Answers to Test 3

1. Let F (x; y) =
Z
y2
arctan t dt. Find Fx (2; 3) and Fy (2; 3).
x
Use the fundamental theorem of calculus to compute Fx (x; y) = arctan x
and Fy (x; y) = 2y arctan y 2 . So Fx (2; 3) = arctan 2 and Fy (2; 3) =
6 arctan 9.
2. Suppose x2 2y 3 + 3z 4 = 6 de…nes z as a function of x and y. Use implicit di¤erentiation to …nd zx and zy at each of the two points (1; 1; 1)
and (1; 1; 1).
x
@
(x2 2y 3 + 3z 4 ) = 2x + 12z 3 zx , so zx =
.
0=
@x
6z 3
@
y2
0=
(x2 2y 3 + 3z 4 ) = 6y 2 + 12z 3 zy , so zy = 3 .
@y
2z
At the point (1; 1; 1) we have zx = 1=6, and at the point (1; 1; 1)
we have zx = 1=6.
At the point (1; 1; 1) we have zy = 1=2, and at the point (1; 1; 1) we
have zy = 1=2.
Did you notice that the point (1; 1; 1) is not on the surface x2
3x4 = 6? My bad.
2y 3 +
3. Verify Clairaut’s theorem for u = ln (x2 y 2 ).
2y
2x
and uy =
.
ux = 2
2
2
x
y
x
y2
4xy
0 (x2 y 2 ) 2x ( 2y)
So uxy =
=
and
2
(x2 y 2 )
(x2 y 2 )2
uyx =
0 (x2 y 2 ) 2y2x
4xy
=
. They are equal.
2
(x2 y 2 )
(x2 y 2 )2
4. Let u = sin xyz. Find uxyz .
ux = yz cos xyz.
uxy = z cos xyz + yz ( sin xyz) xz = z cos xyz
xyz 2 sin xyz.
uxyz = cos xyz + z ( sin xyz) xy (xy2z sin xyz + xyz 2 (cos xyz) xy) =
cos xyz 3xyz sin xyz x2 y 2 z 2 cos xyz.
5. Laplace’s equation in two dimensions is
@ 2u @ 2u
+
=0
@x2 @y 2
Which of the following functions satisfy Laplace’s equation? Show your
calculations.
(a) u = 3x + 5y
ux = 3 and uy = 5, so uxx = 0 and uyy = 0. Satis…es Laplace’s
equation.
(b) u = x2 + y 2
ux = 2x and uy = 2y, so uxx = 2 and uyy = 2. Thus uxx + uyy =
4 6= 0. Doesn’t satisfy Laplace’s equation.
(c) u = x3 3xy 2
ux = 3x2 3y 2 and uy = 6xy, so uxx = 6x and uyy =
Satis…es Laplace’s equation.
6x.
(d) u = x3 y xy 3
ux = 3x2 y y 3 and uy = x3 3xy 2 , so uxx = 6xy and uyy =
Satis…es Laplace’s equation.
6xy.
(e) u = x2 y xy 2
ux = 2xy y 2 and uy = x2 2xy, so uxx = 2y and uyy = 2x.
Thus uxx + uyy = 2y 2x 6= 0. Doesn’t satisfy Laplace’s equation.
(f) u = (x y)5
ux = 5 (x y)4 and uy = 5 (x y)4 , so uxx = 20 (x y)3 and
uyy = 20 (x y)3 . Thus uxx + uyy = 40 (x y)3 =
6 0. Doesn’t
satisfy Laplace’s equation.
2

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