panel: Assign Name
b=∂2∂x2 a+∂2∂y2 a→assign
Replace x and y with r cost and r sint, respectively, and simplifying
Expression palette: Evaluation template Press the Enter key.
Context Panel: Simplify≻Assuming Positive
Context Panel: Simplify≻Assuming Real Range Complete dialog as per figure to the right.
Context Panel: Right-hand Side
Context Panel: Expand≻Expand
bx=a|f(x)x=r cost,y=r sint
D1,1⁡f⁡r⁢cos⁡t,r⁢sin⁡t+D2,2⁡f⁡r⁢cos⁡t,r⁢sin⁡t=D1,1⁡F⁡r2⁢cos⁡t2+r2⁢sin⁡t2,arctan⁡r⁢sin⁡t,r⁢cos⁡t⁢r⁢cos⁡tr2⁢cos⁡t2+r2⁢sin⁡t2−D1,2⁡F⁡r2⁢cos⁡t2+r2⁢sin⁡t2,arctan⁡r⁢sin⁡t,r⁢cos⁡t⁢sin⁡tr⁢cos⁡t2⁢1+sin⁡t2cos⁡t2⁢r⁢cos⁡tr2⁢cos⁡t2+r2⁢sin⁡t2−D1⁡F⁡r2⁢cos⁡t2+r2⁢sin⁡t2,arctan⁡r⁢sin⁡t,r⁢cos⁡t⁢r2⁢cos⁡t2r2⁢cos⁡t2+r2⁢sin⁡t23/2+2⁢D1⁡F⁡r2⁢cos⁡t2+r2⁢sin⁡t2,arctan⁡r⁢sin⁡t,r⁢cos⁡tr2⁢cos⁡t2+r2⁢sin⁡t2−D1,2⁡F⁡r2⁢cos⁡t2+r2⁢sin⁡t2,arctan⁡r⁢sin⁡t,r⁢cos⁡t⁢r⁢cos⁡tr2⁢cos⁡t2+r2⁢sin⁡t2−D2,2⁡F⁡r2⁢cos⁡t2+r2⁢sin⁡t2,arctan⁡r⁢sin⁡t,r⁢cos⁡t⁢sin⁡tr⁢cos⁡t2⁢1+sin⁡t2cos⁡t2⁢sin⁡tr⁢cos⁡t2⁢1+sin⁡t2cos⁡t2+2⁢D2⁡F⁡r2⁢cos⁡t2+r2⁢sin⁡t2,arctan⁡r⁢sin⁡t,r⁢cos⁡t⁢sin⁡tr2⁢cos⁡t3⁢1+sin⁡t2cos⁡t2−2⁢D2⁡F⁡r2⁢cos⁡t2+r2⁢sin⁡t2,arctan⁡r⁢sin⁡t,r⁢cos⁡t⁢sin⁡t3r2⁢cos⁡t5⁢1+sin⁡t2cos⁡t22+D1,1⁡F⁡r2⁢cos⁡t2+r2⁢sin⁡t2,arctan⁡r⁢sin⁡t,r⁢cos⁡t⁢r⁢sin⁡tr2⁢cos⁡t2+r2⁢sin⁡t2+D1,2⁡F⁡r2⁢cos⁡t2+r2⁢sin⁡t2,arctan⁡r⁢sin⁡t,r⁢cos⁡tr⁢cos⁡t⁢1+sin⁡t2cos⁡t2⁢r⁢sin⁡tr2⁢cos⁡t2+r2⁢sin⁡t2−D1⁡F⁡r2⁢cos⁡t2+r2⁢sin⁡t2,arctan⁡r⁢sin⁡t,r⁢cos⁡t⁢r2⁢sin⁡t2r2⁢cos⁡t2+r2⁢sin⁡t23/2+D1,2⁡F⁡r2⁢cos⁡t2+r2⁢sin⁡t2,arctan⁡r⁢sin⁡t,r⁢cos⁡t⁢r⁢sin⁡tr2⁢cos⁡t2+r2⁢sin⁡t2+D2,2⁡F⁡r2⁢cos⁡t2+r2⁢sin⁡t2,arctan⁡r⁢sin⁡t,r⁢cos⁡tr⁢cos⁡t⁢1+sin⁡t2cos⁡t2r⁢cos⁡t⁢1+sin⁡t2cos⁡t2−2⁢D2⁡F⁡r2⁢cos⁡t2+r2⁢sin⁡t2,arctan⁡r⁢sin⁡t,r⁢cos⁡t⁢sin⁡tr2⁢cos⁡t3⁢1+sin⁡t2cos⁡t22
→assuming positive
D1,1⁡f⁡r⁢cos⁡t,r⁢sin⁡t+D2,2⁡f⁡r⁢cos⁡t,r⁢sin⁡t=D2,2⁡F⁡r,arctan⁡sin⁡t,cos⁡t+D1⁡F⁡r,arctan⁡sin⁡t,cos⁡t⁢r+D1,1⁡F⁡r,arctan⁡sin⁡t,cos⁡t⁢r2r2
