1. No category

Math 243
Spring 2015
(Practice) Final Exam
5/11/2015
Time Limit: 2 hours
Name:
• No calculators or notes are allowed.
• One side of each sheet is blank and may be used as scratch paper.
• Check your answers whenever possible.
• Show your work clearly.
Grade Table (for instructor use only)
Question Points Score
1
15
2
20
3
20
4
15
5
15
6
15
7
15
8
20
9
10
10
15
11
20
12
20
Total:
200
Math 243
(Practice) Final Exam - Page 2 of 16
Common trig function values
x
sin x
cos x
tan x
0
0
1
0
π
6
1
2
π
4
√
5/11/2015
The cross product
If ~u = ha, b, ci and ~v = hp, q, ri, then
~u × ~v = hbr − cq, cp − ar, aq − bpi.
3
2
√1
3
√1
2
π
3
√1
2
√
3
2
1
2
1
√
3
π
2
1
0
undefined
Differential geometry
d~r
~v
• Tˆ =
=
ds
|~v |
The 2nd derivative test
2
H = fxx fyy − fxy
H<0
saddle
H > 0, fxx (or fyy ) < 0
local maximum
H > 0, fxx (or fyy ) > 0
local minimum
H=0
inconclusive
ˆ
ˆ
ˆ = 1 dT = 1 dT
• N
dTˆ dt
κ ds
dt dTˆ dTˆ 1
• κ= =
ds |~v | dt Z
• arclength =
b
|~v (t)| dt
a
Math 243
(Practice) Final Exam - Page 3 of 16
5/11/2015
1. (15 points) Let l be line through (3, 2, 1) that is normal to the plane 2x − y + 2z = −2.
Where does l meet the plane?
Solution: The vector h2, −1, 2i is normal to the plane, therefore parallel to l. So l
can be parametrized as follows:
~r(t) = h3, 2, 1i + th2, −1, 2i.
To see where l passes through the plane, plug the parametrization into the equation
for the plane:
2(3 + 2t) − (2 − t) + 2(1 + 2t) = −2,
and solve it to get t = −8/9. Plugging this value of t back into the parametrization
yields the point h11/9, 26/9, −7/9i.
Math 243
(Practice) Final Exam - Page 4 of 16
5/11/2015
2. For each of the following limits, find it if it exists, otherwise show why it does not.
x3 − y 2
(a) (10 points)
lim
(x,y)→(0,0) x2 + y 2
Solution: When x = 0, the limit becomes
−y 2
= 1.
y→0 y 2
lim
On the other hand, when y = 0, the limit becomes
x3
= lim x = 0,
x→0
x→0 x2
lim
and so the limit doesn’t exist.
(b) (10 points)
lim
(x,y)→(0,0) x2
xy
+ y2 + 1
Solution: Plug in to get 0.
Math 243
(Practice) Final Exam - Page 5 of 16
3. (a) (10 points) Sketch the level curves of z = g(x, y) =
z = 2.
5/11/2015
p
x2 − y at z = 0, z = 1, and
y
4
z=0
2
x
−4
−2
2
z=1
−2
−4
4
z=2
Math 243
(Practice) Final Exam - Page 6 of 16
5/11/2015
(b) (10 points) Sketch the level curves of z = f (x, y) = ex+y at z = 1, z = e, and
z = 1/e.
y
4
2
x
−4
−2
2
4
−2
z=e
z=1
−4
z = 1/e
Math 243
(Practice) Final Exam - Page 7 of 16
5/11/2015
4. (15 points) Suppose that the acceleration of a particle is given by ~a(t) = ht, 0, t2 i and
we know that its initial velocity ~v (0) = h1, 1, 0i and its initial position ~r(0) = h0, 1, 0i.
Find ~r(t).
Solution:
~r(t) =
t3
t4
+ t, t + 1,
6
12
Math 243
(Practice) Final Exam - Page 8 of 16
5. (15 points) Find the curvature of the curve
~r(t) = ht, t2 , 2t + 1i
at the point (2, 4, 5).
Solution: ~v (t) = h1, 2t, 2i, so |~v (t)| =
√
5 + 4t2 . Then
1
Tˆ(t) = √
h1, 2t, 2i,
5 + 4t2
and
dTˆ
1
4t
=√
h0, 2, 0i −
h1, 2t, 2i,
2
dt
(5 + 4t2 )3/2
5 + 4t
which when t = 2 is equal to the vector
2
h−4, 5, −8i.
213/2
The curvature is
√
2
1 2
5
h−4, 5, −8i = 3/2 .
κ=
3/2
|~v (2)| 21
21
5/11/2015
Math 243
(Practice) Final Exam - Page 9 of 16
5/11/2015
6. (15 points) Find a vector equation for the line that is normal to the surface
z = ln(x2 + y 2 )
at the point (1, 0, 0).
Solution: The given surface is the level surface where g(x, y, z) = ln(x2 +y 2 )−z = 0,
so we take its gradient:
2x
2y
~
∇g =
,
,
−1
= h2, 0, −1i.
x2 + y 2 x2 + y 2
(1,0,0)
(1,0,0)
So an equation for the normal line is
~r(t) = h1, 0, 0i + th2, 0, −1i.
Math 243
(Practice) Final Exam - Page 10 of 16
5/11/2015
7. (15 points) The equation
x − y 2 + z 3 + sin(xyz) = 3
defines a surface in 3-dimensional space. Find an equation for the tangent plane to this
surface at the point (2, 0, 1).
