Solution to HW 4, Ma 1c Prac 2016
Remark : every function appearing in this homework set is sufficiently nice–
at least C 3 following the jargon from the textbook–we can apply all kinds of
theorems from the textbook without worrying too much about the quality of
functions.
Problem 3.4.4: Find the extrema of f (x, y) = x − y subject to the constraint
g(x, y) = x2 − y 2 = 2.
Solution: Note that ∇g(x, y) = (2x, −2y), which is nonsingular everywhere
away from the origin, so g(x, y) = x2 −y 2 = 2 is a sufficiently well-behaved levelcurve. Now, we wish to find all points (x, y) ∈ C := {x2 − y 2 = 2} which are
extrema for f , and to classify those points. Setting ∇f (x, y) = λ∇g(x, y) for λ ∈
R and (x, y) ∈ C, we see that as ∇f (x, y) = (1, −1) and ∇g(x, y) = (2x, −2y),
we obtain alongside our constraint on the curve the system of equations:
2λx = 1
−2λy = −1
2
x − y 2 = 2.
Now, noting that the first two equations above imply λx = 21 and λy = 12 ,
one must have λ 6= 0. Solving for x and y, we obtain that any extremum of
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f (x, y) = x−y along x2 −y 2 = 2 must have x = 2λ
= y, and hence be of the form
1
1
1 2
1 2
(x, y) = ( 2λ , 2λ ). However, any such point must have ( 2λ
) − ( 2λ
) = 0 6= 2,
and consequently does not lie on our curve C. Thus, the function f (x, y) = x−y
does not have any extrema along the curve defined by x2 − y 2 = 2. This makes
sense since x2 − y 2 = 2 defines a hyperbola, which is noncompact (it is certainly
closed, but is not bounded).
Probldem 3.4.6: Find the extrema of f (x, y, z) = x + y + z subject to the
constraints g1 (x, y, z) = x2 − y 2 = 1 and g2 (x, y, z) = 2x + z = 1.
Solution: We wish to find the extrema of f along the curve
S = {(x, y, z) ∈ R3 |x2 − y 2 = 1 and 2x + z = 1}
= {(x, y, z) ∈ R3 |g1 (x, y, z) = 1 and g2 (x, y, z) = 1}
= {(x, y, z) ∈ R3 |x2 − y 2 = 1} ∩ {(x, y, z) ∈ R3 |2x + z = 1}
(which is clearly a curve, as the intersection of two transverse surfaces), for
which we will use the method of Lagrange multipliers on multiple constraints.
1
Thus, we will attempt to find a point (x, y, z) ∈ S satisfying the equation
∇f (x, y, z) = λ1 ∇g1 (x, y, z) + λ2 ∇g2 (x, y, z)
for some λ1 , λ2 ∈ R. Noting that ∇f (x, y, z) = (1, 1, 1), that ∇g1 (x, y, z) =
(2x, −2y, 0), and that ∇g2 (x, y, z) = (2, 0, 1), we may rewrite the above condition as
(1, 1, 1) = (2λ1 x + 2λ2 , −2λ1 y, λ2 ).
In total, we must solve the following system of five equations:
2λ1 x + 2λ2 = 1
−2λ1 y = 1
λ2 = 1
2
x − y2 = 1
2x + z = 1.
Since λ2 = 1, the first equation reduces to 2λ1 x+2 = 1, and hence to 2λ1 x = −1,
so as the second equation may be written as 2λ1 y = −1, we in fact have that
2λ1 x = 2λ1 y. Furthermore, the second equation immediately implies that λ1 6=
0. Thus, we may cancel 2λ1 from both sides to obtain that x = y. However,
this immediately implies that any extremum may not be on S, as it implies
that x2 − y 2 = x2 − x2 = 0 6= 1. Thus, f has no extrema when subject to the
constraints x2 − y 2 = 1 and 2x + z = 1 (i.e. f has no extrema when restricted
to S).
Problem 3.4.12: Use the method of lagrange multipliers to find the absolute
maximum and minimum values of f (x, y) = x2 + y 2 − x − y + 1 on the unit disk.
