Tuesday, October 19, 2010
1st Midterm Exam MTH 164 (Multivariable Calculus)
1. (20 points) Consider the surface given by the equation
x2 − 2y 2 + z 2 + yz = 2
(a) Find an equation of the tangent plane to the surface at the point P (2, 1, −1).
Answer:
The equation is a level set of g(x, y) = x2 − 2y 2 + z 2 hence the gradient of this function
is normal to the surface. ∇g = �2x, −4y + z, 2z + y� and ∇g(2, 1, −1) = �4, −5, −1�.
A plane is defined by a point and its normal vector, so with X = �x, y, z� the tangent
plane is defined by ∇g · (X − (2, 1, −1)) = 0 or 4x − 5y − z = 4
If you use coordinates �dx, dy, dz� = �x − 2, y − 1, z + 1� relative to the coordinate
system whose origin is at (2, 1, −1) then you get the slightly simpler equation of a
plane through the this new origin 4d − 5dy − dz = 0.
(b) If the point Q(2.1, 0.9, k) lies near P (2, 1, −1) and on the surface, use linear approximation to approximate the value of k.
Answer:
On the tangent plane we have 4(.1) − 5(−.1) − dz = 0 hence .9 = dz = z + 1 or
z = −.1. Hence the z coordinate of the corresponding point on the surface will be
z = −.1 + error and we are assuming that the changes are small enough so that the
error term is negligible and our linear approximation is fairly accurate.
Another approach is to assume that z = z(x, y) is a function of x and y defined
implicitly by the equation above. Then using implicit differentiation on the equation
∂z
∂z
∂z
∂z
above we can calculate that ∂x
(2, 1) = 4 and ∂y
(2, 1) = −5 so dz = ∂x
dx + ∂y
dy =
4(.1) − 5(−.1) = .9 and z = −.1 as before.
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Tuesday, October 19, 2010
1st Midterm Exam MTH 164 (Multivariable Calculus)
(c) Find the maximum rate of change of the height of the surface at P (2, 1, −1) above the
xy plane and the direction in the xy-plane at which it occurs.
Answer:
dz = 4dz − 5dy = �4, −5� · �dx, dy� = ��4, −5�� ��dx, dy�� cos(θ) So the maximum
rate of a change (for a unit vector) will be when the vector is parallel to �4, −5�. The
√
optimal direction is �u = √�4,−5�
and the rate of change is 16 + 25.
16+25
A second approach follows the second method in part (b) above.
∂z
∂z
dz = ∂x
dx + ∂y
dy where the partial derivatives were calculated
∂z ∂z
ferentiation. Then start rewriting this as dz = � ∂x
, ∂y � · �dx, dy�
�4, −5� · �dx, dy� = ��4, −5�� ��dx, dy�� cos(θ) and finally continue
late the direction of maximum increase.
First observe that
using implicit dif= ∇z · �dx, dy� =
as before to calcu-
These two approaches say exactly the same thing, but they use slight different notation.
Both notations are widely used so it’s important to get used to translating between
the two.
(d) Find the directional derivative of z = f (x, y) at P (2, 1, −1) in the direction �v =
�4, 2, 0�.
Answer:
The best answer is Dv f = �4, −5� · √�4,2�
16+4)
√
= 6/ 20 where we have replaced �v by the
parallel unit vector.
Answer:
2. (5 points) A million years ago, an alien species built a vertical tower on a horizontal
plane. When they returned, they discovered that the ground had tilted so that the measurements of 3 points on the ground gave coordinates of P (0, 0, 0), Q(2, 1, 0) and R(0, 2, 1).
By what angle does the tower now deviate from the vertical? (Leave answer in terms of
arccosine.)
Answer:

î

A normal vector to the plane is given by N = det 2
0
its angle with the vertical we have cos(θ) = �0, 0, 1�
√
cos−1 (4/ 21)

