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MATH 2263 Name (Print): Summer 2009 Student ID: Exam 2: Solutions July 8, 2009 Signature: Time limit: 55 minutes
This exam contains 10 pages (including this cover page and a scratch page) and 7 problems. Check to make sure you have all 10 pages. Enter all requested information at the top of this page, and put your initials on the top of every page, in case the pages become separated. No calculators or note-sheets are allowed.
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1 15 pts 2 15 pts 3 15 pts 4 10 pts 5 10 pts 6 15 pts 7 20 pts Total 100 pts
- (15 pts) Find and classify all the local maxima, local minima, and saddle points of
f (x, y) = 4xy โ x^4 โ y^4.
Ans: We first must find all critical points of f. To do this we have to solve the system
fx(x, y) = 4y โ 4 x^3 = 0 (1) fy(x, y) = 4x โ 4 y^3 = 0. (2)
Solving Equation (1) for y we have y = x^3. We substitute this into Equation (2) to obtain
4 x โ 4(x^3 )^3 = 0, 4 x โ 4 x^9 = 0, 4 x(1 โ x^8 ) = 0.
This has solutions when x = 0 and when x = ยฑ1. We then use y = x^3 to find y, giving three critical points: (0, 0), (1, 1), and (โ 1 , โ1). For each critical point, we have to calculate the discriminant, which involves the second derivatives:
fxx(x, y) = โ 12 x^2 , fyy(x, y) = โ 12 y^2 , fxy(x, y) = 4, D(x, y) = fxxfyy โ (fxy)^2 = 144x^2 y^2 โ 16.
We then have
D(0, 0) = โ 16 < 0 , D(1, 1) = 144 โ 16 > 0 , D(โ 1 , โ1) = 144 โ 16 > 0 , fxx(1, 1) = โ 12 < 0 , fxx(โ 1 , โ1) = โ 12 < 0 ,
which tells us that (0, 0) is a saddle point and (1, 1) and (โ 1 , โ1) are both local maximums.
- (15 pts) Sketch the region of integration, reverse the order of integration, and evaluate the integral (^) โซ 8
0
โ (^3) x
y^4 + 1
dy dx.
Ans: โซ (^8)
0
โ (^3) x
y^4 + 1
dy dx =
0
โซ (^) y 3
0
y^4 + 1
dx dy
0
y^3 y^4 + 1
dy [u = y^4 + 1, du = 4y^3 dy]
0
u
du
ln |u|
17
1
ln 17.
(Note: A sketch is also required, and will be added later.)
- (10 pts) Using a double integral, find the area of the region bounded by the curves
y = 1 โ x and y = 1 โ x^2.
(Be sure to sketch the region.)
Ans:
A(D) =
D
dA
0
โซ (^1) โx 2
1 โx
dy dx
0
y|^1 โx
2 1 โx dx
=
0
(1 โ x^2 ) โ (1 โ x) dx
0
โx^2 + x dx
x^3 3
x^2 2
1
0 = โ
(Note: A sketch is also requires, and will be added later.)
- (15 pts) Let F (x, y, z) = z โ xey^ cos(z) + 1.
(a) Find the gradient โF (x, y, z).
Ans: We have โF (x, y, z) = ใfx, fy, fz ใ = ใโey^ cos z, โxey^ cos z, 1 + xey^ sin zใ.
(b) Find the directional derivative of F at the point (1, 0 , 0) in the direction of v = iโ 2 j+4k.
Ans: Because v is not a unit vector we must find the unit vector in the same direction:
u =
v |v|
11 + (โ2)^2 + 4^2
v =
Then the directional derivative is
DuF (1, 0 , 0) = โF (1, 0 , 0) ยท u
= ใโ 1 , โ 1 , 1 ใ ยท
(c) Find the equation of the tangent plane to the surface z + 1 = xey^ cos(z)
at the point (1, 0 , 0).
Ans: We can write this surface as
z + 1 = xey^ cos(z) z + 1 โ xey^ cos z = 0 F (x, y, z) = 0, so the equation of the tangent plane to the surface is given by
Fx(x 0 , y 0 , z 0 )(x โ x 0 ) + Fy(x 0 , y 0 , z 0 )(y โ y 0 ) + Fz (x 0 , y 0 , z 0 )(z โ z 0 ) = 0 Fx(1, 0 , 0)(x โ 1) + Fy(1, 0 , 0)(y โ 0) + Fz (1, 0 , 0)(z โ 0) = 0 โ1(x โ 1) โ 1(y) + 1(z) = 0 โx โ y + z + 1 = 0.
- (20 pts) Use the method of Lagrange multipliers to find the maximum and minimum values of the function f (x, y) = 2x^2 โ 8 y + 5, subject to the constraint x^2 + 4y^2 = 4. (You may assume that both a maximum and minimum value exist.)
Ans: We need to solve the equations
4 x = 2ฮปx (1) โ8 = 8ฮปy (2) x^2 + 4y^2 = 4. (3)
We would like to divide by x in Equation (1). Before we can do that we must see what happens if x = 0. In the case that x = 0, Equation (2) tells us nothing, and Equation (3) tells us
x^2 + 4y^2 = 4 4 y^2 = 4 y^2 = 1 y = ยฑ 1.
So (0, 1) and (0, โ1) are points where a maximum or minimum might occur. In the case the x 6 = 0, we can divide by 2x in Equation (1), giving ฮป = 2. Then Equation (3) gives
โ8 = 8ฮปy โ8 = 16y
โ
= y.
We can then use this value for y in Equation (3) to yield
x^2 + 4y^2 = 4 x^2 + 4(โ 1 /2)^2 = 4 x^2 + 1 = 4 x = ยฑ
Therefore (
3 , โ 1 /2) and (โ
3 , โ 1 /2) are two more points where the maximum may occur.
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