Solution: Let D ⊂ R2 denote the unit disk, and let DO denote its interior,
the open disk. To find all the extremal values of f on D, it suffices to check for
extrema on DO , and then check for extrema on ∂D = S 1 = {(x, y)|x2 + y 2 = 1}
using the method of lagrange multipliers. First, note that ∇f (x, y) = (2x −
1, 2y − 1), so, setting this equal to zero, we see that (2x − 1, 2y − 1) = 0 implies
that (x, y) = ( 21 , 12 ) is an extremum of f on D. Since there are no more possible
zeros of the equations 2x − 1 = 0 and 2y − 1 = 0, this is the only extremum on
DO .
Now, we turn our attention to S 1 . Noting that S 1 as we have defined it is
a smooth level curve of the function g(x, y) = x2 + y 2 , we may use the method
of Lagrange multipliers to find extrema. Note that ∇g(x, y) = (2x, 2y). Any
extrema (x, y) ∈ S 1 of f on S 1 must satisfy ∇f (x, y) = λ∇g(x, y). Thus,
rewriting this, we obtain
(2x − 1, 2y − 1) = (2λx, 2λy).
2
Thus, we obtain the following system of equations:
2x − 1 = 2λx
2y − 1 = 2λy
x2 + y 2 = 1.
Rearranging the first two equations, we get that 2(1 − λ)x = 1 = 2(1 − λ)y.
Thus, one must have that 1 − λ 6= 0, so we may cancel 2(1 − λ) from both sides.
From this we see that we must have x = y. Thus,
x2 + y 2 = 2x2 = 1,
so x = y = ± √12 . Thus, S 1 contributes two further possible extrema to f
evaluated on D, namely, ±( √12 , √12 ) ∈ D. Evaluating f at each of these points,
we get
1
1
1 1
1
1 1
f ( , ) = ( )2 + ( )2 − − + 1 =
2 2
2
2
2 2
2
√
1
1
1
1
1
1
f ( √ , √ ) = ( √ )2 + ( √ )2 − √ − √ + 1 = 2 − 2
2
2
2
2
2
2
√
1
1 2
1 2
1
1
1
f (− √ , − √ ) = (− √ ) + (− √ ) + √ + √ + 1 = 2 + 2,
2
2
2
2
2
2
so ( 21 , 12 ) is an absolute minimum of f on D, with value f ( 21 , 12 ) = 12 and
(− √12 , − √12 ) is an absolute maximum of f on D, with value f (− √12 , − √12 ) =
√
2 + 2.
Problem 3.4.23: Find the absolute maximum and minimum values of f (x, y, z) =
x + y − z on the unit ball B = {(x, y, z) ∈ R3 |x2 + y 2 + z 2 ≤ 1}.
Solution: Taking B O to be the open unit ball, we proceed analogously
to the previous exercise. Noting that ∇f (x, y, z) = (1, 1, −1), we see that the
equation ∇f (x, y, z) = 0 has no solutions on B O .
Thus, we turn our attention to ∂B = S 2 = {(x, y, z) ∈ R3 |x2 + y 2 + z 2 = 1}.
Clearly, S 2 as we have defined it is a smooth level set of g(x, y, z) = x2 + y 2 +
z 2 , and so we may apply the method of Lagrange multipliers. Any extremum
(x, y, z) ∈ S 2 for f must satisfy ∇f (x, y, z) = λ∇g(x, y, z) for some λ ∈ R.
Noting that ∇g(x, y, z) = (2x, 2y, 2z), we may rewrite the above condition as
(1, 1, −1) = (2λx, 2λy, 2λz),
which leaves us with the following system of equations:
2λx = 1
2λy = 1
2λz = −1
2
2
x + y + z 2 = 1.
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Now, the first three equations clearly imply both that 2λx = 2λy = −2λz = 1
and that λ 6= 0, so, cancelling by 2λ, we get that x = y = −z. Thus, one has
from the last equation that
x2 + y 2 + z 2 = x2 + x2 + (−x)2 = 3x2 = 1,
so one has that x = y = −z = ± √13 . Thus, the absolute extrema of f on D
are ( √13 , √13 , − √13 ) and (− √13 , − √13 , √13 ). Evaluating f at each of these points
yields
√
1
1
1
1
1
3
1
f ( √ , √ , − √ ) = √ + √ − (− √ ) = √ = 3
3
3
3
3
3
3
3
√
1
1
1
1
1
1
3
f (− √ , − √ , √ ) = (− √ ) + (− √ ) − √ = √ = − 3.
