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Fall 2006
Group G1
University of Illinois at Urbana-Champaign
Math 380
Graded Homework III
Due Friday, September 29.
2
1. Find
√ the directional derivative of the mapping f dened by f (x, y) = xy + ln(x + 1) in the direction of
3 1
u=(
, ).
2 2
2x
Correction. One has ∇f (x, y) = (y +
, x) ; since u is a unit vector, the derivative of f in the direction
2
x +1
of u is simply ∇f.u, in other words
√
√
x
3
3x
y+ 2
+ .
∇u f (x, y) =
2
x +1 2
2. Given unit vectors u = (ux , uy ) and v = (vx , vy ), and a function z = z(x, y) with continuous second-order
partial derivatives, nd a formula for the mixed second order directional derivative ∇u ∇v z .
∂z
∂z
By denition, we have ∇v z = ∇z.v = vx + vy .
∂x
∂y
Applying the same denition, we obtain ∇u ∇v z = ∇(
∇u ∇ v z = (
∂z
∂z
vx +
vy ).u. Therefore,
∂x
∂y
∂2z
∂2z
∂2z
∂2z
∂2z
∂2z
∂2z
∂2z
vx +
vy ,
vx + 2 vy ).(ux , uy ) =
vx ux +
vy ux +
vx uy + 2 vy uy .
2
2
∂x
∂x∂y
∂y∂x
∂y
∂x
∂x∂y
∂y∂x
∂y
3. Show that y is dened implicitly as a function of x in the neighborhood of the point P in the following
equations :
π
• x cos(xy) = 0, P = (1, ) ;
2
• xy + log(xy) = 1, P = (1, 1) (log stands for the natural logarithm).
Use implicit dierentiation to compute y 0 (1) and y 00 (1) in the rst case, and y 0 (1) in the second case.
2
Correction. The implicit function theorem tells us that, if F : R
→ R and λ ∈ R, then the equation
∂F
F (x, y) = λ denes implicitly y as a function of x in a neighborhood of a solution P as soon as
is
∂y
continuous and dierent from 0 at P .
∂F
∂F
π
• In the rst case we have F (x, y) = x cos(xy), so that
= −x sin(xy), which implies that
(1, ) = −1.
∂y
∂y
2
Therefore the hypothesis of the theorem is satised, and y is dened implicitly as a function of x near P .
Implicit dierentiation yields (cos(xy) − y sin(xy))dx − x sin(xy)dy =0. Since we are looking for y 0 (1), we may
π
notice that close to (1, ) the equation x cos(xy) = 0 means that cos(xy) = 0, so that implicit dierentiation
2
simply gives −y sin(xy)dx − x sin(xy)dy = 0, or
π
2
dy
y
y0
y
y
= y 0 (x) = − . Thus y 00 (x) = − + 2 = 2 2 close to P ,
dx
x
x
x
x
and this gives y 0 (1) = − , and y 00 (1) = π . (Notice that, in that case, one can actually solve the equation : in
π
a neighborhood of P , one simply has y =
; dierentiating this equality gives for y 0 (1) and y 00 (1) the values
2x
we obtained earlier).
∂F
1
∂F
• In this case F (x, y) = xy + log(xy), so
= x + . Therefore,
(1, 1) = 2 6= 0, so the equation denes
∂y
y
∂y
implicitly y as a function of x near P .
Implicit dierentiation gives this time y 0 (x) = −
y+
x+
1
x
1
y
, or y 0 (x) =
y 2 (x)x + y(x)
. This shows that y 0 (1) = 1.
y(x)x2 + x
(Bonus, just to show how atrocious iterated implicit dierentiations can get) If we were asked to compute also
y 00 (1), then we would have to take a deep breath, then write that
y 00 =
(2yy 0 x + y 2 + y 0 )(yx2 + x) − (y 2 x + y)(x2 y 0 + 2xy + 1)
(2 + 1 + 1)(1 + 1) − (1 + 1)(1 + 2 + 1)
, so y 00 (1) =
=0.
(yx2 + x)2
(1 + 1)2
4. Show that 2xy − z + 2xz 3 = 5 can be solved implicitly for z as a function of x near (1, 2, 1).
∂2z
Compute the rst-order partial derivatives of z at (1, 2), as well as 2 (1, 2).
∂y
Correction. We have this time an equation of the type F (x, y, z) = λ ; for this to dene z implicitly near
∂F
be continuous and dierent from 0 at that point. A direct computation yields
(1, 2, 1) , it is enough that
∂z
∂F
∂F
that
= −1 + 6xz 2 , so that
(1, 2, 1) = 5 6= 0. Therefore the equation denes z implicitly as a function
∂z
∂z
of (x, y) near (1, 2, 1). Implicit dierentiation leads to 2ydx + 2xdy − dz + 2z 3 dx + 6xz 2 dz = 0, so that
(2y + 2z 3 )dx + 2xdy
. This enables us to obtain
(2y + 2z 3 )dx + 2xdy = (1 − 6xz 2 )dz , or dz =
1 − 6xz 2
∂z
2y + 2z 3
=
,
∂x
1 − 6xz 2
From (*) we obtain that
∂z
2x
=
(∗)
∂y
1 − 6xz 2
6
∂z
2
∂z
(1, 2) = − and
(1, 2) = − . We also get
∂x
5
∂y
5
∂z
12xz ∂y
∂2z
= 2x
.
∂y 2
(1 − 6xz 2 )2
Given the values computed above, this yields
12(− 25 )
48
∂2z
(1,
2)
=
2
=−
.
∂y 2
25
125
5. Determine whether the function f : R3 → R3 dened by f (x, y, z) = (ex cos(y) + z, x sin(y) sin(z), xz cos(y)),
π
admits a dierentiable inverse g near (1, , π). If so, give the value of the Jacobian determinant of g at the
2
π
point f (1, , π) = (π, 0, 0).
2
The consequence of the implicit function theorem seen in class (called the local inversion theorem)
π
implies that we only need to check whether the Jacobian determinant of f vanishes or not at (1, , π). The
2
Jacobian matrix of f at (x, y, z) is
Correction.

ex cos(y)
−ex sin(y)
1
Jf(x,y,z) = sin(y) sin(z) x cos(y) sin(z) −x sin(y) cos(z)
z cos(y)
−xz sin(y)
x cos(y)


0 −e 1
This implies that Jf(1, π2 ,π) = 0 0 1. The determinant of this matrix is 0, which proves that f does not
0 0 0
π
admit a dierentiable inverse g near f (1, , π) = (π, 0, 0).
2
