Solutions

Calculus II
Math 116
Homework 7
Due Friday Mar. 28
(1) Evaluate the double integral
Z Z
f (x, y)dA
R
for the function f (x, y) and the region R.
(a) f (x, y) = 2 + x; R is the rectangle defined by 1 ≤ x ≤ 2 and 0 ≤ y ≤ 1.
Solution.
Z
2
1
Z
2
Z
[2y + xy]10 dx
(2 + x)dydx =
1
1
0
2
Z
=
1
2
(2 + x)dx = 2x + 12 x2 1
= (4 + 2) − (2 + 12 )
= 72 .
4
(b) f (x, y) =
xy
;
1+y 2
R is the rectangle defined by −2 ≤ x ≤ 2 and 0 ≤ y ≤ 1.
Solution.
Z
2
Z
−2
1
0
xy
dydx =
1 + y2
Z
2
−2
2
Z
=
−2
[ 12 x ln |1 + y 2 |]10 dx
ln(2)
xdx
2
( ln(2)
(2)2 )
4
=
= 0.
−
=
2
ln(2) 2 x
4
−2
ln(2)
( 4 (−2)2 )
4
(c) f (x, y) = 2x + 4y; R is bounded by x = 1, x = 3, y = 0, and y = x + 1.
Solution.
Z
3
Z
x+1
Z
3
[2xy + 2y 2 ]x+1
0 dx
(2x + 4y)dydx =
1
0
1
Z
=
=
=
3
Z
2
(2x(x + 1) + 2(x + 1) )dx =
1
( 34 (3)3 + 3(3)2
188
≈ 62.67.
3
3
2
(4x + 6x + 2)dx =
1
4 3
x
3
3
+ 3x + 2x
2
1
+ 2(3)) − ( 34 (1)3 + 3(1)2 + 2(1))
4
(d) f (x, y) = x + y; R is bounded by x = 0, x =
1
√
y and y = 4.
2
Solution.
Z
4
√
Z
y
4
Z
0
0
√
0
4
Z
1
y
2
=
0
= (4 +
=
y
[ 12 x2 + xy]0 dy
(x + y)dxdy =
+
3
y 2 dy
2
(32))
5
1 2
y
4
=
4
+
5
2 2
y 5
0
− (0)
84
.
5
4
2
(e) f (x, y) = xey ; R is bounded by x = 0, y = x2 and y = 4.
Solution.
Z 2Z
4
Z
y2
4
Z
√
y
xe dydx =
xe dxdy =
x2
0
Z
y2
0
Z
=
0
0
4
0
4
√y
1 2
dy
x
e
2
y2
0
4
2
2
y
1
1 y ye
e
dy
=
2
4
0
= 14 e16 − 14 .
4
(2) Find the volume of the solid bounded above by the surface z = f (x, y) and below by
the plane region R:
(a) f (x, y) = 2x + y; R is the triangle bounded by y = 2x, y = 0 and x = 2.
Solution.
2
Z
2x
Z
Z
2
[2xy + 21 y 2 ]2x
0 dx
0
2
Z 2
=
6x2 dx = 2x3 (2x + y)dydx =
V olume =
0
0
0
0
= 16.
4
(b) f (x, y) = ex+2y ; R is the triangle with vertices (0, 0), (1, 0), and (0, 1).
Solution.
Z
1
Z
V olume =
1−x
x+2y
e
0
Z
1
dydx =
0
0
Z
=
0
1−x
1 x+2y e
2
dx
0
1
( 12 e2−x − 12 ex )dx = [− 12 e2−x − 12 ex ]10
= [− 21 e1 − 12 e1 ] − [− 21 e2 − 21 ]
= −e + 12 e2 + 12 .
4
3
(3) Find the average value of the function f (x, y) = xy over the plane region bounded by
y = x, y = 2 − x, and y = 0.
Solution. The region we’re integrating over is a triangle with a base of length 2 and a
height of 1. Thus the area of the triangle is 1. Then the average value of f (x, y) over
this region is just the integral
Z 2 Z 2−x
Z 1Z x
xydydx.
xydydx +
1
0
0
0
This is a bit long. We can simplify our work by changing the order of integration so
that
Z 1 Z 2−y
Z 2 Z 2−x
Z 1Z x
xydxdy.
xydydx =
xydydx +
0
y
0
0
1
0
Then
Z
1
Z
2−y
Z
1 2 x y
2
xydxdy =
0
y
0
Z
=
0
=
=
2−y
1
dy
y
1
[ 12 (2
[− 32 y 3
1
.
3
2
− y) y −
+
1 3
y ]dy
2
Z
1
=
(−2y 2 + 2y)dy
0
y 2 ]10
4
(4) Find the maximum and minimum values of the function f (x, y) = ex−y subject to the
constraint x2 + y 2 = 1.
