394
CHAPTER 43
43.17
For the function/of Problem 43.16, find the rate of change of/at (4,1,0) along the normal line to the plane
3(jc — 4) — (y — 1) + 2z = 0 in the direction of increasing x.
A vector parallel to the normal line to the given plane is (3, -1,2) and it is pointing in the direction of
(3,-1,2). Since V/=(8,8,-8), the rate of
increasing x. The unit vector in this direction is u =
change is V/-u = 8(l, 1, -!)•
(3,-1,2) =
(0) = 0.
43.18 For functions f(x, y, z) and g(x, y, z), prove V(/ + g) =Vf + Vg.
?/ + Vg = (/,,/,,/J + (g,,g,,gJ = ((/ + g),,(/ + g) y ,(/ + g)J=V(/ + g).
43.19 Prove V( /g) = / Vg + g Vf.
^(/g) = (fa + Lg, fgy + fyg, fg: + f2g) = (fg,, fgy, fg, ) + (/.g, fyg, L g) = / ?g + g Vf.
43.20
Prove V(f) = nf"'
l \f.
The proof is by induction on n. The result is clearly true when n — \. Assume the result true forn. Then
(by Problem 43.19)
(by the inductive hypothesis)
Thus, the identity also holds for n + 1.
43.21
Prove
By Problem 43.19,
Now solve for V(//g).
43.22
If z =f(x, y) has a relative maximum or minimum at a point (x 0 , y 0 ), show that Vz = 0 at (x 0 , y 0 ).
We wish to show that both partial derivatives of z vanish at (x g , y a ). The plane y = y 0 intersects the
surface z=f(x, y) in a curve z=f(x, y0) that has a relative maximum (or minimum) at x = xa. Hence,
dz/dx = Q at (x 0 , y a ). Similarly, the plane x - x 0 intersects the surface z = f(x, y) in a curve z =
f(xo< y) tnat nas a relative maximum (or minimum) at y = y0. Hence, the derivative (*o> >"o)- Similar results hold for functions/of more than two variables.
43.23 Assume that f(x, y) has continuous second partial derivatives in a disk containing the point (*„, y 0 ) inside it.
Assume also that (x0, y0) is a critical point of/, that is, fx = fy = 0 at (jc0, y0). Let A =/„/,.,. - (f,y)2 (the
Hessian determinant). State sufficient conditions for/to have a relative maximum or minimum at (*„, y 0 ).
Case 1. Assume A > 0 at (*0, y0). (a) If /„ + fyy < 0 at (x0, y0), then /has a relative maximum at
(*o> .Xo)- (&) M f**+fyy>0 at (*o> >o). tnen / nas a relative minimum at (x0, y0). Case 2. If A<0, /
has neither a relative maximum nor a relative minimum at (x 0 , y 0 ). Case 3. If A = 0, no conclusions can be
drawn.
43.24
Using the assumptions and notation of Problem 43.23, give examples to show that case 3, where A = 0, allows
no conclusions to be drawn.
Each of the functions /,(*, y) = x
4 + y
4 , f 2 (x, y) = -(x
4 + y
4 ), and /,(*, y) = x
3 - y
3
vanishes at
(0,0) together with its first and second partials; hence A(0,0) = 0 for each function. But, at (0,0), /, has a
relative minimum, / 2 has a relative maximum, and / 3 has neither [/,(*, 0) = x
3
takes on both positive and
negative values in any neighborhood of the origin].
43.25
Find the relative maxima and minima of the function f(x, y) = 2x + 4y — x
2 — y
2 - 3.
f, = 2-2x, fy=4-2y. Setting /, =0, £ = 0, we have jr = l, y = 2. Thus, (1,2) is the only
critical point. /x,=0, /„ =-2, and fyy = -2. So, A=/^/vv -(fxy)2 = 4>0. Since /„+/„ =-4<0,
there is a relative maximum at (1,2), by Problem 43.23.
