Sample Midterm, MATH 2A03/2X03, Spring 2014
Ruipeng Shen
May 23, 2014
The duration of test is 90 minutes. Total marks = 30. Thus you have an average of 3
minutes to spend on each mark.
1. Consider the function
y3
, if (x, y) 6= (0, 0);
f (x, y) =
x2 + y 2

0,
if (x, y) = (0, 0).


[2] (a) Find the gradient ∇f (x, y) for (x, y) 6= (0, 0).
∂f
[2] (b) Find the partial derivative
(0, 0) by the definition
∂y
∂f
f (0, h) − f (0, 0)
(0, 0) = lim
.
h→0
∂y
h
[2] (c) Determine whether the partial derivative fy (x, y) is continuous at (0, 0).
Solution
(a) We have
−2xy 3
;
(x2 + y 2 )2
3y 2 (x2 + y 2 ) − 2y · y 3
3x2 y 2 + y 4
fy =
=
.
(x2 + y 2 )2
(x2 + y 2 )2
fx =
Thus the gradient is given by
−2xy 3
3x2 y 2 + y 4
,
(x2 + y 2 )2 (x2 + y 2 )2
.
(b) We have
h3
−0
2
2
∂f
f (0, h) − f (0, 0)
h−0
(0, 0) = lim
= lim 0 + h
= lim
= 1.
h→0
h→0
h→0
∂y
h
h
h
(c) We need to check whether the following limit holds
3x2 y 2 + y 4
= 1.
(x,y)→(0,0) (x2 + y 2 )2
lim
1
This is not true, since the limit along x-axis is
3x2 02 + 04
= 0 6= 1.
x→0 (x2 + 02 )2
lim fy (x, 0) = lim
x→0
2. Assume f (x) is a C 2 function of one variable, whose value is always positive. Let us
define
h(u, v) = (u + v)f (u + v).
[2] (a) Find the second derivative huu in terms of derivatives of f by the chain rule.
[3] (b) Determine the direction in which the function has the maximal rate of increase at
the point (0, 0). Does this depend on specific f (x)?
Solution
Let us first calculate the partial derivatives.
∂
∂
hu =
(u + v) f (u + v) + (u + v)f 0 (u + v) (u + v)
∂u
∂u
0
= f (u + v) + (u + v)f (u + v)
∂
∂
(u + v) f (u + v) + (u + v)f 0 (u + v) (u + v)
hv =
∂v
∂v
0
= f (u + v) + (u + v)f (u + v)
∂
∂
∂
0
huu = f (u + v) (u + v) +
(u + v) f 0 (u + v) + (u + v)f 00 (u + v) (u + v)
∂u
∂u
∂u
0
0
00
= f (u + v) + f (u + v) + (u + v)f (u + v)
= 2f 0 (u + v) + (u + v)f 00 (u + v).
(a) We have huu = 2f 0 (u + v) + (u + v)f 00 (u + v).
(b) The direction is given exactly by the gradient
∇h(0, 0) = (f (0) + 0, f (0) + 0) = f (0) · (1, 1).
Thus the direction could be given by the vector (1, 1), which does not depend on f (x) as
long as f (x) > 0.
3. Let g(x, y) be a function defined on R2 by
f (x, y) = x3 − y 3 + 3xy.
[2] (a) Find all critical points by the first derivative test.
[3] (b) Classify the critical points you found in part (a) into local maxima, local minima
and saddle points by the second derivative test.
Solution
(a) In order to find the critical points, we consider the equation system
fx = 3x2 + 3y = 0;
fy = −3y 2 + 3x = 0.
2
This means we have y = −x2 and x = y 2 . Plugging y = −x2 into the second identity, we
have x = (−x2 )2 = x4 . Thus we have either x = 0 or x = 1. Using y = −x2 , we have two
critical points, namely (0, 0) and (1, −1).
(b) Now let us apply the second derivative test on each critical point to determine whether
it is a local maximum, minimun or saddle point. We need to calcualte the second derivatives
and determinant first.
fxx = 6x;
fxy = 3;
fyy = −6y;
D = fxx fyy − (fx y)2 = −36xy − 9.
We have D(0, 0) = −9 < 0, thus the first critical point (0, 0) is a saddle point.
We have D(1, −1) = 27 > 0 and fxx (1, −1) = 6 > 0, thus the second critical point (1, −1)
is a local minimum.
4. Let c(t) = (eat cos t, eat sin t) with t ∈ [0, 2π] be a spiral .
[3] (a) Find the unit tangent vector T(t), principal unit normal vector N(t) and the curvature κ(t).
Rt
[2] (b) Calculate the arc length function s(t) = 0 kc0 (τ )kdτ and reparametrize the curve
by its arc length s.
