The University of British Columbia
Midterm Examination - June 10, 2015
Closed book examination
Time: 60 minutes
Last Name
First
Student Number
Signature
MATH 200 253 (Circle one)
Special Instructions:
No memory aids, calculators, or electronic devices of any kind are allowed on the test.
Show all your work; little or no credit will be given for a numerical answer without the
correct accompanying work. Numerical answers should be left in calculator-ready form,
unless otherwise indicated. If you need more space than the space provided, use the back of
the previous page. Where boxes are provided for answers, put your final answers in them.
Rules governing examinations
• Each candidate must be prepared to produce, upon request, a
UBCcard for identification.
• Candidates are not permitted to ask questions of the invigilators,
except in cases of supposed errors or ambiguities in examination
questions.
• No candidate shall be permitted to enter the examination room
after the expiration of one-half hour from the scheduled starting
time, or to leave during the first half hour of the examination.
• Candidates suspected of any of the following, or similar, dishonest practices shall be immediately dismissed from the examination
and shall be liable to disciplinary action.
(a) Having at the place of writing any books, papers
or memoranda, calculators, computers, sound or image players/recorders/transmitters (including telephones), or other memory aid devices, other than those authorized by the examiners.
(b) Speaking or communicating with other candidates.
(c) Purposely exposing written papers to the view of other candidates or imaging devices. The plea of accident or forgetfulness
shall not be received.
• Candidates must not destroy or mutilate any examination material; must hand in all examination papers; and must not take any
examination material from the examination room without permission of the invigilator.
Page 1 of 7 pages
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June 10, 2015
Math 200-253
Name:
[8] 1. Consider the function f (x, y) =
Page 2 of 7 pages
√
3
x · ey .
a) (2 marks) Is the function f (x, y) differentiable at the point (1, 0)? Justify your answer.
b) (3 marks) Now imagine an ant crawling on the surface with equation z = f (x, y),
standing in the point p = (1, 0, 1). Use differentials to estimate the change of height if
the ant moves to the point on the graph with x and y coordinates (0.97, 0.03).
c) (3 marks) What’s the (horizontal) direction of steepest ascent for the ant in the point
p = (1, 0, 1)? Your answer should be a vector with 2 components, not necessarily of
length 1.
If the ant goes in that direction, at what angle will it be initially climbing above the
horizontal? Leave your answer in term of arctan.
Solution
a) The partial derivatives of f (x, y) are
1 y
e
fx = √
3
3 x2
and
fy =
√
3
xey .
They are defined and continuous around (1, 0) (precisely for all points with x 6= 0), so
f (x, y) is differentiable in (1, 0).
b) The differential of f (x, y) is
