MATH 32A: FINAL REVIEW
JOE HUGHES
1. Parametrization and vector-valued functions
1. Find a parametrization of the hyperbola x2 − y 2 = 1 using trig functions.
Solution: Recall that sec2 t = 1 + tan2 t, which can be rewritten as sec2 t − tan2 t = 1. So
take x = sec t and y = tan t.
The last thing to figure out is the range of t. As t goes from 0 to π2 , sec t increases from 1
to +∞, and tan t increases from 0 to +∞. Then as t increases from π2 to π, sec t increases
from −∞ to −1, and tan t increases from −∞ to 0.
As t increases from π to 3π
2 , sec t decreases from −1 to −∞ and tan t increases from 0 to
∞, and finally as t increases from 3π
2 to 2π, sec t decreases from ∞ to 1 and tan t increases
from −∞ to 0. So the interval [0, 2π] is all that’s needed for the parametrization.
2. Parametrize the intersection of the paraboloids z = x2 + y 2 and z = 8 − x2 − y 2 .
Solution: Eliminating z, we get
x2 + y 2 = 8 − x2 − y 2
or 2x2 + 2y 2 = 8, so x2 + y 2 = 4. Therefore a parametrization is given by
r(t) = h2 cos t, 2 sin ti
3. Find a parametrization of Bernoulli’s lemniscate (x2 + y 2 )2 = x2 − y 2 . (Hint: try
t2 = x2 + y 2 ).
Solution: If we set t2 = x2 + y 2 , then the equation of the lemniscate becomes
t4 = x2 − y 2 = x2 + y 2 − 2y 2 = t2 − 2y 2
Therefore
t2 − t4
2
Similarly, we can also write the equation of the curve as
y=
t4 = x2 − y 2 = 2x2 − x2 − y 2 = 2x2 − t2
so
x2 =
t2 + t4
2
1
2
JOE HUGHES
This leaves us with two issues: how to choose the signs of the square roots, and determining
what interval t should lie in.
For the first issue, observe that if a point (x, y) lies on the curve, then so do the points
(x, −y), (−x, y), and (−x, −y). Therefore it’s enough to parametrize the curve in the first
quadrant, because then the rest of the curve is determined by the above symmetries. So
take
r
r
t2 + t4
t2 − t4
y=
x=
2
2
For the second issue, note that the curve passes through the origin, and crosses the x-axis
at the point (1, 0). The point (0, 0) corresponds to t = 0, and the point (1, 0) corresponds
to t = 1. So we should take t in the interval [0, 1].
2. Arc Length
1. Find the arc length parametrization of r(t) = het cos t, et sin t, et i for t in (−∞, ∞).
Solution: The tangent vector is
r(t) = het cos t − et sin t, et sin t + et cos t, et i = het cos t, et sin t, et i + h−et sin t, et cos t, 0i
Call the first vector-valued function on the right a(t), and the second b(t). Then a(t)·b(t) =
0 for all t, hence by the Pythagorean theorem
||r(t)||2 = ||a(t)||2 + ||b(t)||2 = 3e2t
The arc length function is then
Z t
√ Z
0
s(t) =
||r (u)|| du = 3
−∞
t
eu du =
√
3et
−∞
which has inverse
s
t = ln √
3
Therefore the arc length parametrization is
D s
s s
s s E
r1 (s) = √ cos ln √ , √ sin ln √ , √
3
3
3
3
3
√ t
If t ranges from −∞ to ∞, then s = 3e ranges from 0 to ∞.
