9-9:50am Midterm Exam 1

Fall 2012 Math 0240

100 points total Solutions

1. (10 points) Evaluate the limit, if it exists. If it does not exist explain why.

x sin y
lim
(x,y)→(0,0) x2 − 2y 2

x sin y 0
Solution: Path 1: x = 0, y → 0 lim 2 2
= lim = 0.
(x,y)→(0,0) x − 2y y→0 −2y 2

x sin y x sin x x sin x

Path 2: y = x, x → 0 lim 2 2
= lim 2 2
= − lim
(x,y)→(0,0) x − 2y x→0 x − 2x x→0 x2
sin x
= − lim = −1.
x→0 x
Two paths give two different results for the limit. Hence, the limit does not exist.

√
x
2. (a) (10 points) Explain why the function f (x, y) = is differentiable at the point (4, 2).
y
√
1 x
Solution: fx (x, y) = √ , fy (x, y) = − 2 . Both functions exist near (4, 2) and are
2y x y
continuous at (4, 2). Hence f (x, y) is differentiable at (4, 2). (See theorem 8, page 620).

√
x
(b) (10 points) Using the linear approximation of f (x, y) = approximate the number
√ y
3.98
.
1.99
Solution: The linearization of f (x, y) at (4, 2) is
1
L(x, y) = f (4, 2) + fx (4, 2)(x − 4) + fy (4, 2)(y − 2), where f (4, 2) = 1, fx (4, 2) = ,
8
1
fy (4, 2) = − . Then
2
1 1
L(x, y) = 1 + (x − 4) − (y − 2)
8 2
√
3.98 √ √ 1 1
= f ( 3.98, 1.99) ≈ L( 3.98, 1.99) = 1 + (3.98 − 4) − (1.99 − 2)
1.99 8 2

1
    
1 2 1 1 1 2 1
=1+ − − − =1− + =1 .
8 100 2 100 400 400 400

3. (10 points) Find parametric equations and symmetric equations for the line of intersection
of the planes x − 2y + z = 1 and x + z = 0.

Solution: The corresponding normal vectors are n¯1 = h1, −2, 1i and n¯2 = h1, 0, 1i. The
directional vector of the line is h−1, 0, 1i since
 
 ī j̄ k̄ 
 
 1 −2 1  = h−2, 0, 2i
 
 
 1 0 1 

and h−1, 0, 1i is a constant multiple of h−2, 0, 2i.

If x + z = 0 then x − 2y + z = −2y + x + z = −2y = 1 and y = −1/2. Take x = 1, then
z = −1. Hence the point (1, −1/2, −1) satisfies equations of both planes and is on the
line. The vector equation of the line hx, y, zi = h1, −1/2, 1i + th−1, 0, 1i gives
parametric equations x = 1 − t, y = −1/2, z = −1 + t and
x−1 z+1
symmetric equations = , y = −1/2 or x = −z, y = −1/2.
−1 1
x−a z+a
(General form of the symmetric equations is = , y = −1/2, where a and b
−b b
are arbitrary constants).

4. (a) (10 points) Find the gradient of f (x, y) = sin(y ln x).

y
Solution: fx (x, y) = cos(y ln x) · , fy (x, y) = cos(y ln x) · ln x
x
y
¯ = h cos(y ln x), cos(y ln x) · ln xi.
∇f
x

(b) (10 points) Evaluate the gradient at the point P (1, −2).

Solution: ∇f ¯ (1, −2) = h −2 cos(−2 · 0), cos(−2 · 0) · 0i = h−2, 0i. (Recall ln 1 = 0).
1

(c) (10 points) Find the rate of change of f at P in the direction of the vector ū =
h √15 , − √25 i.
Solution: ¯ (1, −2) · ū = h−2, 0i · h √1 , − √2 i = − √2 .
The rate of change is ∇f 5 5 5

2
5. (10 points) Find the curvature of

r̄(t) = et cos t ī + et sin t j̄ + t k̄

at the point (1, 0, 0).

