Math 202
Exam 2
Instructions
1. Do NOT write your answers on these sheets. Nothing written on the test papers will be graded.
2. Please begin each section of questions on a new sheet of paper.
3. Please do not write answers side by side.
4. Please do not staple your test papers together.
5. Limited credit will be given for incomplete or incorrect justification.
6. All answers must be exact. Numeric approximations will be marked incorrect.
Questions
1. Limits (6 each)
(a) Determine if the following limit exists
x2 − 2x − 2y − y 2
(x,y)→(1,−1) x2 − 2x + 2 + 2y + y 2
lim
(1, t)
12 − 2 − 2t − t2
lim
=
t→−1 12 − 2 + 2 + 2t + t2
−1 − 2t − t2
lim
=
t→−1 1 + 2t + t2
lim −1 =
t→−1
(t, m(t − 1) − 1)
t − 2t − 2[m(t − 1) − 1] − [m(t − 1) − 1]2
lim 2
=
t→1 t − 2t + 2 + 2[m(t − 1) − 1] + [m(t − 1) − 1]2
(1 − m2 )(t − 1)2
lim
=
t→1 (1 + m2 )(t − 1)2
1 − m2
=
lim
t→1 1 + m2
−1.
2
The limit does not exist, because each line has a different limit.
1
1 − m2
.
1 + m2
Exam 2: Surfaces
2
(b) Test the following limit along y = mx3 . Explain why the limit cannot exist.
x3 y
(x,y)→(0,0) x6 + y 2
lim
(t, mt3 )
t3 (mt3 )
=
lim 6
t→0 t + (mt3 )2
mt6
lim 6
=
t→0 t + m2 t6
mt6
lim 6
=
t→0 t (1 + m2 )
m
=
lim
t→0 1 + m2
m
1 + m2
Each cubic has a different limit, therefore the limit does not exist.
Exam 2: Surfaces
3
2. Derivatives 2
y
g(x, y) = 1+x
2.
(a) (3) Calculate ∇g
∇g(x, y) =
−2xy 2
2y
,
2
2
(1 + x ) 1 + x2
(b) (3) Calculate the derivative of g at (3, 4) in the direction (1, 1).
1
√ (1, 1).
2
24 4
∇g(3, 4) =
− ,
.
25 5
1
24 20
1
− +
∇g(3, 4) · √ (1, 1) = √
25 25
2
2
−4
√ .
=
25 2
d~ =
(c) (4) If Guido is traveling along the path y =
x = 1?
√
x on the surface g(x, y) what is his direction at the point
√
t
t, t,
.
1 + t2
1 (1 + t2 ) − 2t2
P 0 (t) =
1, √ ,
(1 + t2 )2
2 t
1
1 − t2
=
1, √ ,
.
2 t (1 + t2 )2
1
P 0 (1) =
1, , 0 .
2
P (t)
=
(d) (4) Write an equation for the plane tangent to g(x, y) at (1, 2).
P (t, s)
=
(1, 0, −2)t + (0, 1, 2)s + (1, 2, 2).
0
=
(2, −2, 1) · (X − (1, 2, 2)).
Exam 2: Surfaces
4
3. Extrema (6 each)
(a) Find and identify extrema for f (x, y) = x3 − 12xy + 8y 3 .
δf
δx
δf
δy
2
δ f
δx2
δ2 f
δy 2
δ2 f
δxδy
=
3x2 − 12y.
=
24y 2 − 12x.
=
6x.
=
48y.
=
−12.
=
0.
x − 4y
=
0.
2
=
4y.
3x2 − 12y
2
x
24y 2 − 12x =
2
2y − x =
0.
0.
2
=
x.
2(x2 /4)2
=
x.
4
2x /16
=
x.
4
=
8x.
2y
x
x4 − 8x =
3
x(x − 8)
=
x =
0.
0.
0, 2.
(2y 2 )2
=
4y.
4
=
0.
3
y(y − 1)
=
0.
4y − 4y
y
=
0, 1.
fx (0, 1)
=
−12.
fx (2, 0)
=
12.
D(0, 0)
=
0(0) − (−12)2
<
0.
=
12(48) − (−12)2
>
0.
D(2, 1)
(0, 1) and (2, 0) are not critical values. (0, 0) is a saddle point and (2, 1) is a minimum.
Exam 2: Surfaces
5
(b) Find the highest and lowest points on p(x, y) = 4x + 6y − 7 constrained by 4x2 + 9y 2 = 72.
4
1
2x
6
1
3y
1
2x
2x
=
λ8x.
=
λ.
=
λ18y.
=
λ.
=
1
.
3y
3y.
2
(2x) + 9y
2
=
72.
2
2
=
72.
2
2
=
72.
18y 2
=
72.
2
=
4.
y
=
±2.
2
4x + (3y)
=
72.
4x2 + (2x)2
4x2 + 4x2
= 72.
= 72.
(3y) + 9y
9y + 9y
y
2
=
8x2
=
72.
2
=
9.
x
= ±3.
x
p(−3, −2)
p(3, −2)
p(3, 2)
p(−3, 2)
The minimum is (−3, −2) and the maximum is (3, 2).
= −31.
= −7.
=
17.
= −7.