3. Curvature
1. Find the curvature κ(t) of r(t) = hcos t, sin t, et i.
Solution: The first derivative is
r0 (t) = h− sin t, cos t, et i
MATH 32A: FINAL REVIEW
3
which has length
||r0 (t)|| =
p
1 + e2t
The second derivative is
r00 (t) = h− cos t, − sin t, et i
so
i
j
k
0
00
t
r (t) × r (t) = − sin t cos t e = et (cos t + sin t)i + et (sin t − cos t)j + k
− cos t − sin t et = het cos t, et sin t, 1i + het sin t, −et cos t, 0i
These vectors are perpendicular, so as in the arc length problem above it follows that
||r0 (t) × r00 (t)||2 = e2t (cos2 t + sin2 t) + 1 + e2t (sin2 t + cos2 t) = 2e2t + 1
Therefore the curvature is
√
κ(t) =
2e2t + 1
3
(e2t + 1) 2
√
2. Find the osculating circle to the graph of y = x at x = 4.
Solution: There are a couple ways of parametrizing the curve, but I think it’s easiest to
view y as the parameter, and set
r(t) = ht2 , ti
for t ≥ 0. Then
r0 (t) = h2t, 1i
r00 (t) = h2, 0i
The point x = 4 corresponds to t = 2, and
r0 (2) × r00 (2) = h4, 1, 0i × h2, 0, 0i = h0, 0, 2i
which has length 2. Since ||r0 (2)|| =
√
17, it follows that
2
κ(2) = 3
17 2
3
Therefore the radius of the osculating circle is 1722 . To find the center, note that the
point r(2) = (4, 2) lies on the circle, and that r0 (2) = h4, 1i is tangent to the curve at this
point.
Hence the inward normal is either h1, −4i or h−1, 4i, and drawing the curve r(t) shows
that the inward normal is h1, −4i. This is not yet a unit vector, so divide by the length to
get
1
N = √ h1, −4i
17
4
JOE HUGHES
Therefore the center of the osculating circle is
h4, 2i +
3
D 25
E
1
17 2
· √ h1, −4i =
, −32
2
2
17
so the circle has parametrization
3
D 25 17 32
E
17 2
c(t) =
+
cos θ, −32 +
sin θ
2
2
2
4. Level curves and graphs
1. Match the functions to the corresponding level curves (see the end for the level curves):
a. f (x, y) = ex
xy
.
x2 +y 2 +1
2 −y
b. f (x, y) = x3 − y 3
c. f (x, y) = ln(x2 + 2y 2 )
d. f (x, y) =
2
Solution: If ex −y = c, then c > 0 and x2 − y = ln(c). Therefore y = x2 − ln(c), which is a
parabola. Therefore (a) matches Figure 3.
Similarly, if ln(x2 + 2y 2 ) = c, then x2 + 2y 2 = ec , which is an ellipse. So (c) matches Figure
1.
The last two are a bit more difficult. But note that the contours in Figure 2 are symmetric
about the line y = x, as is the function in (d). Similarly, the contours in Figure 4 are
symmetric about the line y = −x, as is the function in (b). So (d) matches Figure 2 and
(b) matches Figure 4.
Another way to get the last two is to look specifically at the contour f (x, y) = 0. In the
case of (b), it is the line y = x, while for (d) it is the two coordinate axes.
2. Match the functions to the corresponding graphs (see the end for the graphs):
a. f (x, y) = sin(x) sin(y)
cos(x).
b. f (x, y) =
1
|x|+|y|+1
c. f (x, y) =
|x|
|x|+|y|+1
d. f (x, y) =
Solution: (d) is the easiest function to start with: f (x, y) = cos(x) has no dependence on
y, so it matches Figure 7. And (a) matches Figure 5 because this is the only other graph
which is periodic.
For the last two, note that the function in (c) is zero whenever x = 0, which matches
Figure 6. So then (b) matches Figure 8.
5. Limits and continuity
1. Evaluate the following limits:
a. lim(x,y)→(1,1)
sin(x2 −y 2 )
cos(xy) .
