MATH 253
Sections 501-503
EXAM 1
Spring 1998
P. Yasskin
36
1 x2 y2
z Ax By C
Problems 1 – 3: Find the plane tangent to the the graph of the function fŸx, y at the point Ÿx, y Ÿ1, 2 . Write the equation of the plane in the form
and find the values of A, B and C in problems 1, 2 and 3:
36
36
6
fŸ1, 2 1 12 22
1 x2 y2
f x Ÿx, y "362 2x 2
f x Ÿ1, 2 "36 2 2 12 "2
1x y
11 2
"
36 2y
f y Ÿx, y f y Ÿ1, 2 "36 2 2 22 "4
1 x2 y2
11 2
z fŸ1, 2 f x Ÿ1, 2 Ÿx " 1 f y Ÿ1, 2 Ÿy " 2 6 " 2Ÿx " 1 " 4Ÿy " 2 "2x " 4y 16
fŸx, y A "2
B "4
C 16
1. (3 points) A a. "4
correctchoice
b. "2
c. 0
d. 2
e. 4
2. (3 points) B correctchoice
a. "4
b. "2
c. 0
d. 2
e. 4
3. (3 points) C a. "4
b. 3
c. 6
d. 9
correctchoice
e. 16
36
represents the height of a mountain and
1 x2 y2
you are at the point Ÿx, y Ÿ1, 2 , in what direction should you walk to go directly
down hill?
a. Ÿ"4, "2 f Ÿ"2, "4 points up hill.
b. Ÿ"2, "4 c. Ÿ4, 2 d. Ÿ2, 4 correctchoice
" f Ÿ2, 4 points down hill.
e. None of these
4. (3 points) If the function fŸx, y 1
Problems 5 – 7: Find the plane tangent to the the graph of the equation x e z z e x y 2 at
the point Ÿx, y, z Ÿ0, 1, 2 . Write the equation of the plane in the form z Ax By C
and find the values of A, B and C in problems 5, 6 and 7:
f Ÿe z zy e x y , zx e x y , x e z e x y fŸx, y, z x e z z e x y
f
Ÿe 2 2 e 0 , 0, e 0 Ÿe 2 2 e 0 , 0, 1 N
Ÿ0,1,2 2
NX NP
Ÿe 2 x z 2
5. (3 points) A a. "2 " e
b. 2 e
correctchoice
c. "2 " e 2
2
d. 2 " e
e. 0
6. (3 points) B a. "2 " e
b. 2 e
c. "2 " e 2
d. 2 " e 2
correctchoice
e. 0
7. (3 points) C correctchoice
a. 2
b.
z
X Ÿx, y, z P Ÿ0, 1, 2 " Ÿe 2 2 x 2
A
" Ÿe 2 2 B0
C2
"e
c. 1
e
d. 2
e
e. e 2
8. (5 points) Below is the contour plot of a function fŸx, y . If you start at the point
f, draw the
v
Ÿ2, 5 and move along a curve whose tangent vector is always curve in the plot.
6
6
4
4
2
2
y
-2
0
2
4
x
6
8
-2
0
-2
-2
-4
-4
2
4
x
6
8
The curve starts at Ÿ2, 5 ends at Ÿ.5, 1 and is perpendicular to each level curve
(circle). The curve should be drawn in the same plot.
2
9. (12 points) Find all critical points of the function fŸx, y 1 2xy " x 2
" 19 y 3 and
classify each as a local maximum, a local minimum or a saddle point.
f x 2y " 2x 0
f y 2x " 1 y 2 0
®
xy
2y " 1 y 2 0
y 2" 1y
3
3
3
So y 0 or y 6 and x y.
So the critical points are Ÿ0, 0 and Ÿ6, 6 .
f yy " 2 y
D f xx f yy " f xy2 Ÿ"2 " 2 y " Ÿ2 2 4 y " 4
f xx "2
f xy 2
3
3
3
x y f xx D
0
"2 "4
saddle
since D 0
6 6 "2 4 local maximum since D 0 and f xx 0
0 0
10. (12 points) Find the point on the paraboloid z " 1 x 2
2
the point Ÿ1, 2, 1 .
" 12 y 2 0 which is closest to
Minimize the square of the distance from the general point Ÿx, y, z to the point Ÿ1, 2, 1 :
f Ÿx " 1 2 Ÿy " 2 2 Ÿz " 1 2
subject to the constraint that the general point Ÿx, y, z is on the paraboloid
g z " 1 x2 " 1 y2 0
2
2
Method 1: Lagrange Multipliers.
f Ÿ2Ÿx " 1 , 2Ÿy " 2 , 2Ÿz " 1 g Ÿ"x, "y, 1 f 5g
2 x"1
2Ÿx " 1 "5x
5 Ÿ "x "2 2x
#1
2 y"2
2Ÿy " 2 "5y
5 Ÿ "y "2 4y
#2
2Ÿz " 1 5
5 2Ÿz " 1 2z " 2
#3
2 4
y
x
2
Equate #3 and #1:
2z x
Plug both into the constraint: 1x
y 2x
z 1x
1 " 12 x 2 " 12 Ÿ2x 2 0
x
x 3 2
y2 3 2
z 3 5
5
5
2
Method 2: Eliminate a Variable.
2
f Ÿx " 1 2 Ÿy " 2 2 1 x 2 1 y 2 " 1
z 1 x2 1 y2
2
2
2
2
1
1
1
2
2
2
f x 2Ÿx " 1 2
x y " 1 Ÿx 0
x 1 y2 " 1
2
2
2
2
1
1
1
2
2
2
f y 2Ÿy " 2 2
x y " 1 Ÿx 0
x 1 y2 " 1
2
2
2
2
2
1
y 2x
Equate #1 and #2:
x y
5 x2 1
Plug back into #1: 1 x 2 1 Ÿ2x 2 " 1 "1 1x
x
2
2
2
Equate #1 and #2:
x
3
2
5
y2
3
2
5
z 1 x 2 1 y 2 1 x 2 1 Ÿ2x 2
2
2
2
2
5 x2
2
x3 2
5
2Ÿx " 1 "1 2x
2 y " 2 " Ÿ
"1 2y
"
x3 2
5
5 x2 5
2
2
3
2
5
2
3
1
x
2
y
#1
#2
5
2
3