z = x 2 − 2xy + y 2 − 2 at x = 1, y = −1, z = 2
Let:
u=
Compute ux1 and ux4
q
x12 + x22 + x32 + x42
Let:
u=
q
x12 + x22 + x32 + x42
Compute ux1 and ux4
SOLUTION:
1
ux1 = (x12 + x22 + x32 + x42 )−1/2 (2x1 ) =
2
Let:
u=
q
x12 + x22 + x32 + x42
Compute ux1 and ux4
SOLUTION:
1
x1
ux1 = (x12 + x22 + x32 + x42 )−1/2 (2x1 ) = q
2
x12 + x22 + x32 + x42
And ux4 is found in a similar way:
Let:
u=
q
x12 + x22 + x32 + x42
Compute ux1 and ux4
SOLUTION:
1
x1
ux1 = (x12 + x22 + x32 + x42 )−1/2 (2x1 ) = q
2
x12 + x22 + x32 + x42
And ux4 is found in a similar way:
ux4 =
Let:
u=
q
x12 + x22 + x32 + x42
Compute ux1 and ux4
SOLUTION:
1
x1
ux1 = (x12 + x22 + x32 + x42 )−1/2 (2x1 ) = q
2
x12 + x22 + x32 + x42
And ux4 is found in a similar way:
x4
ux4 = q
x12 + x22 + x32 + x42
(88) The paraboloid z = 6 − x − x 2 − 2y 2 intersects the plane
x = 1 in a parabola. Find the parametric equations for the tangent
line to this parabola at the point (1, 2, −4), and graph.
(88) The paraboloid z = 6 − x − x 2 − 2y 2 intersects the plane
x = 1 in a parabola. Find the parametric equations for the tangent
line to this parabola at the point (1, 2, −4), and graph.
SOLUTION: The parabola is found by setting x = 1
z = 4 − 2y 2
⇒
zy = −4y
The derivative at y = 2 is −8. The tangent line therefore has
direction
(88) The paraboloid z = 6 − x − x 2 − 2y 2 intersects the plane
x = 1 in a parabola. Find the parametric equations for the tangent
line to this parabola at the point (1, 2, −4), and graph.
SOLUTION: The parabola is found by setting x = 1
z = 4 − 2y 2
⇒
zy = −4y
The derivative at y = 2 is −8. The tangent line therefore has
direction (we’re not moving in x, just in y ): h0, 1, −8i, and the
tangent line is therefore:
(88) The paraboloid z = 6 − x − x 2 − 2y 2 intersects the plane
x = 1 in a parabola. Find the parametric equations for the tangent
line to this parabola at the point (1, 2, −4), and graph.
SOLUTION: The parabola is found by setting x = 1
z = 4 − 2y 2
⇒
zy = −4y
The derivative at y = 2 is −8. The tangent line therefore has
direction (we’re not moving in x, just in y ): h0, 1, −8i, and the
tangent line is therefore:
x =1 y =2+t
z = −4 − 8t
A Special Function
( 3
x y −xy 3
if (x, y ) 6= (0, 0)
x 2 +y 2
f (x, y ) =
0 if (x, y ) = (0, 0)
A Special Function
( 3
x y −xy 3
if (x, y ) 6= (0, 0)
x 2 +y 2
f (x, y ) =
0 if (x, y ) = (0, 0)
fx (x, y ) =
y (x 4 + 4x 2 y 2 − y 4 )
(x 2 + y 2 )2
fy (x, y ) =
x(x 4 − 4x 2 y 2 − y 4 )
(x 2 + y 2 )2
Recall: If (x, y ) 6= (0, 0), then
f (x, y ) =
x 3 y − xy 3
x2 + y2
Does fx (0, 0) exist? (Use the definition!)
fx (0, 0) = lim
h→0
f (0, h) − f (0, 0)
=
h
Recall: If (x, y ) 6= (0, 0), then
f (x, y ) =
x 3 y − xy 3
x2 + y2
Does fx (0, 0) exist? (Use the definition!)
fx (0, 0) = lim
h→0
0−0
f (0, h) − f (0, 0)
=0
= lim
h→0
h
h
Recall: If (x, y ) 6= (0, 0), then
f (x, y ) =
x 3 y − xy 3
x2 + y2
Does fx (0, 0) exist? (Use the definition!)
fx (0, 0) = lim
h→0
Similarly, fy (0, 0) = 0.