MATH 32A: FINAL REVIEW
5
Solution: Both the numerator and denominator are continuous functions, and the denominator is non-zero at (1, 1), so we can simply plug the point in:
sin(x2 − y 2 )
sin(0)
=
=0
cos(xy)
cos(1)
(x,y)→(1,1)
lim
y sin x
x sin y .
b. lim(x,y)→(0,0)
Solution: Using limit laws, this limit can be split up into two one-variable limits:
y sin x h
sin x ih
y i
lim
= lim
lim
=1·1=1
x→0 x
y→0 sin y
(x,y)→(0,0) x sin y
c. lim(x,y)→(0,0)
x5 −y 5
.
x3 −x2 y+xy 2 −y 3
Solution: I think the easiest way to evaluate this limit is to start by factoring the numerator
and denominator:
x5 − y 5 = (x − y)(x4 + x3 y + x2 y 2 + xy 3 + y 4 )
and
x3 − x2 y + xy 2 − y 3 = (x − y)(x2 + y 2 )
so
x5 − y 5
x4 + x3 y + x2 y 2 + xy 3 + y 4
=
lim
x2 + y 2
(x,y)→(0,0) x3 − x2 y + xy 2 − y 3
(x,y)→(0,0)
lim
Now switch to polar coordinates:
x4 + x3 y + x2 y 2 + xy 3 + y 4
x2 + y 2
(x,y)→(0,0)
lim
= lim
r→0+
r4 (cos4 θ + cos3 θ sin θ + cos2 θ sin2 θ + cos θ sin3 θ + sin4 θ)
r2
= lim r2 (cos4 θ + cos3 θ sin θ + cos2 θ sin2 θ + cos θ sin3 θ + sin4 θ) = 0
r→0+
by the squeeze theorem, because the term in parentheses is always between −5 and 5.
3
3
2. For each real number a, define fa (x) = x2x+y+y2 +a . For which values of a can f (x, y) be
extended to a continuous function on all of R2 ?
Solution: If a > 0, then x2 + y 2 + a > 0 for all (x, y), so fa (x, y) is a continuous function.
On the other hand, if a < 0, then x2 + y 2 + a = 0 for any point (x0 , y0 ) such that
x20 + y02 = −a. Such a point cannot be the origin because a 6= 0, so lim(x,y)→(x0 ,y0 ) fa (x, y)
does not exist and fa cannot be extended to a continuous function on R2 .
6
JOE HUGHES
That leaves the case a = 0. In this case x2 + y 2 = 0 when x = y = 0, but by using polar
coordinates it follows that
r3 (cos3 θ + sin3 θ)
x3 + y 3
=
lim
=0
r2
(x,y)→(0,0) x2 + y 2
r→0+
lim
by the squeeze theorem. Therefore f0 (x, y) can be extended to a continuous function on
R2 by setting f (0, 0) = 0.
6. Partial derivatives
1. Show that there is no function f (x, y) such that fx = y sin x and fy = sin x.
Solution: If such a function existed, then we would have fxy = sin x and fyx = cos x. Since
these functions are continuous and not equal, this contradicts Clairaut’s theorem. So no
such function f exists.
2. Find all functions f (x, y) such that fx = 2xy and fy = x2 + 1.
Solution: The antiderivative of 2xy with respect to x is x2 y + g(y), where g(y) is a function
only of y. Then differentiating x2 y + g(y) with respect to y shows that
x2 + 1 = fy = x2 + g 0 (y)
Therefore g(y) = y + C for a constant C, so f (x, y) = x2 y + y + C.
3. Two functions u(x, y) and v(x, y) are said to satisfy the Cauchy-Riemann equations
if ux = vy and uy = −vx . Suppose that two level curves u(x, y) = c1 and v(x, y) = c2
intersect at a point P , and that ∇uP 6= 0, ∇vP 6= 0. Show that the level curves u(x, y) = c1
and v(x, y) = c2 are perpendicular at P .
Solution: Recall that ∇uP is perpendicular to the level curve of u at P as long as ∇uP 6= 0,
and similarly for v.
Therefore if we can show that ∇uP and ∇vP are perpendicular, then it will follow that the
level curves are perpendicular (this is because there are only two perpendicular directions
in R2 ).
