Homework 5: Solutions

Homework 5: Solutions

3.1.32:

f (x, y) =

xy(x2-y2) x2+y2

(x, y) (0, 0)

0

(x, y) = (0, 0)

Note: f is continuous, (by computing lim(x,y)(0,0) of the formula above, e.g. using polar coorinates).

(a) Find fx and fy when (x, y) (0, 0):

Away from (0, 0), f can be differentiated using the formula defining it,

as:

f (x, y)

x

=

(x2

+

y2)[y(x2

-

y2) + 2x2y] (x2 + y2)2

-

2x2y(x2

-

y2),

f y

(x,

y)

=

(x2

+ y2)[x(x2

-

y2) - 2xy2] (x2 + y2)2

- 2xy2(x2

-

y2).

(b) Show fx(0, 0) = fy(0, 0) = 0:

We must use the limit definition of derivative here:

f (0, 0)

=

lim

f (h, 0) - f (0, 0)

=

lim

0

h2

-0

=

0.

x

h0

h

h0 h

f (0, 0)

=

lim

f (0, h) - f (0, 0)

=

lim

0

h2

-0

=

0.

y

h0

h

h0 h

Notice that the limits at (0, 0) of the partials computed in part (a) (e.g. using polar coordinates) are in fact 0, so the first partials are continuous, and f is in fact C1.

(c) Show fyx(0, 0) = 1 and fxy(0, 0) = -1:

We must use the limit definitions here as well:

2f (0, 0)

xy

=

lim

h0

fy(h, 0) - fy(0, 0) h

=

lim

h0

h2(h3) - h(h2)2

0

=

1,

1

2f (0, 0)

yx

=

lim

h0

fx(0, h) - h

fx(0, 0)

=

lim

h0

h2(0 - h3) h(h2)2

=

-1.

(d) What happened?

The theorem on the equality of mixed partials only applies when the function is C2. While this function is C1 as shown in part (b), it is not C2 as can be shown by computing fyx(x, y) for (x, y) (0, 0) and taking a limit as (x, y) (0, 0):

2f

lim

(x,y)(0,0)

xy

(x,

y)

=

lim

(x,y)(0,0)

8(x2 - y2)x2y2 + (x2 - y2)(x2 + y2)2 (x2 + y2)3

.

This limit doesn't exist, (e.g. using polar coordinates), and the function is not C2.

3.2.2: L R2 R linear, so L(x, y) = ax + by:

(a) Find the first-order Taylor approximation for L:

Since L is linear, and since the first-order approximation gives the best linear function at a point that fits the function, it should be L itself. Let's verify this, at any point (x0, y0):

L(x0, y0) = ax0 + by0, Lx(x0, y0) = a, Ly(x0, y0) = b.

So the first-order Taylor approximation is: L(x, y) L(x0, y0) + Lx(x0, y0)(x - x0) + Ly(x0, y0)(y - y0),

L(x, y) (ax0 + by0) + a(x - x0) + b(y - y0) = ax + by = L(x, y). Note: You may also write this as: L((x0, y0)+h) (ax0+by0)+ah1+bh2 = a(x0+h1)+b(y0+h2) = L((x0, y0)+h),

which tells you the same thing!

2

(b) Find the second-order Taylor approximation for L: We note: fxx = fxy = fyx = fyy 0, so if we plug into the formula, the second-order approximation to L(x, y) is also just L(x, y) itself.

(c) What about higher approximations?: Since all higher derivatives are 0, these will all be the function L itself.

3.2.4: Find the second-order Taylor formula for f at x0, y0:

f (x, y)

=

1+

1 x2

+

y2

=

(1

+ x2

+

y2)-1

,

where

x0

=

0, y0

=

0

We compute: f (0, 0) = 1

fx(0, 0) = -2x(1 + x2 + y2)-2 (0,0) = 0 fy(0, 0) = -2y(1 + x2 + y2)-2 (0,0) = 0

fxx(0, 0) = [-2(1 + x2 + y2)-2 + 8x2(1 + x2 + y2)-3] (0,0) = -2 fxy(0, 0) = fyx(0, 0) = 8xy(1 + x2 + y2)-3 (0,0) = 0 fyy(0, 0) = [-2(1 + x2 + y2)-2 + 8y2(1 + x2 + y2)-3] (0,0) = -2

With this done, we write down the second order Taylor formula:

f ((0,

0)

+

(x,

y))

=

1

-

2 x2 2

-

2 y2 2

+

R2((0,

0),

(x,

y))

Or briefly: f (x, y) = 1 - x2 - y2 + R2(x, y)

3.2.10: Find the second-order Taylor approximation for f at (1, 2):

f (x, y) = x cos(y) - y sin(x)

We compute: f (1, 2) = cos(2) - 2 sin() = 1,

fx(1, 2) = [cos(y) - y cos(x)] (1,2) = cos(2) - 2 cos() = 1 + 2, fy(1, 2) = [-x sin(y) - sin(x)] (1,2) = -2 sin(2) - sin() = 0,

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fxx(1, 2) = 2y sin(x) (0,0) = 22 sin() = 0, fxy(1, 2) = [- sin(y) - cos(x)] (1,2) = - sin(2) - cos() = , fyx(1, 2) = , fyy(1, 2) = -2x cos(y) (1,2) = -2 cos(2) = -2.

With this done, we write down the second-order Taylor approximation:

1 f (x, y) 1 + (1 + 2)(x - 1) +

(x - 1)(y - 2) + (y - 2)(x - 1) - 2(y - 2)2

2

1 + (1 + 2)(x - 1) + (x - 1)(y - 2) - 2 (y - 2)2. 2

You may also write this as:

f

((1,

2)

+

h)

1

+

(1

+

2)h1

+

h1h2

-

2 2

h22.

3.3.4: Find critical points and determine type:

f (x, y) = x2 + y2 + 3xy,

f (x, y) = (2x + 3y, 2y + 3x). f is 0 = (0, 0) only when (x, y) = (0, 0), so it's the only critical point.

fxx 2, fxy 3, fyx 3, fyy 2.

The Hessian matrix (anywhere, and in particular at (0, 0)) is:

23 32

,

with determinant D = 4 - 9 = -5 < 0, so (0, 0) is a saddle point.

3.3.10: Find critical points and determine type:

f (x, y) = y + x sin(y),

f (x, y) = (sin(y), 1 + x cos(y)).

4

Now, sin(y) = 0 when y = n, for all integers n.

Then setting 1 + x cos(n) = 1 + (-1)nx = 0, we get that critical points are: (x, y) = ((-1)n+1, n) for n Z.

At these points:

fxx 0, fxy = cos(n) = (-1)n, fyx = cos(n) = (-1)n, fyy = -x sin(y) ((-1)n+1,n) = -(-1)n+1 sin(n) = 0.

So the Hessian matrix at ((-1)n+1, n) is:

0 (-1)n (-1)n 0

,

with determinant D = -(-1)2n = -1 < 0, so all these points are saddle points.

3.3.12: Find critical points and determine type:

f (x, y) = (x - y)(xy - 1), f (x, y) = ((xy - 1) + y(x - y), -(xy - 1) + x(x - y)). For f = 0 get two equations to solve:

(xy - 1) + y(x - y) = 0, -(xy - 1) + x(x - y) = 0. Adding the two, we get, x2 - y2 = 0, so x = ?y.

Now if x = -y, the first equation yields: -x2 - 1 - x2 - x2 = 0, or, 3x2 + 1 = 0,

which is impossible for real x.

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