Solution
(a) This part is straight-forward computation
c0 (t) =(eat (a cos t − sin t), eat (a sin t + cos t));
p
kc0 (t)k = e2at (a cos t − sin t)2 + e2at (a sin t + cos t)2 ;
p
=eat a2 + 1
T(t) =
c0 (t)
1
=√
(a cos t − sin t, a sin t + cos t);
0
kc (t)k
a2 + 1
1
T0 (t) = √
(−a sin t − cos t, a cos t − sin t);
a2 + 1
kT0 (t)k =1
T0 (t)
1
=√
(−a sin t − cos t, a cos t − sin t);
0
kT (t)k
a2 + 1
kT0 (t)k
κ(t) =
=(a2 + 1)−1/2 e−at .
kc0 (t)k
N(t) =
The answers are given below
• The unit tangent vector is
√ 1
(a cos t
a2 +1
• The principal unit normal vector is
− sin t, a sin t + cos t);
√ 1
(−a sin t
a2 +1
• The curvature is κ(t) = (a2 + 1)−1/2 e−at .
3
− cos t, a cos t − sin t);
(b) We have
Z
t
0
kc (τ )kdτ =
s(t) =
0
Z tp
√
aτ
a2 + 1e dτ =
0
a2 + 1(eat − 1)
.
a
Its inverse is t = a−1 ln(1 + a(a2 + 1)−1/2 s), s ∈ [0, a−1 (a2 + 1)1/2 (e2aπ − 1)]. Thus we can
reparametrize the curve by
as
1
as
1
as
as
c(s) =
1+ √
cos
, 1+ √
sin
;
ln 1 + √
ln 1 + √
a
a
a2 + 1
a2 + 1
a2 + 1
a2 + 1
Here s ∈ [0, a−1 (a2 + 1)1/2 (e2aπ − 1)].
[3 mark] 5. Write down the second-order Taylor formula of a C 2 function f (x, y) at the
point (0, 1).
f (x, y) = ln(1 + xy) + sin(x2 )
Solution
We first calculate the partial derivatives
y
fx =
+ 2x cos(x2 );
1 + xy
x
fy =
;
1 + xy
−y 2
fxx =
+ 2 cos(x2 ) − 4x2 sin(x2 );
(1 + xy)2
1
fxy =
;
(1 + xy)2
−x2
fyy =
.
(1 + xy)2
Thus we have
f (x, y) =f (0, 1) + fx (0, 1)(x − 0) + fy (0, 1)(y − 1)
1
+
fxx (0, 1)(x − 0)2 + 2fxy (0, 1)(x − 0)(y − 1) + fyy (0, 1)(y − 1)2 + R2
2
1
=0 + x + 0 + [x2 + 2x(y − 1) + 0] + R2
2
1 2
=x + x + x(y − 1) + R2 .
2
[1 mark each] 6. Determine whether the following statements are true or false. Simply give
your answer in one word. No argument, reasoning or proof is required.
(a)
x3 y
= 0.
(x,y)→(0,0) (x2 + y 2 )3/2
lim
Answer: True
By the inequality |x| ≤
x3 y
(x2 + y 2 )3/2 ≤
p
x2 + y 2 , we have
!3
|x|
p
|y| ≤ |y|.
x2 + y 2
4
Thus we have
−|y| ≤
x3 y
≤ |y|
(x2 + y 2 )3/2
As a result, we know the limit is zero by squeeze theorem (pinching theorem).
(b) Any subset of R2 must have at least one boundary point.
Answer: False The whole plane R2 and the empty set do not have any boundary point.
(c) If F, G : R2 → R2 are two C 1 vector-valued functions, then their composition F ◦ G is
necessarily of class C 1 .
Answer: True Since both two functions are continuous, we know the composition is
continous. In addition, the chain rule claims the composition is still differentiable, and its
derivative can be given by
(DF ◦ G) · DG,
which is continuous by basic properties of continuity. In summary, the composition is
continuous and has continuous derivative. Thus it is of class C 1 .
(d) The parametric equation c(t) = (cos t − sin t, 2 cos t + 2 sin t) represents an ellipse.
Answer: True
The pair (x, y) = (cos t − sin t, 2 cos t + 2 sin t) satisfies the equation
x2 y 2
+
= 1.
2
8
(e) A function defined on an open domain can never have an absolute maximum.
Answer: False The function f (x, y) = −x2 − y 2 defined in an open disk {(x, y)|x2 + y 2 <
1}has an absolute maximum 0 at the origin.
(f) If F : R2 → R2 is a differentiable vector field, then DF(1, 3) is a vector in R2 .
Answer: False
The derivative DF(1, 3) is a 2 by 2 matrix.
5