Midterm Exam II
Math 1321 - Accelerated Engineering Calc II
April 4, 2016
Answer each question completely in the area below. Show all work and explain your reasoning. If
the work is at all ambiguous, it is considered incorrect. No phones, calculators, or notes are
allowed. Anyone found violating these rules or caught cheating will be asked to leave
immediately. Point values are in the square to the left of the question. If there are any other
issues, please ask the instructor.
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well as agree to abide University Honor Code:
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Solutions
Question
Points
1
10
2
10
3
15
4
10
5
15
6
15
7
15
8
10
Total:
100
Score
Potentially Useful Information
x = ρ sin φ cos θ,
y = ρ sin φ sin θ,
z = ρ cos φ,
ρ2 = x 2 + y 2 + z 2 .
Math 1321: Exam 2
10
April 4, 2016
1. While doing math for fun, Chris looks for a particular continuous function f (x, y ) and discovers
that it must satisfy:
∂f
∂f
=y
and
= x 2.
∂x
∂y
Immediately, he panics, knowing no such function f (x, y ) can exist. Why is this the case?
Solution: This problem is deliberately tricky. The key is to take another set of derivatives, first
if we take the y derivative of fx :
∂ ∂f
= 1,
fxy =
∂y ∂x
and then, if we take the x derivative of fy :
fy x
∂
=
∂x
∂f
∂y
= 2x.
Note, that both fxy and fy x are continuous functions that are NOT equal to each other. This
violates Clairaut’s theorem, which says that if these two partials are continuous, they must be
equal. Thus, no such f can exist.
2/12
/10 pts
Math 1321: Exam 2
10
April 4, 2016
2. A skier is on a mountain with altitude described by
z = 100 − 4x 2 − 3y 2 .
The skier is located at the point with xy coordinates (1, 1) and wants to ski downhill along the
steepest possible path. In which direction should the skier begin skiing?
Solution: This is modified very slightly from a practice exam problem and a homework
problem.
The key fact to remember with this problem: at a point, the gradient describes the direction
of the greatest rate of change. In other words, ∇f produces the largest (positive) directional
derivative, and therefore, −∇f produces the largest descent at a point. Thus, we just need to
compute −∇f (1, 1).
∇f = h∂f /∂x, ∂f /∂y i = h−8x, −6y i,
and therefore
−∇f (1, 1) = −h8x, −6y ix=1,y =1 = h8, 6i.
Although we could compute the unit vector in this direction (as this is not one), it still describes
a direction, meaning this suffices as an answer.
3/12
/10 pts
Math 1321: Exam 2
15
April 4, 2016
3. Find all the critical points of f (x, y ) = x 2 y − x 2 − 2y 2 and classify each as a local minimum,
maximum, or saddle point.
Solution: This is taken directly from the practice exam and is nearly identical to a previous
quiz.
To find the critical points, we find where the gradient ∇f = 0. This is actually a vector equation,
saying that both fx = 0 and fy = 0. Computing each of these, we have:
∂f
= 2xy − 2x = 2x(y − 1),
∂x
∂f
= x 2 − 4y .
∂y
Setting fx = 0, we see two possibilities: x = 0 or y = 1, then plugging these conditions into
fy = 0, we get the three critical points are: (0, 0), (2, 1), (−2, 1).
We now classify the equilibrium by the second derivative test which says we must examine the
quantity
2
D = fxx fy y − fxy
,
where fxx = 2y − 2,
fxy = 2x,
fy y = −4,
meaning we have
D = (2y − 2)(−4) − (2x)2 = 8 − 8y − 4x 2 .
We then see, for the point (0, 0), D = 8 > 0 and fxx = −2 < 0, meaning that this is a local
maximum.
For both (2, 1) and (−2, 1), we have D = −16 < 0 and therefore these are both saddle points.
4/12
/15 pts
Math 1321: Exam 2
10
April 4, 2016
4. Minimize the function:
f (x, y ) = −x 2 + y 2
subject to the constraint
g(x, y ) = x 2 + y = 0.
Solution: This problem is a constrained minimization, meaning that we should use Lagrange
multipliers, which say: the minimum occurs when
∇f = λ∇g,
for some unknown scalar λ. Thus, we implicitly have three unknowns: x, y , λ and three equations:
fx = λgx ,
fy = λgy ,
g(x, y ) = x 2 + y = 0.