df =
√
∂f
∂f
1 y
3
dx +
dy = √
e
dx
+
xey dy
3
2
∂x
∂y
3 x
and in the point (1, 0) we get
1
df = dx + dy.
3
By using the approximation given by the differential df at the point (1, 0), the variation is
1
f (0.97, 0.03) − f (1, 0) ≈ df = (−0.03) + 0.03 = 0.02.
3
c) The direction of steepest ascent is given by the gradient ∇f (1, 0) = h 31 , 1i.
The directional
√
10
derivative√in the direction of the gradient is its magnitude |∇f (1, 0)| = 3 , and the angle
is arctan 310 .
June 10, 2015
Math 200-253
Name:
Page 3 of 7 pages
[10] 2. Given the surface S in space with equation x3 z 2 + xy = 4:
a) (2 marks) Find an equation of the tangent plane of the surface S at the point (1, 0, 2).
b) (3 marks) Assuming that the equation of the surface S defines z implicitly as a function
of x and y around the point (1, 0, 2), find the value of ∂z/∂x at the point (x, y) = (1, 0).
Your answer should be a number.
c) (2 marks) A person hiking on the surface S is in the point (1, 0, 2) and wants to
move in a direction in which the air humidity H(x, y, z) = xyz stays constant. Which
direction should he head to? Your answer should be a vector with 3 components, and
not necessarily of length 1.
d) (3 marks) The air temperature is a function T (H, P ) = P 4 + P H of the humidity
∂T
H(x, y, z) = xyz and pressure P (x, y, z) = 2x2 y + xz. Find the partial derivative
∂x
at the point (1, 0, 2).
Solution
a) The surface S is a level surface of the function g(x, y, z) = x3 z 2 + xy. One normal vector
to the tangent plane of S at (1, 0, 2) is given by the gradient ∇g(1, 0, 2).
We have ∇g = h3x2 z 2 + y, x, 2x3 zi, and ∇g(1, 0, 2) = h12, 1, 4i.
An equation of the tangent plane is then
12(x − 1) + y + 4(z − 2) = 0.
b) By differentiating both sides of g(x, y, z) = 4 with respect to x we get
3x2 z(x, y)2 + 2x3 z(x, y)zx (x, y) + y = 0
and in the point (1, 0), using z(1, 0) = 2, this becomes
12 + 4zx (1, 0) + 0 = 0
so zx (1, 0) = −3.
c) The person has to move in a direction which is both on the tangent plane to the surface
S and orthogonal to the gradient of H at (1, 0, 2).
We have
∇H = hyz, xz, xyi
so
∇H(1, 0, 2) = h0, 2, 0i
and one direction which is orthogonal to both ∇H(1, 0, 2) and the normal vector to the
tangent plane ∇g(1, 0, 2) is the cross product ~v = ∇g(1, 0, 2) × ∇H(1, 0, 2).
We find
~v = h12, 1, 4i × h0, 2, 0i = h−8, 0, 24i.
d) We can use the chain rule
∂T ∂H
∂T ∂P
∂T
=
+
.
∂x
∂H ∂x
∂P ∂x
We have
∂T
∂H
∂T
∂P
∂H
∂x
∂P
∂x
=P
= 4P 3 + H
= yz
= 4xy + z
and since P (1, 0, 2) = 2 and H(1, 0, 2) = 0, at (1, 0, 2) we get
∂T
(1, 0, 2) = P (1, 0, 2) = 2
∂H
∂T
(1, 0, 2) = 4(P (1, 0, 2))3 + H(1, 0, 2) = 32
∂P
∂H
(1, 0, 2) = 0
∂x
∂P
(1, 0, 2) = 2
∂x
and in conclusion
∂T
∂T
∂H
∂T
∂P
(1, 0, 2) =
(1, 0, 2)
(1, 0, 2) +
(1, 0, 2)
(1, 0, 2)
∂x
∂H
∂x
∂P
∂x
= 2 · 0 + 32 · 2 = 64.
Alternatively you can write T directly as a function of x, y, z and compute the partial derivative normally.
June 10, 2015
Math 200-253
Name:
Page 5 of 7 pages
[7] 3. Consider the lines given by parametric equations