Solution: At the point (1, 0, 0) t = 0. The curvature is

|r̄0 (0) × r̄00 (0)|

κ=
|r̄0 (0)|3

r̄0 (t) = het (cos t − sin t), et (cos t + sin t), 1i,
r̄00 (t) = het (cos t−sin t−sin t−cos t), et (cos t+sin t+cos t−sin t), 0i = h−2et sin t, 2et cos t, 0i
r̄0 (0) = h1, 1, 1i, r̄00 (0) = h0, 2, 0i,
√ √ √
|r̄0 (0) × r̄00 (0)| = |h−2, 0, 2i| = 8 = 2 2, |r̄0 (0)| = 3,
√ r  3/2 √
2 2 8 2 2 6
κ= √ = = =
3 3 27 3 9

6. (10 points) Use implicit differentiation to find zx and zy if

yz = ln(x + 2z)

∂ ∂ 1 + 2zx
Solution: (yz) = (ln(x + 2z)), yzx = , y(x + 2z)zx = 1 + 2zx ,
∂x ∂x x + 2z
1
(xy + 2yz − 2)zx = 1, zx = .
xy + 2yz − 2
∂ ∂ 2zy
(yz) = (ln(x + 2z)), z + yzy = , xz + 2z 2 + (xy + 2yz)zy = 2zy ,
∂y ∂y x + 2z
xz + 2z 2
(xy + 2yz − 2)zy = −xz − z 2 , zy = − .
xy + 2yz − 2

7. (10 points) Using Lagrange multipliers find the maximum and minimum values of the
function f (x, y) = 4x2 + 9y 2 subject to the constraint xy = 4

Solution: Under the constraint xy = 4 x can be made arbitrary large (while y is

small). Therefore, there is no upper bound for the function f (x, y) and its maximum
never attains. Let’s find minimum using Lagrange multipliers. Denote g(x, y) = xy.

3
 ¯ (x, y) = λ∇g(x,
Then using ∇f ¯ y), g(x, y) = 4 we get the system of equations

8x = λy
18y = λx
xy = 4

Multiplying the first line by the second we obtain 8 · 18xy = λ2 xy or 8 · 18 = λ2 .

(Note: xy = 4 and hence x 6= 0 and y 6= 0). Then λ2 = 8 · 18 = 16 · 9 = (4 · 3)2 and
8
λ = −12 or λ = 12. From the first line y = x.
λ
2 2
Case λ = −12. Then y = − x and the last line gives − x2 = 4, no solutions.
3 3
2 2 2 √ √
Case λ = 12. Then y = x and the last line gives x = 4, x2 = 6, x = − 6 or x = 6.
3 3
√ 2√ √ 2√
There are two solutions of the system (x, y) = (− 6, − 6) and (x, y) = ( 6, 6).
3 3
√ 2√ √ 2√ 4
f ( 6, 6) = f (− 6, − 6) = 4 · 6 + 9 · · 6 = 48 is the minimum value of f (x, y).
3 3 9

bonus problem [10 points extra] A cardboard box without a lid is to have a volume of 32, 000
cm3 . Find the dimensions that minimize the amount of cardboard used.

Solution: (See also the example 5, page 648). Denote V = 32, 000 and let x be the width, y
be the length, and z be the hight. Then xyz = V and area is A = xy + 2xz + 2yz.
V 2V 2V
Then z = and A(x, y) = xy + + is a function of two variables which we have
xy y x
to minimize. Note that x > 0, y > 0.
2V 2V
Ax = y − 2 = 0, Ay = x − 2 = 0 ⇒ 2V = x2 y = xy 2 ⇒ x = y. Then x3 = 2V and
√ x y
3
x = y = 2V .
√ √
CP is ( 3 2V , 3 2V ).
4V 4V
Axx = 3 , Ayy = 3 , Axy = 1. At CP Axx = 2V , Ayy = 2V .
x y
Then D = 4V 2 − 1 > 0, Axx = 2V > 0. Therefore, A(x, y) attains a local minimum at the CP
which is the absolute minimum.
√
3 √ 32, 000
2V = 3 64, 000 = 40. Hence x = y = 40 cm and z = = 20 cm.
40 · 40