We see that
∇f = h−2x, 2y i,
∇g = h2x, 1i,
so our system becomes:
−2x = λ2x,
2y = λ,
x 2 + y = 0.
The first equation is only true when either x = 0 or λ = −1.
If we try the first possibility, x = 0, plugging this into our third equation, we have y = 0 as well
and from the second equation, λ = 0. Thus, the value of f (0, 0) = 0 is a possible minimizer.
Exploring the other possibility: λ = −1, we see that this yields
this into
√ y = −1/2. Plugging
√
x 2 + y = 0, we get two values of x that produce this: x = 1/ 2 and x = −1/ 2. When you
plug either of these into f (x, y ), you get −1/4, which clearly is smaller than 0. Thus, these two
values produce the minimization.
5/12
/10 pts
Math 1321: Exam 2
April 4, 2016
5. Consider the following double integral:
ZZ
Z
f (x, y ) dA =
D
5
10
1Z x
(1 + 2y ) dy dx
x2
0
(a) Rewrite (set up, but do not compute) this integral in the opposite integration order, that
is: set up the equivalent integral of the form
ZZ
f (x, y ) dx dy .
(b) Using either setup, compute the actual value of the integral.
Solution: This problem is modified slightly from a previous quiz and was almost explicitly
mentioned to be on the exam.
(a) On the quiz, you were just simply asked to draw the region D, which we saw to be:
1.2
1.0
0.8
y =x
0.6
y = x2
0.4
0.2
0.2
0.4
0.6
0.8
1.0
1.2
In the setup in the original problem, we have the slices going vertically (describing the y
bounds), but now, we can reformulate this by taking the slices horizontally. In this case, we
√
see that x is bounded between y and y , in which case, y varies between 0 and 1, meaning
we have:
Z Z √
Z Z
1
1
x
y
(1 + 2y ) dy dx =
x2
0
(1 + 2y ) dx dy .
0
y
(b) It doesn’t matter which form you take to do the integral. In the off chance you weren’t able
to get part (a), we’ll demonstrate the calculation for the integral provided by the problem.
We first compute the integral of the inside:
Z x
x
(1 + 2y ) dy = y + y 2 y =x 2 = (x + x 2 ) − x 2 − x 4 = x − x 4 .
x2
And now the integral of the outside:
Z
0
1
x2 x5
−
(x − x ) dx =
2
5
4
6/12
x=1
=
x=0
1 1
− .
2 5
/15 pts
Math 1321: Exam 2
April 4, 2016
6. Consider the surface
z = f (x, y ) = x 2 + 3xy − y 2 .
10
5
(a) Find the equation for the tangent plane at the point (x, y ) = (2, 3).
(b) Using your answer from part (a), approximate the value of z when (x, y ) = (2.05, 2.96).
Solution: This is taken verbatim from a quiz.
x
2
4
0
-2
50
z
0
-50
4
2
y
0
-2
(a) We know that the equation of the tangent plane at a point (x0 , y0 , z0 ) on z = f (x, y )
satisfies
z − z0 = fx (x0 , y0 )(x − x0 ) + fy (x0 , y0 )(y − y0 ).
Thus, we need to compute each partial derivative, fx and fy , where we think of the other
variable remaining constant, which leads to:
fx (x, y ) = 2x + 3y ,
fy (x, y ) = 3x − 2y .
We must evaluate these, and the function f (x, y ) itself at the point of interest. First note
that f (2, 3) = 13 = z0 and also
fx (2, 3) = 2(2) + 3(3) = 13,
fy (2, 3) = 3(2) − 2(3) = 0.
Our tangent plane equation is then is simply (note the y terms fall out)
z − 13 = 13(x − 2).
(b) Recall that the tangent plane is described by
z − z0 = fx (x0 , y0 )(x − x0 ) + fy (x0 , y0 )(y − y0 ).
We can interpret this as: near the point (x0 , y0 , z0 ), the value of z = f (x, y ) can be
approximated by the value of z on the tangent plane. We can see this by looking at the
plot of the tangent plane and the true surface above.
