Differentiability Implies Continuity - math.uh.edu

EXERCISES 15.1

Find the gradient. 4. f (x, y) = x−y 1
1. f (x, y) = 3x2 − xy + y. x2 + y2 . 1
2. f (x, y) = Ax2 + Bxy + Cy2. 1
3. f (x, y) = xe xy. 6. f (x, y) = ln (x2 + y2). 1
1
5. f (x, y) = 2xy2 sin (x2 + 1). 8. f (x, y) = Ax + By
7. f (x, y) = e x−y − e y−x. Cx + Dy . F
1
9. f (x, y, z) = x2y + y2z + z2x. 11. f (x, y, z) = x2ye−z. 1
10. f (x, y, z) = x2 + y2 + z2. 1

12. f (x, y, z) = xyz ln (x + y + z).
13. f (x, y, z) = e x+2y cos (z2 + 1).
14. f (x, y, z) = e yz2/x3.
15. f (x, y, z) = sin (2xy) + ln (x2z).
16. f (x, y, z) = x2y/z − 3xz4.

Find the gradient vector at the point P.
17. f (x, y) = 2x2 − 3xy + 4y2 at P(2, 3).
18. f (x, y) = 2x(x − y)−1 at P(3, 1).
19. f (x, y) = ln (x2 + y2) at P(2, 1).

15.1 DIFFERENTIABILITY AND GRADIENT 867

20. f (x, y) = x tan−1 (y/x) at P(1, 1). 38. Derive (15. 1. 5).
39. Let f (x, y) = 1 + x2 + y2.
21. f (x, y) = x sin (xy) at P(1, π/2).
(a) Find the points (x, y), if any, at which ∇f (x, y) = 0.
22. f (x, y) = xye−(x2+y2) at P(1, −1).

23. f (x, y, z) = e−x sin (z + 2y) at P(0, 1 π , 1 π ). (b) Sketch the graph of the surface z = f (x, y).
4 4
(c) What can you say about the surface at the point(s) found
24. f (x, y, z) = (x − y) cos π z at P(1, 0, 1 ). in part (a)?
2

25. f (x, y, z) = x − y2 + z2 at P(2, −3, 4). 40. Repeat Exercise 39 for f (x, y) = 4 − x2 − y2.
41. (a) Show that, if c q h is o(h), then c = 0. HINT:
26. f (x, y, z) = cos (xyz2) at P(π , 1 , −1). First
4 set h = h i, then set h = h j, then h = h k.

c In Exercises 27 and 28, use a CAS to find the gradient of f at

the point P. (b) Show that, if

27. (a) f (x, y) = xy2e−xy; P(0, 2). f (x + h) − f (x) = y q h + o(h)

(b) f (x, y) = sin (2x + y) − cos (x − 2y); P(π/4, π/6).

(c) f (x, y) = x − y ln (x2y); P(1, e). and f (x + h) − f (x) = z q h + o(h),

28. (a) f (x, y, z) = x + y2 − z3; P(1, 2, −3). then y = z.

(b) f (x, y, z) = xy P(1, −2, 3)
x − y + z;
42. Show that, if g is o(h), then
(c) f (x, y, z) = x sin (z ln y); P(1, e2, π/6).

Obtain the gradient directly from Definition 15.1.2 lim g(h) = 0.

h→0

29. f (x, y) = 3x2 − xy + y. 30. f (x, y) = 1 x2 + 2xy + y2. 43. Set 
2

31. f (x, y, z) = x2y + y2z + z2x.  2xy (x, y) = (0, 0)
f (x, y) =  , (x, y) = (0, 0).
32. f (x, y, z) = 2x2y − 1
. x2 + y2
z 0,

Find a function f with the gradient F. (a) Show that f is not differentiable at (0, 0).

33. F(x, y) = 2xy i + (1 + x2) j. (b) In Section 14.6 you saw that the first partials ∂f /∂x and
∂f /∂y exist at (0, 0). Since these partials obviously exist
34. F(x, y) = (2xy + x) i + (x2 + y) j. at every other point of the plane, we can conclude from
Theorem 15.1.3 that at least one of these partials is not
35. F(x, y) = (x + sin y) i + (x cos y − 2y) j. continuous in a neighborhood of (0,0). Show that ∂f /∂x
is discontinuous at (0, 0).
36. F(x, y, z) = yz i + (xz + 2yz) j + (xy + y2) k.

