Math 417 Midterm 2
1. Use the rst order approximation theorem to prove that
ex+2y − 1 − x − 2y
p
= 0.
(x,y)→(0,0)
x2 + y 2
lim
Let f (x, y) = ex+2y . Observe that D1 f (x, y) = ex+2y and D2 f (x, y) = 2ex+2y
are continuous and ∇f (0, 0) = (1, 2). So by the rst order approximation theorem,
Solution.
0=
f (x, y) − f (0, 0) − h∇f (0, 0), (x, y)i
ex+2y − 1 − x − 2y
p
= lim
.
(x,y)→(0,0)
(x,y)→(0,0)
k(x, y)k
x2 + y 2
lim
1
2. Let f : A → R be continuously dierentiable, where A ⊂ Rn is an open set containing
the line segment L from x to x + h. Suppose f (x) = f (x + h). Show that h∇f (z), hi = 0
for some z ∈ L.
Solution.
By the mean value theorem, 0 = f (x + h) − f (x) = h∇f (z), hi for some z ∈ L.
2
3. Consider the function
(
f (x, y) =
x2 y 2
,
x2 +y 2
0,
(x, y) 6= (0, 0)
.
(x, y) = (0, 0)
Show that for (x, y) 6= (0, 0),
D1 f (x, y) =
2xy 4
(x2 + y 2 )2
and
D2 f (x, y) =
2x4 y
.
(x2 + y 2 )2
Then show that D1 f (0, 0) = 0 and D2 f (0, 0) = 0.
Solution.
Observe that for (x, y) 6= (0, 0),
D1 f (x, y) =
2xy 4
(x2 + y 2 )2xy 2 − x2 y 2 (2x)
=
(x2 + y 2 )2
(x2 + y 2 )2
and
f (t, 0) − f (0, 0)
0−0
= lim
= 0.
t→0
t→0
t
t
The formulas for D2 f follow from symmetry of f .
D1 f (0, 0) = lim
3
4. Let f be as in Problem 3.
i) Show that ∇f (0, 0) = 0 and ∇ f (0, 0) =
2
Solution.
0 0
.
0 0
From Problem 3, ∇f (0, 0) = 0 and
D1 f (t, 0) − D1 f (0, 0)
t→0
t
D2 f (0, t) − D2 f (0, 0)
D2 D2 f (0, 0) = lim
t→0
t
D2 f (t, 0) − D2 f (0, 0)
D1 D2 f (0, 0) = lim
t→0
t
D1 f (0, t) − D1 f (0, 0)
D2 D1 f (0, 0) = lim
t→0
t
D1 D1 f (0, 0) = lim
4
0−0
t→0
t
0−0
= lim
t→0
t
0−0
= lim
t→0
t
0−0
= lim
t→0
t
= lim
= 0,
= 0,
= 0,
= 0.
4. (continued) Let f be as in Problem 3.
ii) By letting (x, y) → (0, 0) along the line y = x, show that
f (x, y)
6= 0.
(x,y)→0 k(x, y)k2
lim
What must be true about the second order partial derivatives of f ?
(Hint: think of the second order approximation theorem.)
Solution.
Notice lim(x,y)→(0,0) f (x, y)/k(x, y)k2 6= 0 because
x2 x2
x4
1
f (x, x)
=
lim
=
lim
= .
x→0 (x2 + x2 )2
x→0 4x4
x→0 k(x, x)k2
4
lim
The second order partial derivatives of f cannot all be continuous: if they were, then
f (x, y) − f (0, 0) − h∇f (0, 0), (x, y)i −
0=
lim
D
∇2 f (0, 0)
x
y
E
, (x, y)
k(x, y)k2
(x,y)→(0,0)
=
1
2
f (x, y)
(x,y)→(0,0) k(x, y)k2
lim
by the second order approximation theorem. But we have seen this is false.
5