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MATH2011 SEVERAL VARIABLE CALCULUS
S1 2004
TEST 1 SOLUTIONS
The versions in this test were colour coded. In these solutions,
they are indicated by codes
B = BLUE, G = GREEN, P = PINK, Y = YELLOW .
Also the code ALL indicates a common question to all
versions.
1.a) [ALL] For x = r cos , y = r sin ,
x
r= cos ,
x
= r sin ,
y
r= sin ,
y
= r cos .
b) [ALL] For z = f(x, y),
z
= fx
x
+ fy
y
= r sin fx
+ r cos fy
.
c) [B] For f(x, y) = x2y2,
f
x= 2xy2,
f
y= 2x2y
z
= r sin 2xy2 + r cos 2x2y = 2r4 cos sin3 + 2r4 cos3 sin
= 2r4 cos sin (cos2 sin2 ) = r4 sin2 cos2 =1
2r4 sin4 .
c) [G] For f(x, y) = y2 x2,
f
x= 2x,
f
y= 2y
z
= r sin (2x) + r cos 2y = 2r2 sin cos + 2r2 cos sin
= 4r2 sin cos = 2r2 sin2 .
c) [P] For f(x, y) = x2 y2,
fx
= 2x, fy
= 2y
z
= r sin 2x + r cos (2y) = 2r2 sin cos 2r2 cos sin
= 4r2 sin cos = 2r2 sin2 .
c) [Y] For f(x, y) = x2 + y2,
f
x= 2x,
f
y= 2y
z
= r sin 2x + r cos 2y = 2r2 sin cos + 2r2 cos sin
= 0.
1
2. [B]
g(x, y) =1
1 + x + 2y,
g
x=
1
(1 + x + 2y)2,
g
y=
2
(1 + x + 2y)2,
2g
x2=
2
(1 + x + 2y)3,
2g
g x=
4
(1 + x + 2y)3,
2g
y2=
8
(1 + x + 2y)3,
hence at (0, 0),
g = 1,g
x= 1,
g
y= 2,
2g
x2= 2,
2g
g x= 4,
2g
y2= 8 .
so the Taylor series of g about (0, 0) up to and including the
2nd order terms is:
1 +1
1!
1(x 0) 2(y 0)
+
1
2!
2(x 0)2 + 2 4(x 0)(y 0) + 8(y 0)2
+
= 1 x 2y + x2 + 4xy + 4y2 +
[Not required in test: The Taylor series in full can be found
as:
g(x, y) =1
1 + (x + 2y)=n=0
(1)n(x + 2y)n if |x + 2y| < 1
=n=0
(1)nn
k=0
n
k
2k xnkyk . ]
2. [G]
g(x, y) =1
1 + x 2y,
g
x=
1
(1 + x 2y)2,
g
y=
2
(1 + x 2y)2,
2g
x2=
2
(1 + x 2y)3,
2g
g x=
4
(1 + x 2y)3,
2g
y2=
8
(1 + x 2y)3,
hence at (0, 0),
g = 1,g
x= 1,
g
y= 2,
2g
x2= 2,
2g
g x= 4,
2g
y2= 8 .
so the Taylor series of g about (0, 0) up to and including the
2nd order terms is:
1 +1
1!
1(x 0) + 2(y 0)
+
1
2!
2(x 0)2 + 2 (4)(x 0)(y 0) + 8(y 0)2
+
= 1 x + 2y + x2 4xy + 4y2 +
[Not required in test: The Taylor series in full can be found
as:
g(x, y) =1
1 + (x 2y)=n=0
(1)n(x 2y)n if |x 2y| < 1
=n=0
(1)nn
k=0
n
k
(2)k xnkyk . ]
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2. [P]
g(x, y) =1
1 x 2y,
g
x=
1
(1 x 2y)2,
g
y=
2
(1 x 2y)2,
2g
x2=
2
(1 x 2y)3,
2g
g x=
4
(1 x 2y)3,
2g
y2=
8
(1 x 2y)3,
hence at (0, 0),
g = 1,g
x= 1,
g
y= 2,
2g
x2= 2,
2g
g x= 4,
2g
y2= 8 .
so the Taylor series of g about (0, 0) up to and including the
2nd order terms is:
1 +1
1!
1(x 0) + 2(y 0)
+
1
2!