→assuming real range
D1,1⁡f⁡r⁢cos⁡t,r⁢sin⁡t+D2,2⁡f⁡r⁢cos⁡t,r⁢sin⁡t=D2,2⁡F⁡r,t+D1⁡F⁡r,t⁢r+D1,1⁡F⁡r,t⁢r2r2
→right hand side
D2,2⁡F⁡r,t+D1⁡F⁡r,t⁢r+D1,1⁡F⁡r,t⁢r2r2
= expand
D2,2⁡F⁡r,tr2+D1⁡F⁡r,tr+D1,1⁡F⁡r,t
From this, deduce that fxx+fyy=Frr+Frr+Fttr2.
Maple Solution - Coded
Notational simplifications
These commands allow xr,t and yr,t to be represented by x and y,respectively; and for derivatives to be written with subscripts.
interfacetypesetting=extended:Typesetting:-Suppressrx,y,tx,y:Typesetting:-Settingsuserep=true:
The composition results in fx,y=fxr,t,yr,t=Fr,t=Frx,y,tx,y
Form the sum fxx+fyy
Apply the diff command for differentiation.
Apply the collect command to collect coefficients of like derivatives.
q1≔collectdiffFr,t,x,x+diffFr,t,y,y,D
q1≔rx,x+ry,y⁢D1⁡F⁡r,t+tx,x+ty,y⁢D2⁡F⁡r,t+rx2+ry2⁢D1,1⁡F⁡r,t+2⁢rx⁢tx+2⁢ry⁢ty⁢D1,2⁡F⁡r,t+tx2+ty2⁢D2,2⁡F⁡r,t
Define r=rx,y=x2+y2 and t=tx,y=arctany/x
R≔x2+y2:
T≔arctany/x:
Evaluate the coefficients in the expression for fxx+fyy
simplifydiffR,x,x+diffR,y,y = 1x2+y2
simplifydiffT,x,x+diffT,y,y = 0
simplifydiffR,x2+diffR,y2 = 1
simplify2 diffR,x diffT,x+2 diffR,y diffT,y = 0
simplifydiffT,x2+diffT,y2 = 1x2+y2
Rewrite the expression for fxx+fyy
fxx+fyy=1rFr+Frr+1r2Ftt
In Cartesian coordinates, the expression fxx+fyy is called the Laplacian of f. This example has investigated how to express the Laplacian in polar coordinates, something that is built into Maple.
Obtain the Laplacian in polar coordinates
Make the notational modifications shown at the right.
Typesetting:-Unsuppressrx,y,tx,y:
Typesetting:-SuppressFr,t:
Apply the Laplacian command from the Student VectorCalculus package
expandStudent:-VectorCalculus:-LaplacianFr,t,coords=polarr,t
Frr+Fr,r+Ft,tr2
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