Now the gradients are ∇u = hux , uy i and ∇v = hvx , vy i. Therefore
∇u · ∇v = hux , uy i · hvx , vy i = ux vx + uy vy = vy vx − vx vy = 0
by the Cauchy-Riemann equations, so ∇u and ∇v are perpendicular.
7. Tangent planes
1. Find the tangent plane to the surface x2 +y 2 +z 2 = 3xyz at the point P = (1, 1, 1).
MATH 32A: FINAL REVIEW
7
Solution: This surface can be viewed as the level surface f (x, y, z) = 0 for f (x, y, z) =
x2 + y 2 + z 2 − 3xyz. Therefore a normal vector to the plane is given by ∇fP . And
∇f = h2x − 3yz, 2y − 3xz, 2z − 3xyi
so ∇fP = h−1, −1, −1i. Therefore the equation of the plane is
−x − y − z = h−1, −1, −1i · h1, 1, 1i = −3
or x + y + z = 3.
8. The chain rule
∂f 2
2
1. If f (x, y) is differentiable and x = u+v, y = u−v, show that ( ∂f
∂x ) −( ∂y ) =
∂f ∂f
∂u ∂v .
Solution: By the chain rule,
∂f ∂x ∂f ∂y
∂f
∂f
∂f
=
+
=
+
∂u
∂x ∂u ∂y ∂u
∂x ∂y
Similarly,
∂f
∂f ∂x ∂f ∂y
∂f
∂f
=
+
=
−
∂v
∂x ∂v
∂y ∂v
∂x ∂y
Therefore
∂f
∂f ∂f
∂f ∂f 2 ∂f 2
∂f ∂f
=
+
−
=
−
∂u ∂v
∂x ∂y ∂x ∂y
∂x
∂y
which is what we wanted to show.
2. Let r = hx, y, zi, and define er to be the unit vector in the direction of r. If f (x, y, z) =
F (r) depends only on r = ||r||, then show that ∇f = F 0 (r)er .
Solution: By the chain rule,
∂f
∂r
x
= F 0 (r)
= F 0 (r) p
2
∂x
∂x
x + y2 + z2
and similarly for
∇f =
∂f ∂f
∂y , ∂z .
Therefore
D ∂f ∂f ∂f E
E
D
x
y
z
,
,
= F 0 (r) p
,p
,p
= F 0 (r)er
2
2
2
2
2
2
2
2
2
∂x ∂y ∂z
x +y +z
x +y +z
x +y +z
A function f having this property is called radial, because it only depends on the radial
distance of a point from the origin.
3.
a. Show that the Laplacian in polar coordinates is given by ∆ = frr +
1
f
r2 θθ
+ 1r fr .
8
JOE HUGHES
Solution: Since x = r cos θ and y = r sin θ, using the chain rule shows that
∂f ∂x ∂f ∂y
∂f
∂f
∂f
=
+
=
cos θ +
sin θ
∂r
∂x ∂r
∂y ∂r
∂x
∂y
Differentiating again using the chain rule, we get
h ∂ 2 f ∂x
h ∂ 2 f ∂x ∂ 2 f ∂y i
∂2f
∂ 2 f ∂y i
=
cos
θ
+
+
sin
θ
+ 2
∂r2
∂x2 ∂r
∂y∂x ∂r
∂x∂y ∂r
∂y ∂r
=
∂2f
∂2f
∂2f
2
cos
θ
+
sin2 θ
(2
cos
θ
sin
θ)
+
∂x2
∂x∂y
∂y 2
(the use of Clairaut’s theorem is not a problem because we can assume f is well-behaved).
