1. Let w = f (x, y) where x = u + v and y = uv.
a) Find
∂w
∂u
b) Find
∂2 w
∂u∂v .
and
∂w
∂v
in terms of partial derivatives of w with respect to x and
y.
a) First use the chain rule for 1 variable namely
∂w ∂x ∂w ∂y
∂w ∂w
∂w
=
+
=
+
v
∂u
∂x ∂u
∂y ∂u
∂x
∂y
(1)
∂w ∂x ∂w ∂y
∂w ∂w
∂w
=
+
=
+
u
∂v
∂x ∂v
∂y ∂v
∂x
∂y
(2)
and also
b) For this part what we need to do is realize that
must differentiate each object carefully. Start with
this is h(u, v) =
∂w
∂x (x, y).
Now we must take
∂h
∂u
∂2w
∂u∂v
∂w
∂x .
=
∂ ∂w
∂u ∂v
thus we
We know that really
and use the chain rule for this.
Using the chain rule this equals
∂ ∂w
∂ 2 w ∂x
∂ 2 w ∂y
∂h
=
=
+
.
∂u
∂u ∂x
∂x2 ∂u ∂y∂x ∂u
(3)
Then evaluating the first partials we get that this equals
∂ 2w
∂2w
∂ ∂w
+
=
v
2
∂u ∂x
∂x
∂y∂x
Now if we take the right hand side of (2) if we want to take
(4)
∂
∂u
∂w
∂y u
then we
must use both the product rule and the chain rule.
Using the product rule we get that
∂w
∂ ∂w ∂ ∂w u+
u =
∂u ∂y
∂u ∂y
∂y
(5)
Now to get the last ingredient in our recipe for chain rule fun we must use the
1
chain rule again, This time on the function g(u, v) =
∂w
∂y (x, y).
This gives us
the following
∂g
∂ ∂w
∂ 2 w ∂x ∂ 2 w ∂y
=
=
+
∂u
∂u ∂y
∂x∂y ∂u
∂y 2 ∂u
(6)
then adding in the first partials we get that this equals
∂2w
∂2w
v
+
∂x∂y
∂y 2
(7)
Taking all of the ingredients together and mixing gives us that
∂2w
∂w
∂2w
∂2w
∂2w
∂2w
+
uv +
=
v
+
u
+
2
2
∂u∂v
∂x
∂y∂x
∂x∂y
∂y
∂y
If conditions on f are nice enough we have that
∂2 w
∂x∂y
=
∂2 w
∂y∂x
(8)
and thus that
∂2w
∂w
∂2w
∂2w
∂2w
+
uv +
=
(u + v) +
2
∂u∂v
∂x
∂x∂y
∂y 2
∂y
(9)
Now let us do some double integrals
Problem 26 asks us to calculate the mass and the center of mass for the
lamina bounded by x = y 2 and x = 4 with the given density ρ(x, y) = y + 3.
We are in the following region enclosed in red
2
y
2
1
x
1
2
3
4
5
6
7
8
9
−1
−2
Let us take M the center of mass given on this region by
M=
ZZ
ρ(x, y)dA =
R
4
Z
Z
√
x
√
− x
0
(y + 3)dxdy =
Z
4
0
√
y2
x
+ 3y|−√x
2
(10)
Taking (10) we get that this equals
Z 4
Z 4
√
√
√ 3
x
x
6 xdx = 4x 2 |40 = 32.
( + 3 x) − ( − 3 x) dx =
M=
2
2
0
0
(11)
Then we have that for Mx we have that
Mx =
ZZ
R
yρ(x, y)dA =
Z
4
0
Z
√
x
√
− x
(y 2 + 3y)dydx =
Z 4 3
3 √x y
+ y 2 |−√x dx (12)
3 2
0
reducing (12) we find that
Mx =
Z
0
4
4 5 4
128
2 3
x 2 dx =
x 2 |0 =
3
15
15
3
(13)
For My we then similarly compute that
My =
ZZ
xρ(x, y)dA =
R
Z
0
4
Z
√
x
√
− x
(xy + 3x)dydx =
Z
4
x
0
√
y2
x
+ 3xy|−√x dx (14)
2
expanding (14) we find that it equals
My =
Z 4
0
Z
(x + 3x ) − (x − 3x ) dx =
2
3
2
2
3
2
0
4
3
6x 2 dx =
12 5 4
384
x 2 |0 =
5
5
M
(15)
x
Putting (11), (13) and (15) we have that the center of mass is ( My , M
M ) =
4
( 12
5 , 5 ).
Hopefully we all understand lamina and double integrals better now.
4