2(x 0)2 + 2 4(x 0)(y 0) + 8(y 0)2
+
= 1 + x + 2y + x2 + 4xy + 4y2 +
[Not required in test: The Taylor series in full can be found
as:
g(x, y) =1
1 (x + 2y)=n=0
(x + 2y)n if |x + 2y| < 1
=n=0
nk=0
n
k
2k xnkyk . ]
2. [Y]
g(x, y) =1
1 x + 2y,
g
x=
1
(1 x + 2y)2,
g
y=
2
(1 x + 2y)2,
2g
x2=
2
(1 x + 2y)3,
2g
g x=
4
(1 x + 2y)3,
2g
y2=
8
(1 x + 2y)3,
hence at (0, 0),
g = 1,g
x= 1,
g
y= 2,
2g
x2= 2,
2g
g x= 4,
2g
y2= 8 .
so the Taylor series of g about (0, 0) up to and including the
2nd order terms is:
1 +1
1!
1(x 0) 2(y 0)
+
1
2!
2(x 0)2 + 2 (4)(x 0)(y 0) + 8(y 0)2
+
= 1 + x 2y + x2 4xy + 4y2 +
[Not required in test: The Taylor series in full can be found
as:
g(x, y) =1
1 (x 2y)=n=0
(x 2y)n if |x 2y| < 1
=n=0
nk=0
n
k
(2)k xnkyk . ]
3
3. [B] Maximise and minimise
h(x, y) = x2 + y2 6x 8y 50, subject to (x, y) = x2 + y2 25 = 0
.
With F(x ,y ,) = h(x, y) (x, y), then F =
0 gives (*)
2x 6 = 2x (1)
2y 8 = 2y (2)
x2 + y2 = 25 (3)
(* equivalently, the Lagrange Multiplier Thm. states local
extrema for h on = 0 occur atcritical points of h for some constant
, on = 0, i.e. h = and = 0.),
The simplest solution is to eliminate by taking y(1) and x(2),
obtaining two expressionsfor 2xy,
y(2x 6) = x(2y 8) so 2xy 6y = 2xy 8x, 6y = 8x or y =4x
3
x2 +
4x
3
2= 25 from (3), i.e. x2 +
16x2
9=
25x2
9= 25, x2 = 9,
x = 3, y = 4 (same signs)
and h(3, 4) = 32 + 42 6 3 8 4 50 = 25 18 32 50 = 75
and h(3,4) = (3)2 + (4)2 6 (3) 8 (4) 50 = 25 + 18 + 32 50 =
25
Hence on = 0, h has a maximum value of 25 (at (3,4)) and a
minimum value of75 (at
(3, 4))
A slower solution is to use (1) and (2) to express x and y in
terms of , substitute into (3) tofind an equation for , solve for
and hence find x and y. This method gives
x =3
1 , y =
4
1
3
1
2+
4
1
2= 25
or 25 = 25(1 )2 or ( 2) = 0
= 0 (x, y) = (3, 4) etc.
and = 2 (x, y) = (3,4) etc.
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3. [G] Maximise and minimise
h(x, y) = x2 + y2, subject to (x, y) = x2 + y2 8x 6y 75 = 0
.
With F(x ,y ,) = h(x, y) (x, y), then F =
0 gives (*)
2x = (2x 8) (1)
2y = (2y 6) (2)
x2 + y2 8x 6y = 75 (3)
(* equivalently, the Lagrange Multiplier Thm. states local
extrema for h on = 0 occur atcritical points of h for some constant
, on = 0, i.e. h = and = 0.),
The simplest solution is to eliminate by taking (2y 6) (1) and
(2x 8) (2), obtainingtwo expressions for (2x 8)(2y 6),
2x(2y 6) = 2y(2x 8) so 4xy 12x = 4xy 16y,
12x = 16y or y =3x
4
x2 +
3x
4
2 8x 6
3x
4= 75 from (3),
i.e.25x2
16
50x
4= 75,
x2
16
2x
4= 3
i.e. x2 8x 24 = 0, i .e. (x 12)(x + 4) = 0
(x, y) = (9, 12) or (4,3)
and h(9, 12) = 92 + 122 = 81 + 144 = 225
and h(4,3) = (4)2 + (3)2 = 25
Hence on = 0, h has a maximum value of 225 (at (9, 12)) and a
minimum value of 25 (at(4,3))
A slower solution is to use (1) and (2) to express x and y in
terms of , substitute into (3) tofind an equation for , solve for
and hence find x and y. This method gives
x =4
1 , y =
3
1
4
1
2+
3
1
2 8
4
1
6
3
1
= 75
or 252 + 50(1 ) 75(1 )2 = 0
2 + 2 22 3 + 6 32 = 0
42 8 + 3 = 0
i.e (2 1)(2 3) = 0
=1
2
(x, y) = (4,3) etc.
