HW 4 Due 7/21
1. Compute second order Taylor Expansion of :
(a) f (x, y) = ex+y at (0, 0)
All of the partials are equal to f so at (0, 0) are 1, so the expansion is:
f (x, y) = 1 + x + y + 12 (x2 + y 2 + 2xy) + R2 (x, y).
(b) f (x, y) = e−x
2 −y 2
cos(xy) at (0, 0).
f (0, 0) = 1
fx (0, 0) = −2xe−x
fxx (0, 0) = −2e
2 −y 2
−x2 −y 2
cos(xy) − ye−x
2 −y 2
sin(xy)|(0,0) = 0
−x2 −y 2
∂
e
cos(xy) − 2x( ∂x
∂ −x
fxy (0, 0) = −2x( ∂y
e
2 −y 2
cos(xy)) − e−x
2 −y 2
∂ −x
cos(xy)) − y( ∂x
e
2 −y 2
2 −y 2
∂ −x
sin(xy) − y( ∂y
e
sin(xy))|(0,0) = −2
sin(xy))|(0,0) = 0
Since the function is unchanged by changing x and y, fy (0, 0) = 0 and fyy (0, 0) = −2.
So the expansion is f (x, y) = 1 − x2 − y 2 + R2 (x, y).
(c) f (x, y) = x cos(πy) − y sin(πx) at (1, 2).
f (1, 2) = 1
fx (1, 2) = cos(πy) − πy cos(πx)|(1,2) = 1 + 2π
fy (1, 2) = −πx sin(πy) − sin(πx)|(1,2) = 0
fxx (1, 2) = π 2 y sin(πx)|(1,2) = 0
fyy (1, 2) = −π 2 x cos(πy)|(1,2) = −π 2
fxy (1, 2) = −π sin(πy) − π cos(πx)|(1,2) = π
so f (1 + h1 , 2 + h2 ) = 1 + (1 + 2π)h1 + 21 (−π 2 h22 + 2πh1 h2 ) + R2 (h1 , h2 ) or
f (x, y) = 1 + (1 + 2π)(x − 1) + 21 (−π 2 (y − 2)2 + 2π(x − 1)(y − 2)) + R2 (x, y).
(d) f (x, y, z) = y 2 x + cos z − ezx at (0, 0, 0).
f (0, 0, 0) = 0
fx (0, 0, 0) = y 2 − zezx |0,0,0 = 0
fy (0, 0, 0) = 2yx|0,0,0 = 0
fz (0, 0, 0) = − sin z − xezx |0,0,0 = 0
fxx = −z 2 ezx |0,0,0 = 0
fxy (0, 0, 0) = 2y|0,0,0 = 0
fxz (0, 0, 0) = −exz − zxexz |0,0,0 = −1
fyy (0, 0, 0) = 2x|0,0,0 = 0
fyz (0, 0, 0) = 0
fzz (0, 0, 0) = − cos z − x2 ezx |0,0,0 = −1.
So f (x, y, z) = −xz − 21 z 2 + R2 (x, y, z)
1
2. Let L(x, y) = ax + by + c, Q(x, y) = ax2 + bxy + cy 2 + d.
(a) Compute the degree 2 Taylor expansion of each at (0, 0), and at (1, 1).
A Taylor expansion is the best polynomial approximation, these are already polynomials
so their Taylor expansions are just the same function.
(b) What about the degree n Taylor expansion at (0, 0)?
Same reason in (a) the n degree Taylor expansions are just the same function.
3. Approximate sin(π/6 + 0.1) cos(π/3 + 0.02) with a second order Taylor expansion.
Expand f (x, y) = sin(x) cos(y) around x = π/6, y = π/3:
√
√
1
and
f (π/6+h1 , π/3+h2 ) ≈ 41 +√ 43 h1 − 43 h2 + 12 (− 14 h21 − 32 h1 h2 − 14 h22 ), then plug in h1 = 10
3
309
1
h2 = 50 ... something like 50 + 1250 = 0.281841..., pretty close to actual value: 0.281807...
(This is probably a better problem if h1 = h2 = 0.1)
4. Classify the critical points of
(a) f (x, y) = e1+x
∇f = e1+x
2 +y 2
2 +y 2
(2x, 2y) = ~0 ⇒ x = y = 0.
fxx (0, 0) = 2e, fxy (0, 0) = 0, fyy (0, 0) = 2e, so it is a local minimum.
(b) f (x, y) = y + x sin y
∇f = (sin y, 1 + x cos y) = ~0, then sin y = 0 ⇒ cos y = ±1 and 0 = 1 + x cos y = 1 ± x ⇒
x = ∓1.