3
3
3
3
3
3
3
Thus, ( √13 , √13 , − √13 ) is an absolute maximum of f on D with value f ( √13 , √13 , − √13 ) =
√
3 and (− √13 , − √13 , √13 ) is an absolute minimum of f on D with value f (− √13 , − √13 , √13 ) =
√
− 3.
Problem 4.2.3: Find the arc length of the curve c(t) = (sin 3t, cos 3t, 2t3/2 ) on
the interval 0 ≤ t ≤ 1.
Solution: Recall that if we write the curve c(t) = (x(t), y(t), z(t)), the arc
length of c along the interval [a, b] is given by
Lba (c)
Z
b
0
Z
b
p
||c (t)||dt =
=
a
x0 (t)2 + y 0 (t)2 + z 0 (t)2 dt.
a
Now, given our particular choice of curve c(t) = (sin 3t, cos 3t, 2t3/2 ), we see that
√
c0 (t) = (3 cos 3t, −3 sin 3t, 3 t).
Thus, simplifying along the way, one may calculate the arc-length of c from 0
to 1 as
Z 1
Z 1q
Z 1
√
√
L10 (c) =
||c0 (t)||dt =
(3 cos 3t)2 + (−3 sin 3t)2 + (3 t)2 dt = 3
1 + tdt
0
0
0
√
= 2[(1 + t)3/2 ]10 = 2(23/2 − 1) = 4 2 − 2.
√ √
Problem 4.2.10: Compute the length of the curve c(t) = (log( t), 3t, 23 t2 )
for 1 ≤ t ≤ 2.
Solution: Recall that if we write the curve c(t) = (x(t), y(t), z(t)), the
length of c with a ≤ t ≤ b is given by
Lba (c)
Z
b
0
Z
||c (t)||dt =
=
a
b
p
a
4
x0 (t)2 + y 0 (t)2 + z 0 (t)2 dt.
√ √
Now, given our particular choice of curve c(t) = (log( t), 3t, 32 t2 ), we see that
c0 (t) = (
2t √
3, 3t).
,
Thus, simplifying along the way, one may calculate the arc-length of c from 1
to 2 as
Z 2r 2
Z 2
Z 1r
1
(6t + 1)2
2
0
2
L1 (c) =
+
3
+
9t
dt
=
dt
||c (t)||dt =
4t2
4t2
1
1
0
Z 2
1
1
1
1
(6t + )dt = [3t2 + log t]21 = (9 + log 2).
=
2 1
t
2
2
Problem 4.3.15: Show that the curve c(t) = (e2t , log |t|, 1t ), t 6= 0 is a flow-line
of the vector field F (x, y, z) = (2x, z, −z 2 ).
Solution: To show that a given curve c : R → Rn is a flowline to a given
vector field F : Rn → Rn , one need simply show that c satisfies the functional
equation c0 (t) = F (c(t)) for all t where c is defined and all c(t) where F is
defined.
In our case, the curve we are given is c(t) = (e2t , log |t|, 1t ), t 6= 0 and the
vector field we are given is F (x, y, z) = (2x, z, −z 2 ). From here, we may simply
carry out the following string of computations for any t 6= 0:
1
1
1
c0 (t) = (2e2t , , − 2 ) = F (e2t , log |t|, ) = F (c(t)).
t
t
t
Thus, one indeed has that c is a flow-line of F .
Problem 4.3.17: Show that the curve c(t) = (sin t, cos t, et ), is a flow-line
of the vector field F (x, y, z) = (y, −x, z).
Solution: To show that a given curve c : R → Rn is a flowline to a given
vector field F : Rn → Rn , one need simply show that c satisfies the functional
equation c0 (t) = F (c(t)) for all t where c is defined and all c(t) where F is
defined.
In our case, the curve we are given is c(t) = (sin t, cos t, et ) and the vector
field we are given is F (x, y, z) = (y, −x, z). From here, we may simply carry out
the following string of computations for any t ∈ R:
c0 (t) = (cos t, − sin t, et ) = F (sin t, cos t, et ) = F (c(t)).
Thus, one indeed has that c is a flow-line of F .
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