37. Find (a) ∇(ln r), (b) ∇( sin r), (c) ∇(er), where
r = x2 + y2 + z2.

PROJECT 15.1 Points Where ∇ f = 0

In this project we find points (x, y) where ∇f = 0. Then we 4xy = 0 and 2y2 − 2x2 − 2 = 0

investigate the behavior of the surface z = f (x, y) at these points. From the first equation, we get x = 0 or y = 0. Setting x = 0
in the second equation yields y = ±1; setting y = 0 in the
In this context, recall the significance of f (x) = 0 for a function second equation yields x2 = −1, which has no solutions. Thus
∇f (x, y) = 0 only at (0, 1) and (0, −1). The figure shows the
f of a single variable. graph of the surface z = f (x, y). Note that f has a maximum at
(0, −1) and a minimum at (0, 1).
For example, let
z
f (x, y) = −2y .

x2 + y2 + 1

Then, ∂f 4xy ∂f 2y2 − 2x2 − 2
= ,=

∂x (x2 + y2 + 1)2 ∂y (x2 + y2 + 1)2

4xy 2y2 − 2x2 − 2
and ∇f (x, y) = i + j.

(x2 + y2 + 1)2 (x2 + y2 + 1)2

Setting ∇f (x, y) = 0, we get the pair of equations

4xy = 0. 2y2 − 2x2 − 2 = 0
(x2 + y2 + 1)2 (x2 + y2 + 1)2
x
which are satisfied iff y

868 CHAPTER 15 GRADIENTS; EXTREME VALUES; DIFFERENTIALS

Problem 1. Let f (x, y) = 5x , Problem 2. Repeat Problem 1 with
f (x, y) = ( sin x)( sin y), 0 ≤ x ≤ 4π, 0 ≤ y ≤ 4π.
x2 + y2 + 1
− 5 ≤ x ≤ 5, −5 ≤ y ≤ 5. Problem 3. Repeat Problem 1 with
f (x, y) = 4xye− (x2+y2), −2 ≤ x ≤ 2, −2 ≤ y ≤ 2.
(a) Find the points (x, y), if any, at which ∇f (x, y) = 0.
(b) Use a CAS to draw the surface z = f (x, y) and the level Problem 4. Repeat Problem 1 with
f (x, y) = (x2 + 4y2)e1−(x2+y2), −2 ≤ x ≤ 2, −2 ≤ y ≤ 2.
curves. Compare with the results of part (a).

(c) Investigate the behavior of f at the points found in part (a).

∗SUPPLEMENT TO SECTION 15.1

PROOF OF THEOREM 15.1.3

We prove the theorem in the two-variable case. A similar argument yields a proof in the
three-variable case, but there the details are more burdensome.

In the first place,

f (x + h) − f (x) = f (x + h1, y + h2) − f (x, y).

Adding and subtracting f (x, y + h2), we have

(1) f (x + h) − f (x) = [ f (x + h1, y + h2) − f (x, y + h2)] + [ f (x, y + h2) − f (x, y)].

By the mean-value theorem for functions of one variable, we know that there are
numbers

0 < θ1 < 1 and 0 < θ2 < 1

such that f (x + h1, y + h2) − f (x, y + h2) = ∂f + θ1h1, y + h2)h1
and (x
∂x

∂f (Exercise 40, Section 4.1)
f (x, y + h2) − f (x, y) = ∂y (x, y + θ2h2)h2.

By the continuity of ∂f /∂x,

∂f + θ1h1, y + h2) = ∂f + 1(h)
∂x (x ∂x (x, y)

where 1(h) → 0 as h → 0. †

†

= ∂f ∂f →
1(h) ∂x (x + θ1h1, y + h2) − (x, y) 0
∂x

since, by the continuity of ∂f /∂x,

∂f ∂f
∂x (x + θ1h1, y + h2) → ∂x (x, y).

15.2 GRADIENTS AND DIRECTIONAL DERIVATIVES 869

By the continuity of ∂f /∂y,

∂f ∂f
(x, y + θ2h2) = (x, y) + 2(h)
∂x ∂x

where 2(h) → 0 as h → 0.

Substituting these expressions in equation (1), we find that

∂f ∂f
f (x + h) − f (x) = ∂x (x, y) + 1(h) h1 + ∂y (x, y) + 2(h) h2

= ∂ f (x, y) + 1(h) (i q h) + ∂f (x, y) + 2(h) (j q h)
∂ x ∂y

= ∂f (x, y) i + 1(h) i qh+ ∂f (x, y) j + 2(h) j qh
∂x ∂y

= ∂ f (x, y) i + ∂f (x, y) j qh+ 1(h) i + 2(h) j q h.
∂ x ∂y

To complete the proof of the theorem we need only show that

(2) [ 1(h) i + 2(h) j] q h = o(h).
From Schwarz’s inequality, |a q b| ≤ ||a|| ||b||, we know that

|[ 1(h) i + 2(h) j] q h| ≤ || 1(h) i + 2(h) j|| ||h||.

It follows that

|[ 1(h) i + 2(h) j] q h| ≤ || 1(h) i+ 2(h) j|| ≤ || 1(h) i||+|| 2(h) j|| = | 1(h)|+| 2(h)|.
||h||
↑ by the triangle inequality

As h → 0, the expression on the right tends to 0. This shows that (2) holds and completes
the proof of the theorem.