So the critical points are (−1, 2kπ), (1, 2kπ + π) for k ∈ Z.
0 ±1
fxx = 0, fxy = cos y, fyy = −x sin y so at critical points, the Hessian is
and all
±1 0
the critical points are saddles.
(c) f (x, y) = (x − y)(xy − 1)
∇f = (xy − 1 + (x − y)y, 1 − xy + (x − y)x) = ~0 ⇒ xy − 1 + (x − y)y = 0 and
1 − xy + (x − y)x = 0. Add up these two equations to get (x − y)(x + y) = 0 ⇒ x = y or
x = −y.
If x = y then plugging into 1 − xy + (x − y)x = 0 gives x2 = 1 ⇒ x = ±1 = y.
If x = −y then plugging into 1 − xy + (x − y)x = 0 gives 3x2 = −1 which has no real
solutions, so the only critical points are (±1, ±1).
2y
2(x − y)
±2 0
The Hessian is
|(±1,±1) =
so both critical points are sad2(x − y)
−2x
0 ∓2
dles.
5. Let f (x, y, z) = x2 + y 2 + z 2 + kyz (k is a constant). Describe the critical points and their
type as k changes.
2
∇f = (2x, 2y + kz, 2z + ky) = ~0 ⇒ x = 0, 2y + kz = 0, 2z + ky = 0 ⇒ −2/kz = y =
−k/2z ⇒ 4z = k 2 z ⇒ k = ±2 or z = 0 (and y = 0).
So if k 6= ±2 the critical point is (0, 0, 0).
If k = ±2 then the critical points are all

2

The Hessian at any point looks like 0
0
2, 4, 2(4 − k 2 ).
points on the line x = 0, y = ∓z.

0 0
2 k , so the determinants of the minors are
k 2
So, for |k| > 2 (0,0,0) is a saddle.
For |k| < 2, (0,0,0) is a local min.
If k = ±2 the test is inconclusive, but in this case the function has the form x2 +(y±z)2 ≥ 0
so that any point on the line of critical points is a local min.
6. Find the greatest perimeter of a rectangle that is inscribed inside the ellipse x2 /a2 +
y 2 /b2 = 1.
We aim to extremize f (x, y) = x + y subject to x2 /a2 + y 2 /b2 = 1 (and x, y > 0).
Solve (1, 1)√= λ(2x/a2 , 2y/b2 ) ⇒ λ 6= 0 and x = a2 /2λ, y = b2 /2λ ⇒ a2 /2λ2 + b2 /4λ2 =
2
2
1 ⇒ λ = ± a2 + b2 /2 ⇒ x = √aa2 +b2 , y = √ab2 +b2 .
√
So the maximum perimeter is 4x + 4y = 4 a2 + b2 .
7. Find the distance from the surface x + y + z 2 = 1 to the origin (0,0,0).
Distance from (x, y, z) to (0,0,0) squared is d2 = x2 + y 2 + z 2 , so extremize d2 subject to
the constraint x + y + z 2 = 1:
(2x, 2y, 2z) = λ(1, 1, 2z) ⇒ 2x = λ, 2y = λ, 2z = λ2z, x + y + z 2 = 1.
The first two say x = y. The third says z = 0 or λ = 1.
If z = 0 then substituting into x + y + z 2 = 1 gives 2x = 1 ⇒ x = y = 21 .
If λ = 1 then substituting in 2x = λy gives x = y =
1
2
⇒ z = 0.
So the closest point is ( 12 , 12 , 0) and the distance is:
√
|( 21 , 12 , 0)| = 1/ 2.
8. A curve c is defined implicitly as satisfying x2 + y 2 = 1 and x2 − xy + y 2 − z 2 = 1. Find
the closest point(s) on the curve to the origin (0, 0, 0).
Since the curve lies on a cylinder, the closest point(s) occur when z = 0 (if there are such
points otherwise they are where |z| is smallest).
Substituting x2 + y 2 = 1 and z = 0 into x2 − xy + y 2 − z 2 = 1 gives xy = 0 i.e. the x and
y axes. But the points are also on the cylinder, so the closest points are (±1, 0), (0, ±1).
3
9. Find and classify the critical point(s) of f (x, y) = x2 + y 2 (x + 1)3 . Are there global
extrema?
Critical points are where 2x + 3y 2 (x + 1)2 = 0 and 2y(x + 1)3 = 0. The second gives 2
cases y = 0 or x = −1. If x = −1 then the first gives −2 = 0 so we must have y = 0, now
the first gives 2x = 0 ⇒ x = 0. So there is only one critical point: (0, 0).
2 0
The Hessian at (0, 0) looks like:
so our only critical point is a local min.