CHAPTER 43
43.17
For the function/of Problem 43.16, find the rate of change of/at (4,1,0) along the normal line to the plane
3(jc — 4) — (y — 1) + 2z = 0 in the direction of increasing x.
A vector parallel to the normal line to the given plane is (3, -1,2) and it is pointing in the direction of
(3,-1,2). Since V/=(8,8,-8), the rate of
increasing x. The unit vector in this direction is u =
change is V/-u = 8(l, 1, -!)•
(3,-1,2) =
(0) = 0.
43.18 For functions f(x, y, z) and g(x, y, z), prove V(/ + g) =Vf + Vg.
?/ + Vg = (/,,/,,/J + (g,,g,,gJ = ((/ + g),,(/ + g) y ,(/ + g)J=V(/ + g).
43.19 Prove V( /g) = / Vg + g Vf.
^(/g) = (fa + Lg, fgy + fyg, fg: + f2g) = (fg,, fgy, fg, ) + (/.g, fyg, L g) = / ?g + g Vf.
43.20
Prove V(f) = nf"'
l \f.
The proof is by induction on n. The result is clearly true when n — \. Assume the result true forn. Then
(by Problem 43.19)
(by the inductive hypothesis)
Thus, the identity also holds for n + 1.
43.21
Prove
By Problem 43.19,
Now solve for V(//g).
43.22
If z =f(x, y) has a relative maximum or minimum at a point (x 0 , y 0 ), show that Vz = 0 at (x 0 , y 0 ).
We wish to show that both partial derivatives of z vanish at (x g , y a ). The plane y = y 0 intersects the
surface z=f(x, y) in a curve z=f(x, y0) that has a relative maximum (or minimum) at x = xa. Hence,
dz/dx = Q at (x 0 , y a ). Similarly, the plane x - x 0 intersects the surface z = f(x, y) in a curve z =
f(xo< y) tnat nas a relative maximum (or minimum) at y = y0. Hence, the derivative (*o> >"o)- Similar results hold for functions/of more than two variables.
43.23 Assume that f(x, y) has continuous second partial derivatives in a disk containing the point (*„, y 0 ) inside it.
Assume also that (x0, y0) is a critical point of/, that is, fx = fy = 0 at (jc0, y0). Let A =/„/,.,. - (f,y)2 (the
Hessian determinant). State sufficient conditions for/to have a relative maximum or minimum at (*„, y 0 ).
Case 1. Assume A > 0 at (*0, y0). (a) If /„ + fyy < 0 at (x0, y0), then /has a relative maximum at
(*o> .Xo)- (&) M f**+fyy>0 at (*o> >o). tnen / nas a relative minimum at (x0, y0). Case 2. If A<0, /
has neither a relative maximum nor a relative minimum at (x 0 , y 0 ). Case 3. If A = 0, no conclusions can be
drawn.
43.24
Using the assumptions and notation of Problem 43.23, give examples to show that case 3, where A = 0, allows
no conclusions to be drawn.
Each of the functions /,(*, y) = x
4 + y
4 , f 2 (x, y) = -(x
4 + y
4 ), and /,(*, y) = x
3 - y
3
vanishes at
(0,0) together with its first and second partials; hence A(0,0) = 0 for each function. But, at (0,0), /, has a
relative minimum, / 2 has a relative maximum, and / 3 has neither [/,(*, 0) = x
3
takes on both positive and
negative values in any neighborhood of the origin].
43.25
Find the relative maxima and minima of the function f(x, y) = 2x + 4y — x
2 — y
2 - 3.
f, = 2-2x, fy=4-2y. Setting /, =0, £ = 0, we have jr = l, y = 2. Thus, (1,2) is the only
critical point. /x,=0, /„ =-2, and fyy = -2. So, A=/^/vv -(fxy)2 = 4>0. Since /„+/„ =-4<0,
there is a relative maximum at (1,2), by Problem 43.23.