ĵ k̂

1 0 = î − 2ĵ + 4k̂. To calculate
2 1
√
· N/�N � = 4/ 21 and hence θ =
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Tuesday, October 19, 2010
1st Midterm Exam MTH 164 (Multivariable Calculus)
3. (10 points) Find the limit, if it exists. Explain briefly how you arrive at your answer
in each case.
(a)
lim
(x,y)→(5,12)
√
2
2
e x +y
Answer:
This function is continuous since it composed from continuous functions so we can
calculate the limit by evaluating at the point. Answer is e13 .
(b)
x2 − y 2
(x,y)→(0,0) x2 + y 2
Answer:
lim
Answer: DNE. The limit along the x axis (y = 0 gives 1 but the limit along the y axis
x = 0 is −1 so there can be no common value for the limit. The limit does not exist.
Answer:
4. (10 points) Consider the vectors �u = �1, 0, 2, 1� and �v = �1, 1, 1, 0�. Write
�v = w
�1 + w
�2
where w
� 1 is parallel to �u and w
� 2 is perpendicular to �u.
Answer:
There are several approaches. One is to use vector projection onto �u to calculate w
� 1 , then
subtract to find w
� 2 . Another is to use w
� 1 = k�u, �u · w
� 2 = 0. Solving the equation above for w
�2
and using these two equations we have �u · (�v − k�u. Since only k is unknown in this equation
we can solve for k and the rest follows.
Answer: w
� 1 = � 12 , 0, 1, 12 � and w
� 2 = � 12 , 1, 0, − 12 �
5. (20 points) Suppose x(t) = et , y(t) = 2 + sin(2t) and z(t) = 2 + cos(4t). If
w(x, y, z) =
evaluate
x y
+
y z
dw
when t = 0.
dt
Answer:
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1st Midterm Exam MTH 164 (Multivariable Calculus)
Tuesday, October 19, 2010
The chain rule in this case:
dw
∂w dx ∂w dy ∂w dz
=
+
+
dt
∂x dt
∂y dt
∂z dt
becomes
dw
1
x
1
y
= et + (− 2 + )(2 cos 2t) + (− 2 (−4 sin 4t)
dt
y
y
z
z
evaluating at t = 0 gives a value of 23 .
6. (10 points) Match the parametric equations with the graphs labeled I-V.

x = t + sin(4t)
(A)
y = t + sin(3t)

x = sin(t + sin(7t))
(C)
y = cos(t)

x = | cos(t)| · cos(t)
(E)
y = | sin(t)| · sin(t)

x = sin(t)
(B)
y = cos(t) − 2 cos(2t)

x = 6 cos(t) + cos(4.5t)
(D)
y = 6 sin(t) − sin(4.5t)
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1st Midterm Exam MTH 164 (Multivariable Calculus)
Tuesday, October 19, 2010
I
II
III
IV
V
Answer:
I:A, II:E, III: C, IV:B, V: D
7. (8 points) Determine the signs of the partial derivatives at the given point for the
function f whose graph is shown.
Page 6 of 8
Tuesday, October 19, 2010
1st Midterm Exam MTH 164 (Multivariable Calculus)
(a) fx
Answer:
negative
(b) fy
Answer:
positive
(c) fxx
Answer:
negative
(d) fyy
Answer:
positive
Answer:
Page 7 of 8
1st Midterm Exam MTH 164 (Multivariable Calculus)
Tuesday, October 19, 2010
8. (8 points) Consider the function
7 − 3xy 2
.
4
Which graph below corresponds to the following traces:
f (x, y) =
(a) The trace for x =
1.5.
(b) The trace for x = −0.4.
(c) The trace for x =
0.2.
(d) The trace for x = −1.4.
I
II
III
IV
Answer:
I:b, II: c, III: d, IV: a
9. (9 points) A quantity z is calculated from x and y using the formula
z = xy 2
Page 8 of 8
Tuesday, October 19, 2010
1st Midterm Exam MTH 164 (Multivariable Calculus)
(a) The quantity x is measured to be 2 ± 0.1 and the quantity y is measured to be 1 ± 0.2,
estimate the maximum error in the calculated value of z.
Answer:
dz = y 2 dx + 2xdy
is the approximating differential dx = ±.1 and dy = ±.2. Hence the maximum error
(assuming the linear approximation is good) would be dz = ±.9
(b) What is the range of possible z values?
Answer:
Using the linear approximation z should lie in the interval (1.1, 2.9). (The linear
estimate is off by about 0.1 at each end so off by between 3 and 10%. That’s not too
bad for an estimate.)
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