Solution.
F (x, y, λ) = ex−y + λ(x2 + y 2 − 1).
Then
Fx = ex−y + 2λx = 0
Fy = −ex−y + 2λy = 0
F λ = x2 + y 2 − 1 = 0
Solving for λ:
−ex−y
2x
ex−y
λ=
2y
λ=
Then
−ex−y
ex−y
=
⇐⇒ − x1 =
2x
2y
1
y
⇐⇒ −x = y.
Then
x2 + x2 − 1 = 0 =⇒ x2 = 12 .
4
Then y 2 = 12 . So the possible extrema are ( √12 , √12 ), (− √12 , − √12 ), (− √12 , √12 ), and
( √12 , − √12 ). However, we know that −x = y. Thus our constrained extrema are
(− √12 , √12 ), and ( √12 , − √12 ). Then (− √12 , √12 ) is our constrained minima with value
e−
√
2
√
and ( √12 , − √12 ) is our constrained maxima with value e 2 .
4
(5) Find the critical point(s) of the functions. Then use the second derivative test to
classify the nature of each of these points, if possible.
2
2
(a) f (x, y) = e2x +y .
Solution. The first partials are
fx = 4xe2x
2 +y 2
2x2 +y 2
fy = 2ye
2
=0
=0
2
Then since e2x +y can never be 0 we have that the only critical point is (0, 0).
The second order partials are
fxx = 4e2x
2 +y 2
+ 16x2 e2x
fyy = 2e2x
2 +y 2
+ 4y 2 e2x
fxy = 8xye2x
2 +y 2
2 +y 2
2 +y 2
Then D(0, 0) = fxx (0, 0)fyy (0, 0) − (fxy (0, 0))2 = (4)(2) − 0 = 8. Then since
D(0, 0) = 8 > 0 and fxx (0, 0) = 4 > 0. The point (0, 0) is a relative minimum of
f with value 1.
4
(b) f (x, y) = x3 + y 2 − 4xy + 17x − 10y + 8.
Solution. The first partials are
fx = 3x2 − 4y + 17 = 0
fy = 2y − 4x − 10 = 0
Solving fy for y we get y = 5 + 2x. Then subsituting into fx we get
3x2 − 4(5 + 2x) + 17 = 0 ⇐⇒ (x − 3)(3x + 1) = 0.
. Thus our critical points are
Thus x = 3 or x = − 31 . Then y = 11 or y = 13
3
1 13
(3, 11) and (− 3 , 3 ). Then the second order partials are
fxx = 6x
fyy = 2
fxy = −4
Then D(3, 11) = fxx (3, 11)fyy (3, 11) − (fxy (3, 11))2 = (18)(2) − (−4)2 = 20. Then
since D(3, 11) = 20 > 0 and fxx (3, 11) = 18 > 0. The point (3, 11) is a relative
minimum of f with value f (3, 11). Then D(− 31 , 13
) = (−2)(2) − (−4)2 < 0, thus
3
(− 13 , 13
) is a saddle point.
4
3
(6) Compute the second-order partial derivatives of f (x, y) = (2x2 + 3y 2 )5 .
5
Solution. The first order partials are
fx = (5)(4x)(2x2 + 3y 2 )4 = 20x(2x2 + 3y 2 )4
fy = (5)(6y)(2x2 + 3y 2 )5 = 30y(2x2 + 3y 2 )4
The second order partials are
fxx = 20(2x2 + 3y 2 )4 + (20x)(4)(4x)(2x2 + 3y 2 )3 = 20(2x2 + 3y 2 )4 + 320x2 (2x2 + 3y 2 )3
fyy = 30(2x2 + 3y 2 )4 = (30y)(4)(6y)(2x2 + 3y 2 )3 = 30(2x2 + 3y 2 )4 + 720y 2 (2x2 + 3y 2 )3
fxy = fyx = (20x)(4)(6y)(2x2 + 3y 2 )3 = 480xy(2x2 + 3y 2 )3
4
(7) Find an equation of the least squares line for the data points (1, 8), (2, 6), (5, 6), (7, 4),
and (10, 1).
Solution. We have the system
(12 + 22 + 52 + 72 + 102 )m + (1 + 2 + 5 + 7 + 10)b = (1)(8) + (2)(6) + (5)(6) + (7)(4) + (10)(1)
(1 + 2 + 5 + 7 + 10)m + 5b = 8 + 6 + 6 + 4 + 1
which when simplified is
179m + 25b = 88
25m + 5b = 25
37
Thus we have 54m = −37, so m = − 54
. Then b =
37
455
f (x) = − 54 x + 54 .
455
.
54
Thus our line is
4

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