(1) Consider the function
f : R2 −→ R3
f (x, y) = (x2 − y 2 , 2xy, e2x+y ).
Find Df ( 21 , −1).
Solution: Recall that Df (x, y) is the matrix formed by placing the
gradient of x2 − y 2 in the first row, the gradient of 2xy in the second row,
and the gradient of e2x+y in the third row. So




2x
−2y 1 2
1
2x 
Df
, −1 =  2y
=  −2 1  .
2
2x+y
2x+y
2e
e
2 1
( 1 ,−1)
2
(2) Given f (x, y) = e2x+5y . Find the 2nd -order Taylor polynomial for f at
(0, 0).
Solution: The formula for the 2nd -order Taylor polynomial is
n X
n
n
X
X
1 ∂ 2f
∂f
(0, 0)hj +
(0, 0)hi hj .
f (x0 + h) ≈ f (x0 ) +
∂x
2
∂x
∂x
j
i
j
i=1 j=1
j=1
where n is the number of variables and ≈ means that the above polynomial
is approximately equal to f (x0 + h) (we are not including the remainder).
In this case, n = 2, x0 = (0, 0) and h = (x, y), where h1 = x and h2 = y.
Therefore,
∂f
∂f
1 ∂ 2f
∂ 2f
∂ 2f
2
2
f (x, y) ≈ f (0, 0)+ (0, 0)x+ (0, 0)y+
(0, 0)x + 2
(0, 0)xy +
(0, 0)y .
∂x
∂x
2 ∂x∂x
∂x∂y
∂y∂y
Now we just need to calculate the derivatives. We find that
∂f
∂f
(0, 0) = 2
(0, 0) = 5
f (0, 0) = 1
∂x
∂y
∂ 2f
∂ 2f
∂ 2f
(0, 0) = 4
(0, 0) = 10
(0, 0) = 25.
∂x∂x
∂x∂y
∂y∂y
Plugging into the formula above we obtain
25
f (x, y) ≈ 1 + 2x + 5y + 2x2 + y 2 + 10xy.
2
(3) Evaluate the iterated integral
Z 3 Z 1
y=0
sin(x2 )dxdy
x= 31 y
by changing the order of integration. Be sure to clearly sketch the region of
integration and indicate how you found the the new limits of integration.
2
Solution: The region that we are integrating over is the region below
the line y = 3x and above the x-axis, between x = 0 and x = 1. If we
integrate first with respect to y, our integral becomes
Z 1 Z 3x
Z 1
Z 1
3
3 3
2
2
sin(x )dydx =
3x sin(x )dx =
sin(u)du = − cos(1),
2 2
0
0
0
0 2
where we made the substitution u = x2 .
(4) Let D be the region bounded by the lines,
x + y = 0,
x + y = 2,
x − y = 0,
x − y = 2.
Evaluate
Z Z
(x + y)ex
2 −y 2
dxdy
D
using the change of variables u = x + y, v = x − y.
Solution: This change of variables is a little different than what we
have encountered so far. If the transformation was x = u + v, y = u − v,
we would view D as the image of D∗ and apply the change of variables
to integrate over D∗ . We would just need to determine D∗ as you did
for the homework problems. However, we are actually substituting u for
x + y and v for x − y, which is more in line with the u-substitution that
we use in single variable calculus. By making this substitution, instead of
integrating over D, we are now integrating over T (D), where T (x, y) =
(x + y, x − y) and T (D) consists of all (u, v) such that u = x + y and
v = x − y where (x, y) ∈ D. We still need to use the change of variables
formula to relate the integral over D to the integral over T (D). We observe
that D = T −1 (T (D)). Therefore, D is the image of T (D) using the inverse
transformation T −1 ! We have that
Z Z
Z Z
x2 −y 2
(x + y)e
dxdy =
ueuv |DT −1 |dvdu
D
T (D)
−1
where |DT | is the absolute value of the determinant of the total derivative of T −1 . To find T −1 , we observe that T is a linear transformation
given by
x
x+y
1 1
x
T
=
=
.
y
x−y
1 −1
y
To find T −1 , we simply find the inverse of the matrix
1 1
T =
.
1 −1
For a given matrix
a b
A=
,
c d
−1
A
1
=
ad − bc
d −b
−c a
.
3
Using this formula,
1/2 1/2
−1
T =
,
1/2 −1/2
1
and |DT −1 | = .
2
Written in nonmatrix form, we observe that T −1 (u, v) = ( 21 u+ 12 v, 12 u− 12 v).
Observe that if we plug in 21 u + 12 v for x and 21 u − 21 v for y in our original
integral, we get the same thing that we get by making the original substitution. This is what we would expect (why?). So our integral becomes
Z Z
1
ueuv dvdu.
2
T (D)
Now we just need to find T (D). To do this, we just find the image of the
vertices of D under T given that T is a linear transformation. The vertices
of D are (0, 0), (1, 1), (2, 0) and (1, −1). T maps these to (0, 0), (2, 0), (2, 2)
and (0, 2). So T (D) is the square where 0 ≤ u ≤ 2 and 0 ≤ v ≤ 2. So our
integral becomes
Z Z
Z
1 2 2 uv
1 2 2u
1
5
ue dvdu =
(e − 1) du = e4 −
2 0 0
2 0
4
4
Note that we could have also rewritten the transformation x = u + v,
y = u − v as a transformation involving x and y to get u = 21 x + 21 y and
v = 21 x − 21 y. This transformation is just the inverse of T and can be
written as
x
1/2 1/2
x
−1
T
=
.
y
1/2 −1/2
y
Once we rewrite our transformation, we can apply the change of variables
as we normally do to rewrite our integral over a domain D∗ such that
T −1 (D∗ ) = D. Given our arguments above, D∗ = T (D). So
Z Z
Z Z
1
1 2
1
1 2
1
1
1
1
x2 −y 2
(x + y)e
(( u + v) + ( u − v))e( 2 u+ 2 v) −( 2 u− 2 v) |DT −1 | dvdu
dxdy =
2
2
2
D
T (D) 2
Z 2Z 2
1
1
5
ueuv dvdu = e4 − .
=
2 0 0
4
4