Similarly,
∂f
∂f ∂x ∂f ∂y
∂f
∂f
=
+
= − r sin θ +
r cos θ
∂θ
∂x ∂θ
∂y ∂θ
∂x
∂y
so differentiating a second time shows that
h ∂ 2 f ∂x
h ∂ 2 f ∂x ∂ 2 f ∂y i ∂f
∂2f
∂ 2 f ∂y i ∂f
=
−r
sin
θ
+
−
r
cos
θ
+
r
cos
θ
+ 2
−
r sin θ
∂θ2
∂x2 ∂θ
∂y∂x ∂θ
∂x
∂x∂y ∂θ
∂y ∂θ
∂y
=
∂2f
∂2f 2
∂f
∂f
∂2f 2 2
2
r
sin
θ
+
(−2r
sin
θ
cos
θ)
+
r cos2 θ −
r cos θ −
r sin θ
2
2
∂x
∂x∂y
∂y
∂x
∂y
Therefore
=
∂2f
1 ∂2f
1 ∂f
+ 2 2 +
2
∂r
r ∂θ
r ∂r
∂2f
∂2f
∂2f
2
2
(cos
θ
+
sin
θ)
+
(sin
θ
cos
θ
−
sin
θ
cos
θ)
+
(sin2 θ + cos2 θ)
∂x2
∂x∂y
∂x2
∂2f
∂2f
=
+
= ∆f
∂x2
∂y 2
as desired.
You may wonder why I chose to think of x, y as functions of r, θ rather than r, θ as functions
of x, y. The reason is that the expressions for x and y in terms of r and θ are easier to
differentiate than the expressions for r and θ in terms of x and y.
b. Use part (a) to show that f (x, y) = tan−1 ( xy ) is harmonic (i.e. ∆f = 0).
p
Solution: Polar coordinates are defined by r = x2 + y 2 and θ = tan−1 ( xy ), so in polar
coordinates f (r, θ) = θ. Therefore fr = frr = fθθ = 0, so
1
1
∆f = frr + 2 fθθ + fr = 0
r
r
and f is harmonic.
MATH 32A: FINAL REVIEW
9
9. Local extrema
1. Find the critical points of the following functions, and determine whether they are
maxima, minima, or neither:
a. f (x, y) = x2 + 3y 4 − 4y 3 − 12y 2 .
Solution: The partial derivatives are
fx = 2x
fy = 12y 3 − 12y 2 − 24y
So a critical point must satisfy x = 0 and 12y 3 − 12y 2 − 24y = 0. Thus either y = 0 or
0 = y 2 − y − 2 = (y + 1)(y − 2), so y = −1 or y = 2. Therefore the critical points are
(0, 0), (0, −1), (0, 2).
The second derivatives are fxx = 2, fxy = 0, and fyy = 36y 2 − 24y − 24. For (0, 0) we have
D(0, 0) = −48, so the origin is a saddle point.
At (0, −1) we get D = 72 and fxx = 2, so this point is a local minimum. And at (0, 2) the
determinant D = 144 and fxx = 2, so this point is also a local minimum.
b. f (x, y) = (y − x2 )(y − 2x2 ).
Solution: The partial derivatives are
fx = −2x(y − 2x2 ) − 4x(y − x2 ) = 8x3 − 6xy = 2x(x2 − 3y)
and
fy = y − 2x2 + y − x2 = 2y − 3x2
The condition fx = 0 implies that either x = 0 or x2 = 3y. If x = 0 then fy = 0 implies
that y = 0 as well.
If x2 = 3y then fy = 0 implies that 2y − 9y = 0, so y = 0 and then x = 0 as well.
Therefore the only critical point is the origin. The second partial derivatives are
fxx = 24x2 − 6y
fyy = 2
fxy = −6x
Therefore D(0, 0) = 0, so the second derivative test is inconclusive.
To determine what’s going on at the origin, look at the level curve f (x, y) = 0. This is
the union of the parabolas y = x2 and y = 2x2 . If we approach the origin through the
region lying above both parabolas (or below both parabolas), then f (x, y) > 0. But if
we approach the origin through the region between the two parabolas, then f (x, y) < 0.
Therefore (0, 0) cannot be a local maximum or minimum for f .