Solution: Let g(x, y, z) = x − y 2 + z 3 + sin(xyz). Then
~
∇g = h1 + yz cos(xyz), −2y + xz cos(xyz), 3z 2 + xy cos(xyz)i(2,0,1)
(2,0,1)
= h1, 2, 3i,
so an equation for the tangent plane is x + 2y + 3z = 5.
Math 243
(Practice) Final Exam - Page 11 of 16
5/11/2015
8. Let
f (x, y) = 2x2 + y 2 − 8x − 20.
(a) (10 points) Classify all local extrema and saddle points (if any) of f .
~ = h4x − 8, 2yi, so the only critical point is (2, 0). fxx = 4 and
Solution: ∇f
fyy = 2, while fxy = 0, so the Hessian is always positive. Since fxx is also
positive, we conclude that (2, 0) is a local minimum.
Math 243
(Practice) Final Exam - Page 12 of 16
5/11/2015
(b) (10 points) Now consider f restricted to the domain x2 + y 2 ≤ 25. Find the maximum and minimum values of f on this restricted domain and where they occur.
Solution: To find max and min values on the boundary, we can parametrize the
circle or use the method of Lagrange multipliers. For the latter, the constraint
~ = h2x, 2yi, which gives us the system of
is g(x, y) = x2 + y 2 − 25 = 0, so ∇g
equations
4x − 8 = λ2x
2y = λ2y
2
x + y 2 = 25
Multiplying the first by y and the second by x gives us the equation 4xy − 8y =
2xy, and so y(x − 4) = 0, which means that either y = 0 or x = 4. From y = 0,
we get the two points (5, 0) and (−5, 0) and from x = 4 we get the points (4, 3)
and (4, −3).
Finally compute the value of f at all the points we have found so far:
x y
f
2 0 -28
5 0 -10
-5 0 70
4 3 -11
4 -3 -11
from which we can see that the minimum value of f is −28, attained at (2, 0)
while the maximum value is 70, attained at (−5, 0).
Math 243
(Practice) Final Exam - Page 13 of 16
5/11/2015
9. (10 points) Find the direction of maximum increase of the function
f (x, y, z) = yx2 + z 2
at the point with the coordinates (−1, 2, −2). Find the directional derivative of f at
that point in this direction.
Solution:
~ ∇f
(−1,2,−2)
= h2xy, x2 , 2zi(−1,2,−2) = h−4, 1, −4i.
The directional derivative is the magnitude of this vector, which is
√
33.
Math 243
(Practice) Final Exam - Page 14 of 16
5/11/2015
10. (15 points) Find a direction in which the derivative of f (x, y) = xy + y 2 at (3, 2) is 0.
Then find a vector equation for the line that passes through (3, 2) that is parallel to this
direction.
Solution:
~
∇f (3,2)
= hy, x + 2yi|(3,2) = h2, 7i.
f increases the fastest in the direction of this vector, and has a directional derivative
of 0 in any direction perpendicular to this vector, for example, in the direction of
h7, −2i. The line is then given by the equation
~r(t) = h3, 2i + th7, −2i.
Math 243
(Practice) Final Exam - Page 15 of 16
5/11/2015
11. (20 points) You are required to construct a closed rectangular box that encloses a volume
of 512 cm3 . What dimensions of the box would minimize its surface area?
Solution: Let x, y, and z represent the dimensions of the box. Then g(x, y, z) =
xyz − 512 = 0 is the constraint and f (x, y, z) = 2xy + 2xz + 2yz is the surface area
we want to minimize. Then
~ = hyz, xz, xyi
∇g
and
~ = h2y + 2z, 2x + 2z, 2x + 2yi.
∇f
We get the system of equations
2y + 2z
2x + 2z
2x + 2y
xyz
= λyz
= λzx
= λxy
= 512
Note that x, y, and z must be positive (they represent the lengths of the sides of
the box, after all). Multiply the first equation by x and the second by y to get
2xy + 2xz = 2xy + 2yz, and so 2xz = 2yz, which means that x = y (we can cancel
z, because we know it is nonzero). Similarly, we can conclude from the second and
third equations that y = z, so x = y = z, which means they all equal 8 cm.
Math 243
(Practice) Final Exam - Page 16 of 16
5/11/2015
12. (20 points) Use the Lagrange multiplier method to find the point (or points) on the
curve 2y = x2 nearest to the point (0, 3). Clearly state the function f (x, y) whose
extrema you wish to find, and the function g(x, y) in the constraint g(x, y) = 0. Explain
why there is no point on the curve that is farthest from (0, 3).
Solution: Let f (x, y, z) = x2 + (y − 3)2 , which the square of the distance between
the point (x, y) and (0, 3), and let g(x, y) = 2y − x2 . Then
~ = h2x, 2y − 6i
∇f
and
~ = h−2x, 2i,
∇g
so we get the system of equations
2x = −λ2x
2y − 6 = λ2
2y = x2
Eliminating λ from the first and second equations gives us 2x = −x(2y − 6), which
can be rearranged to give x(y − 2) = 0. From x = 0, we get the point (0, 0) and from
y = 2, the points (2,
√ 2) and (−2, 2). Of these (2, 2) and (−2, 2) are at the minimum
distance, which is 5. There is no farthest point, because the graph of 2y = x2 is
unbounded.