and =3
2 (x, y) = (12, 9) etc.
5
3. [P] Maximise and minimise
h(x, y) = x2 + y2, subject to (x, y) = x2 + y2 + 6x + 8y 75 = 0
.
With F(x ,y ,) = h(x, y) (x, y), then F =
0 gives (*)
2x = (2x + 6) (1)
2y = (2y + 8) (2)
x2 + y2 + 6x + 8y = 75 (3)
(* equivalently, the Lagrange Multiplier Thm. states local
extrema for h on = 0 occur atcritical points of h for some constant
, on = 0, i.e. h = and = 0.),
The simplest solution is to eliminate by taking (2y + 8) (1) and
(2x + 6) (2), obtainingtwo expressions for (2x + 6)(2y + 8),
2x(2y + 8) = 2y(2x + 6) so 4xy + 16x = 4xy + 12y,
16x = 12y or y =4x
3
x2 +
4x
3
2+ 6x + 8
4x
3= 75 from (3),
i.e.25x2
9+
50x
3= 75,
x2
9+
2x
3= 3
i.e. x2 + 6x 27 = 0, i.e. (x + 9)(x 3) = 0
(x, y) = (3, 4) or (9,12)
and h(3, 4) = 32 + 42 = 25
and h(9,12) = (9)2 + (12)2 = 81 + 144 = 225
Hence on = 0, h has a maximum value of 225 (at (9,12)) and a
minimum value of 25(at (3, 4))
A slower solution is to use (1) and (2) to express x and y in
terms of , substitute into (3) tofind an equation for , solve for
and hence find x and y. This method gives
x =3
1 , y =
4
1
3
1
2+
4
1
2+ 6
3
1
+ 8
4
1
= 75
or 252 + 50(1 ) 75(1 )2 = 0
2 + 2 22 3 + 6 32 = 0
42 8 + 3 = 0
i.e (2 1)(2 3) = 0
=1
2
(x, y) = (3, 4) etc.
and =3
2 (x, y) = (9,12) etc.
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3. [Y] Maximise and minimise
h(x, y) = x2 + y2 + 6x 8y 50, subject to (x, y) = x2 + y2 25 = 0
.
With F(x ,y ,) = h(x, y) (x, y), then F =
0 gives (*)
2x + 6 = 2x (1)
2y 8 = 2y (2)
x2 + y2 = 25 (3)
(* equivalently, the Lagrange Multiplier Thm. states local
extrema for h on = 0 occur atcritical points of h for some constant
, on = 0, i.e. h = and = 0.),
The simplest solution is to eliminate by taking y(1) and x(2),
obtaining two expressionsfor 2xy,
y(2x + 6) = x(2y 8) so 2xy + 6y = 2xy 8x, 6y = 8x or y = 4x
3
x2 +
4x
3
2= 25 from (3), i.e. x2 +
16x2
9=
25x2
9= 25, x2 = 9,
x = 3, y = 4 (opp. signs)
and h(3,4) = 32 + (4)2 + 6 3 8 (4) 50 = 25 + 18 + 32 50 = 25
and h(3, 4) = (3)2 + 42 + 6 (3) 8 4 50 = 25 18 32 50 = 75
Hence on = 0, h has a maximum value of 25 (at (3,4)) and a
minimum value of75 (at
(3, 4))
A slower solution is to use (1) and (2) to express x and y in
terms of , substitute into (3) tofind an equation for , solve for
and hence find x and y. This method gives
x =3
1 , y =
4
1
3
1
2+
4
1
2= 25
or 25 = 25(1 )2 or ( 2) = 0
= 0 (x, y) = (3, 4) etc.
and = 2 (x, y) = (3,4) etc.
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