10
JOE HUGHES
10. Global extrema
1. Find the global extrema of the function f (x, y) = y − x on the domain D = {(x, y) :
y − x2 ≥ 0, x2 + y ≤ 4}.
Solution: The domain D is the region between two parabolas, so is bounded. It is also
closed because it contains the portions of the parabolas that make up its boundary, so f
is guaranteed to have a global max and min on D.
The partial derivatives are fx = −1 and fy = 1, so there are no critical points. Hence it’s
enough to look at the boundary.
√
The parabolas y = x2 and y = 4 − x2 intersect√when x2 =√4 − x2 , so x = ± 2. Thus the
2
2
boundary
√ consists
√ of the parabola y = x for − 2 ≤ x ≤ 2, and the parabola y = 4 − x
for − 2 ≤ x ≤ 2.
When y = x2 , f (x, x2 ) = x2 − x, which has derivative 2x − 1 and critical point x = 12 .
√
√
Adding in the boundary points x = ± 2, our list of points to consider is ( 21 , 14 ), ( 2, 2),
√
and (− 2, 2).
When y = 4 − x2 , f (x, 4 − x2 ) = 4 − x − x2 , which has derivative −(2x + 1). Therefore
f (x, 4 − x2 ) has a critical point when 2x + 1 = 0, so x = − 21 and y = 4 − x2 = 15
4 .
√
√
Therefore the list of candidate points for a max or min is ( 12 , 14 ), ( 2, 2), (− 2, 2), and
(− 21 , 15
4 ). And
1 1 1 1
√
√
1
f ,
= − =−
f ( 2, 2) = 2 − 2
2 4
4 2
4
√
√
17
1 15 15 1
f (− 2, 2) = 2 + 2 f − ,
=
+ =
2 4
4
2
4
√
Since 2 ≈ 1.414, it follows that the minimum of f on D is − 41 at ( 12 , 14 ) and the maximum
1 15
is 17
4 at (− 2 , 4 ).
2
2
2. Show that f (x, y) = (x2 + 2y 2 )e−x −y has a global maximum and a global minimum
on R2 , and find the points at which they occur.
Solution: The global minimum is easy: f (x, y) ≥ 0 for all points (x, y) in R2 , so the global
minimum occurs at the point at which f (x, y) = 0, namely (0, 0).
The global maximum is a bit trickier because R2 is unbounded. But note that f (x, y) is
2
2
very small when x2 + y 2 gets large because of the exponential decay in the term e−x −y .
Therefore there is some constant M such that 0 ≤ f (x, y) ≤ 10−6 if x2 + y 2 ≥ M . (The
number 10−6 could be replaced by any other small positive number).
For this reason, it’s enough to look for the global maximum of f on the disk D = {(x, y) :
x2 + y 2 ≤ M }, which must exist because D is closed and bounded. Moreover, if we can
MATH 32A: FINAL REVIEW
11
find a local maximum (x0 , y0 ) of f inside D for which f (x0 , y0 ) > 10−6 , then we can also
ignore the boundary of D.
So now our problem consists of finding the critical points of f inside D. The partial
derivatives of f are
2 −y 2
+ (x2 + 2y 2 )e−x
2 −y 2
+ (x2 + 2y 2 )e−x
fx = 2xe−x
2 −y 2
(−2x) = 2xe−x
2 −y 2
(1 − x2 − 2y 2 )
2 −y 2
(−2y) = 2ye−x
2 −y 2
(2 − x2 − 2y 2 )
and
fy = 4ye−x
Setting fx = 0 shows that either x = 0 or x2 + 2y 2 = 1, and setting fy = 0 shows that
either y = 0 or x2 + 2y 2 = 2.
We already know that (0, 0) is a global minimum for f , so it can be ignored. If x = 0 and
x2 + 2y 2 = 2, then y = ±1. If y = 0 and x2 + 2y 2 = 1, then x = ±1. Since the equations
x2 + 2y 2 = 2 and x2 + 2y 2 = 1 cannot be simultaneously true, the list of points to consider
is (0, ±1) and (±1, 0).