0−0
f (0, h) − f (0, 0)
=0
= lim
h→0
h
h
If (x, y ) 6= (0, 0), then
fxy =
x 6 + 9x 4 y 2 − 9x 2 y 4 − y 6
= fyx
(x 2 + y 2 )3
Continuous at the origin?
If (x, y ) 6= (0, 0), then
fxy =
x 6 + 9x 4 y 2 − 9x 2 y 4 − y 6
= fyx
(x 2 + y 2 )3
Continuous at the origin?
Reminder- If (x, y ) 6= (0, 0), then:
fx (x, y ) =
y (x 4 + 4x 2 y 2 − y 4 )
(x 2 + y 2 )2
Compute, using the definition:
fxy (0, 0) =
fy (x, y ) =
x(x 4 − 4x 2 y 2 − y 4 )
(x 2 + y 2 )2
Reminder- If (x, y ) 6= (0, 0), then:
fx (x, y ) =
y (x 4 + 4x 2 y 2 − y 4 )
(x 2 + y 2 )2
fy (x, y ) =
Compute, using the definition:
fx (0, h) − fx (0, 0)
=
h→0
h
fxy (0, 0) = lim
x(x 4 − 4x 2 y 2 − y 4 )
(x 2 + y 2 )2
Reminder- If (x, y ) 6= (0, 0), then:
fx (x, y ) =
y (x 4 + 4x 2 y 2 − y 4 )
(x 2 + y 2 )2
fy (x, y ) =
x(x 4 − 4x 2 y 2 − y 4 )
(x 2 + y 2 )2
Compute, using the definition:
fx (0, h) − fx (0, 0)
−h5
= lim 5 = −1
h→0
h→0 h
h
fxy (0, 0) = lim
fyx (0, 0) =
Reminder- If (x, y ) 6= (0, 0), then:
fx (x, y ) =
y (x 4 + 4x 2 y 2 − y 4 )
(x 2 + y 2 )2
fy (x, y ) =
x(x 4 − 4x 2 y 2 − y 4 )
(x 2 + y 2 )2
Compute, using the definition:
fx (0, h) − fx (0, 0)
−h5
= lim 5 = −1
h→0
h→0 h
h
fxy (0, 0) = lim
fyx (0, 0) = lim
h→0
fx (h, 0) − fx (0, 0)
=
h
Reminder- If (x, y ) 6= (0, 0), then:
fx (x, y ) =
y (x 4 + 4x 2 y 2 − y 4 )
(x 2 + y 2 )2
fy (x, y ) =
x(x 4 − 4x 2 y 2 − y 4 )
(x 2 + y 2 )2
Compute, using the definition:
fx (0, h) − fx (0, 0)
−h5
= lim 5 = −1
h→0
h→0 h
h
fxy (0, 0) = lim
fx (h, 0) − fx (0, 0)
h5
= lim 5 =
h→0
h→0 h
h
fyx (0, 0) = lim
Reminder- If (x, y ) 6= (0, 0), then:
fx (x, y ) =
y (x 4 + 4x 2 y 2 − y 4 )
(x 2 + y 2 )2
fy (x, y ) =
x(x 4 − 4x 2 y 2 − y 4 )
(x 2 + y 2 )2
Compute, using the definition:
fx (0, h) − fx (0, 0)
−h5
= lim 5 = −1
h→0
h→0 h
h
fxy (0, 0) = lim
fx (h, 0) − fx (0, 0)
h5
= lim 5 = 1
h→0
h→0 h
h
fyx (0, 0) = lim
Reminder- If (x, y ) 6= (0, 0), then:
fx (x, y ) =
y (x 4 + 4x 2 y 2 − y 4 )
(x 2 + y 2 )2
fy (x, y ) =
x(x 4 − 4x 2 y 2 − y 4 )
(x 2 + y 2 )2
Compute, using the definition:
fx (0, h) − fx (0, 0)
−h5
= lim 5 = −1
h→0
h→0 h
h
fxy (0, 0) = lim
fx (h, 0) − fx (0, 0)
h5
= lim 5 = 1
h→0
h→0 h
h
fyx (0, 0) = lim
Therefore,
fxy (0, 0) 6= fyx (0, 0)
(even though they are equal everywhere else!)