0 2
It seems like since this is the only critical point it should also be a global min however
this is false! In higher dimensions the graph somehow has room to switch over without
gaining another critical point.
The value at (0, 0) is f (0, 0) = 0. But restricted to x = −2 say, the function has the form
4 − y 2 which goes to −∞ as y → ∞, so there is no global min. Also we can see it has no
global max since it goes to infinity along y = 0 as x → ∞ so there is no global max.
10. Find the global extrema (max/min values) of f (x, y) = sin x + cos y.
| sin x|, | cos x| ≤ 1 ⇒ |f (x, y)| ≤ 2 so the global max is 2 and the global min is −2.
11. Find the global extrema (max/min values) of f (x, y) = x2 + xy + y 2 on x2 + y 2 ≤ 1.
The critical point is where 2x + y = 0, 2y + x = 0 ⇒ x = y = 0.
On the boundary Lagrange says to solve: 2x + y = 2λx, 2y + x = 2λy. Subtracting gives
x − y = 2λ(x − y) ⇒ x = y or λ = 21 . If λ = 12 then 2x + y = x ⇒ x = −y. In both cases
√
plugging into x2 + y 2 = 1 gives x = ±1/ 2.
√
√
√
√
So the critical points on the boundary are (±1/ 2, ±1/ 2), (±1/ 2, ∓1/ 2).
Compare the values at all of these candidates:
√
√
√
√
f (0, 0) = 0, f (±1/ 2, ±1/ 2) = 1 + 1/2 = 3/2, f (±1/ 2, ∓1/ 2) = 1 − 1/2 = 1/2.
So the global max is 3/2 and the global min is 0.
12. Find three numbers whose product is 27 and whose sum is maximal.
Extremize x + y + z subject to xyz = 27 gives
(1, 1, 1) = λ(yz, xz, xy) ⇒ yz = xz = xy ⇒ x = y = z = 3.
(This should read minimal)
13. Find three numbers whose sum is 27 and whose product is maximal.
Extremize xyz subject to x + y + z = 27 gives
(yz, xz, xy) = λ(1, 1, 1) ⇒ yz = xz = xy ⇒ x = y = z = 9.
(These last 2 should also say for x, y, z > 0)
14. Find and classify the extrema of f (x, y, z) = x2 + y 2 + z 2 on z ≥ 2 + x2 + y 2 .
f has only one critical point (0, 0, 0) which does not lie in the region z ≥ 2 + x2 + y 2 ,
hence the extrema must lie on the boundary z = 2 + x2 + y 2 i.e. 0 = 2 + x2 + y 2 − z.
4
Lagrange says solve (2x, 2y, 2z) = λ(2x, 2y, −1) ⇒ x = 0 = y or λ = 1. If λ = 1 then
z = −1/2 and 0 = 2 + 1/2 + x2 + y 2 , imposssible with real numbers, hence x = y = 0 and
so z = 2.
Since f is the distance from the origin and this point is the closest on the solid parabaloid
z ≥ 2 + x2 + y 2 this point (0, 0, 2) is where f achieves it’s minimum value of 4 (restricted
to z ≥ 2 + x2 + y 2 ).
15. Find extrema of cos(x2 − y 2 ) subject to x2 + y 2 = 1.
We can use Lagrange, to get x(λ + sin(x2 − y 2 )) = 0 and y(λ − sin(x2 − y 2 )) = 0 and
x2 + y 2 = 1.
One case for the first is x = 0 ⇒ y = ±1 and λ = − sin(1).
One case for the second is y = 0 ⇒ x = ±1 and λ = − sin(1).
So we have critical points (0, ±1) and (±1, 0).
The remaining cases is when x, y are both not 0. Then we can cancel out x, y and
√ subtract
the first 2 eqns to get sin(x2 − y 2 ) = 0 ⇒ x2 − y 2 = 0 ⇒ x = ±y ⇒ x = ±1/ 2.
√
√
√
√
So we have critical points (±1/ 2, ±1/ 2) and (±1/ 2, ∓1/ 2).
cos(x2 − y 2 ) takes the √
values cos(1)
= cos(−1)
√
√ and cos(0)
√ at these critical points, so the
max is 1 (at the (±1/ 2, ±1/ 2) and (±1/ 2, ∓1/ 2) critical points) and the min is
cos(1) (at the (0, ±1) and (±1, 0) critical points).
16. Compute the curvature and torsion of c(t) = (t, t2 , t3 ).
First compute c0 (t) = (1, 2t, 3t2 ) and c00 (t) = (0, 2, 6t) and c000 (t) = (0, 0, 6).
So c0 (t) × c00 (t) = (6t2 , −6t, 2).