Finally, f (0, ±1) = 2e−1 , while f (±1, 0) = e−1 . Therefore f has a global maximum of 2e−1
at the points (0, ±1).
11. Lagrange multipliers
1. Find the extrema of the function f (x, y, z) = xyz subject to the constraint x2 +y 2 +z 2 =
1.
Solution: Any point P which is a max or min for f subject to the constraint g(x, y, z) = 0
must satisfy
hyz, xz, xyi = ∇f = λ∇g = λh2x, 2y, 2zi
The first equation is 2λx = yz. If x = 0 then exactly one of y, z is also zero (they cannot
both be zero because the point (0, 0, 0) does not lie on the unit sphere), so we get the
yz
.
candidate points (0, ±1, 0) and (0, 0, ±1). Otherwise λ = 2x
Similarly, from the equations 2λy = xz and 2λz = xy, we get the additional candidate
xy
points (±1, 0, 0), or else λ = xz
2y and λ = 2z .
Hence assuming that x, y, z 6= 0, we are left with the equations
yz
xz
xy
=
=
2x
2y
2z
Multiplying all three expressions by 2xyz leads to
y 2 z 2 = x2 z 2 = x2 y 2
The first equation implies that y = ±x, and the second implies that y = ±z. Plugging
these points into the constraint equation leads to 3x2 = 1, so x = ± √13 .
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JOE HUGHES
Therefore the candidate points are (± √13 , ± √13 , ± √13 ), where the signs in each coordinate
can vary independently.
At the six points (±1, 0, 0), (0, ±1, 0) and (0, 0, ±1) we have f (x, y, z) = 0.
At the eight points of the form (± √13 , ± √13 , ± √13 ), we have f (x, y, z) =
1
√
3 3
if the number
1
√
of negative signs is even, and f (x, y, z) = − 3 3 if the number of negative signs is odd. So
each of these eight points is a global max or min.
2. Find the point on the curve y = x +
origin.
1
x
in the first quadrant which is closest to the
Solution: It is generally easier to work with distance squared rather than distance, so we
need to minimize f (x, y) = x2 + y 2 subject to the constraint g(x, y) = 0, where g(x, y) =
x + x1 − y.
Using Lagrange multipliers, we get
h2x, 2yi = ∇f = λ∇g = λh1 − x−2 , −1i
The second component implies that λ = −2y, and the first component implies that 2x =
λ(1 − x−2 ), or
λ=
2x
2x3
=
1 − x−2
x2 − 1
Note that x 6= 1, because then 2x = λ(1 − x−2 ) would become 2 = 0, which is a contradiction.
Setting the two expressions for λ equal shows that
y=
x3
1 − x2
and plugging this into the constraint equation y = x + x1 , we get
x2 + 1
x3
=
x
1 − x2
1
so 1 − x4 = x4 . Therefore x = 2− 4 , and y = x +
1
x
1
1
= 2 4 + 2− 4 .
Finally, we need to check that this point is actually a minimum. This is best seen just by
plotting the curve y = x + x1 in the first quadrant:
MATH 32A: FINAL REVIEW
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As the plot suggests, the distance x2 + y 2 goes to ∞ as x → 0, ∞, and there is a unique
minimum.
For a constraint like y = x+ x1 , using Lagrange multipliers is actually unnecessary: we could
instead plug y = x + x1 into f (x, y) and then use single-variable calculus. The advantage
of this approach is that you then have the first derivative test and second derivative test
for functions of one variable at your disposal. On the other hand, not every constraint
equation allows us to solve for one variable in terms of the other, so it’s good to get some
practice with Lagrange multipliers.
12. Figures
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JOE HUGHES
Figure 1
Figure 2
MATH 32A: FINAL REVIEW
Figure 3
Figure 4
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JOE HUGHES
Figure 5
Figure 6
MATH 32A: FINAL REVIEW
Figure 7
Figure 8
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