x
=
1

x = 0
L1 : y = 1
L2 : y = s




z =1+t
z=0
a) (2 marks) Write down the distance between a generic point on L1 and a generic point
on L2 as a function f (s, t) of the two parameters s and t.
b) (3 marks) Set g(s, t) = f (s, t)2 to be the square of the function f . Find the only critical
point of the function g(s, t), and classify it (for example using the second derivative
test).
c) (2 marks) Compute the value of the function f (s, t) on the critical point you found in
b), and give a geometric interpretation of this value and of the critical point.
Solution
a) Using the formula for the distance between two points given their coordinates we find
p
f (s, t) = (−1)2 + (s − 1)2 + (−1 − t)2
p
= 1 + (s − 1)2 + (t + 1)2 .
b) The function g(s, t) is given by g(s, t) = 1 + (s − 1)2 + (t + 1)2 (note that maximizing or
minimizing g(s, t) is equivalent to maximizing or minimizing f (s, t)).
Its partial derivatives are
gs (s, t) = 2(s − 1)
and
gt (s, t) = 2(t + 1)
so critical points are the solutions of
(
2(s − 1) = 0 = 0
2(t + 1) = 0.
The only solution is the point (s, t) = (1, −1).
The second derivatives are gss = 2, gtt = 2 and gst = 0, so the discriminant D(g) = 4 > 0,
and since gss = 2 > 0 we can conclude that the point (1, −1) is a local minimum.
c) The value of the function f (s, t) at the point (s, t) = (1, −1) is 1, which is (basically by
definition as shortest distance) the distance between the two lines. The minimum we
found in b) corresponds to the value of the parameters s, t (1 and −1 respectively) that give
the two points on the two lines that realize the shortest distance.
June 10, 2015
Math 200-253
Name:
Page 6 of 7 pages
[7] 4.
a) (2 marks) Sketch the region D = {(x, y) | x ≥ −1, y ≥ 0, and y ≤ 4 − x2 } in the
plane. Is it closed and bounded?
b) (5 marks) Find the absolute maximum and absolute minimum of the function
f (x, y) = xy − x + 1
in the region D.
Solution
a) Sketch:
Yes, it is closed and bounded.
b) Let’s find the critical points: fx = y − 1 and fy = x, so the only critical point is (0, 1),
which is in the interior of the region D. The value of f at the point is f (0, 1) = 1.
Let’s analyze the boundary: there are three different pieces to analyze:
• The vertical segment: equation is x = −1 and y varies from 0 to 3. The function is
g(y) = f (−1, y) = −y + 1 + 1 = 2 − y, and this doesn’t have critical points in the
segment. At the boundary points we get the values g(0) = 2 and g(3) = −1.
• The horizontal segment: equation is y = 0 and x varies from −1 to 2. The function
is g(x) = f (x, 0) = −x + 1, again with no critical points in the segment, and at the
boundary points g(−1) = 2 and g(2) = −1.
• The piece of parabola: equation is y = 4 − x2 and x ranges from −1 to 2 again. The
function is g(x) = f (x, 4 − x2 ) = x(4 − x2 ) − x + 1 = −x3 + 3x + 1. The critical points
of g are solutions of g 0 (x) = −3x2 + 3 = 0 so x = ±1, and only 1 is in the interior of
the segment. The function there is g(1) = −1 + 3 + 1 = 3, and at the boundary points
of the segment we find g(−1) = 1 − 3 + 1 = −1 and g(2) = −8 + 6 + 1 = −1 (we
already had these last two values from the previous analysis).
In conclusion, the absolute maximum of f (x, y) in the region D is 3 and the absolute minimum is −1 (both of them occur in the boundary).
June 10, 2015
Math 200-253
Name:
Page 7 of 7 pages
[8] 5. Consider the function f (x, y) = x3 ey .
a) (5 marks) Using Lagrange multipliers, find maximum and minimum values of the function f (x, y) on the circle x2 + y 2 = 4.
b) (3 marks) Check that all points (0, y) with zero x-coordinate are critical points of the
function f (x, y). Are they local maxima, local minima or saddle points?
Solution
a) Let’s compute partial derivatives: fx = 3x2 ey , fy = x3 ey , gx = 2x, gy = 2y, so the
Lagrange multiplier system is

2 y

3x e = λ2x
x3 ey = λ2y

 2
x + y2 = 4
If x = 0, then y = ±2, so we have the points (0, ±2), and we can assume x 6= 0.
By dividing the second equation by the first (now this is allowed, since x 6= 0, and if λ = 0
then also x = 0) we find xy = x3 , so x2 = 3y. Plugging this is the constraint gives 3y + y 2 = 4,
√
so (y + 23 )2 = 4 + 49 = 25
with solutions y = 1 and y = −4. If y = 1 we get x = ± 3, and
4
√
the points (± 3, 1), and if y = −4 then x2 = −12 doesn’t have any (real) solution.
By computing the function values we get
f (0, ±2) = 0
√
√
f ( 3, 1) = 3 3 · e
√
√
f (− 3, 1) = −3 3 · e.
√
√
Thus the maximum of f (x, y) on the given circle is 3 3 · e and the minimum is −3 3 · e.
b) Both partial derivatives of f (x, y) vanish if x = 0, so the points of the form (0, y) are all
critical points.
If you try to use the second derivative test, you’ll see that it’s inconclusive.
The points are all saddle points: ey > 0 for every y, so the sign of f (x, y) = x3 ey is the
same as the sign of x. In any small disk around a point (0, y) there are both points (x, y)
with positive x, and points with negative x, so there will be both points where f (x, y) is
positive (so bigger than f (0, y) = 0) and points where f (x, y) is negative (so smaller than
f (0, y) = 0).
The End