Then κ(t) =
And τ (t) =
|c0 (t)×c00 (t)|
|c0 (t)|3
√
=
36t4 +36t2 +4
.
(1+4t2 +9t4 )3/2
det(c0 (t),c00 (t),c000 (t))
|c0 (t)×c00 (t)|2
=
12
36t4 +36t2 +4
=
3
9t4 +9t2 +1
17. Compute the arc length of
√
(a) c(t) = (cos t, sin t, 3t), 0 ≤ t ≤ 2π.
√
c0 (t) = (− sin t, cos t, 3) so the length is:
R 2π 0
R 2π √ 2
R
2 t + 3dt = 2π 2dt = 4π.
|c
(t)|dt
=
sin
t
+
cos
0
0
0
√
t2
(b) c(t) = ( 2 , 2t, log t), 1 ≤ t ≤ 2.
√
c0 (t) = (t, 2, 1/t), so the length is:
R2 0
R2p
R2p
R2
|c (t)|dt = 1 t2 + 2 + 1/t2 dt = 1 (t + 1/t)2 dt = 1 t + 1/tdt = t2 /2 + log t|21 =
1
2 + log 2 − 1/2 = 3/2 + log 2.
18. Let F~ = (F1 , F2 , F3 ) : R3 → R3 be any C 2 vector field and f : R3 → R be any C 2 function.
(a) Verify that ∇ × (∇f ) = ~0 (curl of the gradient is 0).
5
î ĵ
∂ ∂
∇ × (∇f ) = ∂x
∂y
∂f ∂f
∂x ∂y
are equal (f is C 2 ).
k̂ ∂ ∂2f
∂2f
∂2f
∂2f
∂2f
∂2f
~
∂z = ( ∂y∂z − ∂z∂y , ∂z∂x − ∂x∂z , ∂x∂y − ∂y∂x ) = 0 since mixed partials
∂f
∂z
(b) Verify that ∇ · (∇ × F~ ) = 0 (divergence of the curl is 0).
∇ × F~ = ( ∂F3 − ∂F2 , ∂F1 − ∂F3 , ∂F2 − ∂F1 ), so
∂y
∇ · (∇ × F~ ) =
∂ 2 F2
∂z∂x
−
∂ 2 F2
∂x∂z
∂z
∂ ∂F3
(
∂x ∂y
∂z
−
∂x
∂x
∂y
∂F2
∂ ∂F1
) + ∂y
( ∂z
∂z
∂ ∂F2
3
1
− ∂F
) + ∂z
( ∂x − ∂F
)=
∂x
∂y
∂ 2 F3
∂x∂y
2
2
2
∂ F3
∂ F1
∂ F1
− ∂y∂x
+ ∂y∂z
− ∂z∂y
+
= 0 since mixed partials are equal.
19. Sketch the vector field F~ (x, y, z) = (x, y, z) around ~0. Compute it’s curl and divergence.
~ for some vector field G?
~
Could F~ = ∇f for some function f ? Could F~ = ∇ × G
Same questions for F~ (x, y, z) = (−y, x, 0).
1st one (F~ (x, y, z) = (x, y, z)):
it looks like:
∇ · (x, y, z) =
∂
x
∂x
+
∂
y
∂y
∂
∇ × (x, y, z) = ( ∂y
z−
+
∂
z
∂z
∂
∂
y, ∂z
x
∂z
= 3 and
−
∂
∂
z, ∂x
y
∂x
−
∂
x)
∂y
= ~0.
~ for any vector field G
~ (by 8b), but we could have F~ = ∇f for some f in
So F~ 6= ∇ × G
fact, f (x, y, z) = 12 (x2 + y 2 + z 2 ) does the trick.
For the 2nd one (F~ (x, y, z) = (−y, x, 0)):
it looks like:
∂
∇ · (−y, x, 0) = − ∂x
y+
∂
x
∂y
∂
∇ × (−y, x, 0) = ( ∂y
0−
∂
∂
x, − ∂z
y
∂z
+
∂
0
∂z
= 0 and
−
∂
∂
0, ∂x
x
∂x
+
∂
y)
∂y
= (0, 0, 2).
~ for some vector
So F~ 6= ∇f for any function f (by 8a), but we could have F~ = ∇ × G
~ The vector field G
~ = (0, 0, xy) works.
field G.
One of the many theorem’s called Poincare’s theorem says that for vector fields on R3 (it
can be generalized to Rn ) whenever there can be a potiential then there will be one (that
is if ∇ × F~ = ~0 then F~ = ∇f for some f , f is called the potential, or if ∇ · F~ = 0 then
~
F~ = ∇ × G for some G)...
you will learn about this stuff again in 23b...
6