Math 212 Spring 2008: Solutions: HW #4
Instructor: S. Cautis
1. section 2.5, #4
h(x, y) =
e−2x−2y +e2xy
e−2x−2y −e2xy
∂h
∂x
(−2e−2x−2y + 2ye2xy )(e−2x−2y − e2xy ) − (e−2x−2y + e2xy )(−2e−2x−2y − 2ye2xy )
(e−2x−2y − e2xy )2
−2x−2y+2xy
−2x−2y+2xy
4e
+ 4ye
−2x−2y
(e
− e2xy )2
=
=
so
where the second line follows from the first after some simplifying.
On the other hand,
−4uv 2
2u(u2 − v 2 ) − (u2 + v 2 )(2u)
∂f
=
=
∂u
(u2 − v 2 )2
(u2 − v 2 )2
and
Also
∂u
∂x
∂f
4u2 v
2v(u2 − v 2 ) − (u2 + v 2 )(−2v)
= 2
=
2
2
2
∂v
(u − v )
(u − v 2 )2
= −e−x−y and
∂f ∂u ∂f ∂v
+
∂u ∂x ∂v ∂x
=
∂v
∂x
= yexy . Hence
4e−2x−2y+xy
−4e−x−y+2xy
−x−y
(−e
)
+
(yexy )
(e−2x−2y − e2xy )2
(e−2x−2y − e2xy )2
which upon closer inspection is the same as what we calculated above for
∂h
∂x .
2. section 2.5, #8
We use the chain rule to get:
∂f
∂f ∂x ∂f ∂y ∂f ∂z
=
+
+
∂ρ
∂x ∂ρ
∂y ∂ρ
∂z ∂ρ
Now
∂x
∂ρ
= cos θ sin φ and
∂y
∂ρ
= sin θ sin φ while
∂z
∂ρ
= cos φ. Thus we get
∂f
∂f
∂f
∂f
=
(cos θ sin φ) +
(sin θ sin φ) +
(cos φ)
∂ρ
∂x
∂y
∂z
The other two partials are computed similarly.
3. section 2.5, #10
We have
D(f ◦ g)(0, 0) =
=
=
=
Df (g(0, 0))Dg(0, 0)
eu−w
− sin(v + u) + cos(u + v + w)
0
− sin(v + u) + cos(u + v + w)


1 0
1
0
−1
0 0 
− sin(2) + cos(3) − sin(2) + cos(3) cos(3)
0 −1
1
1
− sin(2) + cos(3) − cos(3)
4. section 2.5, #15
We want to compute the tangent vector to f (c(t)) at f (c(0)) = f (0, 0) =
(1, 1). By the chain rule this is
x+y
′ e
ex+y
c1 (t)
′
Df (c(0, 0))c (0, 0) =
|
ex−y −ex−y (0,0) c′2 (t) t=0
1 1
1
=
1 −1
1
2
=
0
where c1 (t) and c2 (t) are the two coordinate functions of c(t).
5. section 2.5, #27
Rx
We write the function h(x) = 0 f (x, y)dy as the composition of two
Ry
g1
g2
functions R −→ R2 −→ R. Here g1 (x) = (x, x) and g2 (x, y) = 0 f (x, t)dt
where t is a dummy variable. Then we apply the chain rule to get
h′ (x) = Dg2 (g1 (x))Dg1 (x)
Now
Z y
∂f
∂g2 ∂g2
,
)=(
(x, t)dt, f (x, y))
∂x ∂y
0 ∂x
Rx
So that Dg2 (g1 (x)) = Dg2 (x, x) = ( 0 ∂f
∂x (x, t)dt, f (x, x)). Meanwhile
1
Dg1 (x) =
1
Dg2 = (
so that
Z
h (x) = (
x
′
0
Z x
∂f
f
1
(x, t)dt, f (x, x))
(x, t)dt + f (x, x)
=
1
∂x
∂x
0
which is exactly what we wanted.
−eu−w
cos(u + v + w)
(
1,
6. section 2.6, #2d
The directional derivative is
(y 2 + 3x2 y, 2xy + x3 )(4,−2)
√ √ √
1/√10
1/√10
= (−92, 48)
= 52/ 10
3/ 10
3/ 10
7. section 2.6, #4b
The normal to the surface is (−2x, −2y, 0) evaluated at (1, 2, 8) which is
(−2, −4, 0) so the tangent plane is
(x − 1, y − 2, z − 8) · (−2, −4, 0) = 0
or
x + 2y − 5 = 0
8. section 2.6, #8
A normal to the surface is (3x2 y 3 , 3x3 y 2 √
+ 1, −1) evaluated at (0, 0, 2) we
gives (0, 1, −1). This √
vector has length 2 so normalizing it we get the
unit normal vector 1/ 2(0, 1, −1).
9. section 2.6, #10
The level surfaces of the function f (x, y, z) = x2 + y 2 + z 2 are spheres
centred at the origin. It is intuitively clear that going away from the
origin in a straight line increases f fastest (in part because the picture is
so symmetric). This direction at a point (x0 , y0 , z0 ) is precisely the vector
(x0 , y0 , z0 ) which is (half of) ∇(f ) = (2x0 , 2y0 , 2z0 ). This verifies theorem
13.
Similarly, the tangent plane to the sphere x2 + y 2 + z 2 = constant at a
point (x0 , y0 , z0 ) is perpendicular to the radius (the line from the origin to
that point). Since this vector is (x0 , y0 , z0 ) it equals (half of) ∇f again.
This verifies theorem 14.