7/12
/15 pts
Math 1321: Exam 2
April 4, 2016
Thus, if we think of moving away slightly from the point (x0 , y0 , z0 ) and calling δx = x − x0 ,
δy = y − y0 and δz = z − z0 , we get the form of the differential:
δz = fx (x0 , y0 )δx + fy (x0 , y0 )δy .
In our particular case, we have that δx = 2.05 − 2 = 0.05 and δy = 2.96 − 3 = −0.04,
meaning we can now approximate the true value of z by:
z − 13 ≈ 13(0.05) + 0(−0.04)
8/12
=⇒
z ≈ 13 + 13(.05) = 13.65.
/0 pts
Math 1321: Exam 2
15
April 4, 2016
7. Ethan, a dear friend of yours has landed himself in jail and calls you begging for $250 to bail him
out. You and another friend, Greg, hatch a plan to come up with the money: go to a casino in
Wendover.
You are a solid gambler, but get nervous when you start making a lot of money, so the probability
density function of your winnings (denoted X) is
(
1
x 0 ≤ x ≤ 100
f1 (x) = 5000
0
otherwise.
Greg is a way more experienced gambler and is definitely going to make more. However, once he
makes $200, the casino assumes he is cheating and kicks him out with nothing, so the probability
density function of Greg’s winnings (denoted Y ) is
(
1
e y /200 0 ≤ y ≤ 200
f2 (y ) = 200(e−1)
0
otherwise.
What’s the probability that you and Greg earn enough (combined) to bail Ethan out of jail?
Set up, but do not compute the integral that describes the answer.
Solution: This is modified from a homework problem and was explicitly mentioned to be
on the exam.
It’s clear from the setup of the problem that the two random variables (your earnings and Greg’s
earnings) are independent, meaning that their joint probability density f (x, y ) is just the product
of the two probability densities. That is:
(
1
xe y /200 0 ≤ x ≤ 100, 0 ≤ y ≤ 200
f (x, y ) = f1 (x)f2 (y ) = 10000(e−1)
0
otherwise.
Thus, to compute the probability that a set of outcomes occurs, we describe this set of outcomes
as a region D in 2D space and then
ZZ
probability outcome is in D =
f (x, y ) dA.
D
We now just need to figure out the region D the problem is describing. The first piece of
information is that you and Greg must make enough money to bail Ethan out. Symbolically, this
corresponds to:
X + Y ≥ 250.
That is, combined, you must make more than $250. We also have two other constraints based
on how much each of you can earn:
X ≤ 100,
Y ≤ 200.
Note, these three inequalities define a region D in the XY plane:
9/12
/15 pts
Math 1321: Exam 2
April 4, 2016
240
220
Y = 200
200
D
180
X = 100
X + Y = 250
160
140
120
20
40
60
80
100
120
140
We could set this up as either type integral. Let’s for instance, take vertical slices. We see the
lower limit of y is always y = 250 − x and the upper limit is y = 200. However, the bounds
of x are a little tricky. The upper bound is x = 100, but the lower bound is the intersection of
x + y = 250 and y = 200 which occurs at x = 50, thus we have:
ZZ
Z 100 Z 200
1
f (x, y ) dA =
xe y /200 dy dx.
10000(e
−
1)
D
50
250−x
If you are curious, the answer ends up being approximately 15%, so Ethan has a reasonable shot
of making it out of the slammer!
10/12
/0 pts
Note:
out sphe
Math 1321: Exam 2
10
April 4, 2016
z
8. Set up, but do not compute the following integral in spherical coordinates: (0, 0, 1)
≈+¥+z@=z
ZZZ
f (x, y , z) dV,
V
z=œ„„„„„
≈+¥
x
y
FIGURE 9
Solution: This exact problem was done in class.
To begin, we first convert the two key surfaces (the sphere, the cone) into spherical coordinates
using the conversions provided on the front page.
For the sphere, we have:
ρ2 = ρ cos φ =⇒ ρ = cos φ.
⇔
For the cone:
p
z = x2 + y2
⇔
q
p
ρ cos φ = ρ2 sin2 φ cos2 θ + ρ2 sin2 φ sin2 θ = ρ sin φ sin2 θ + cos2 θ,
which simplifies to
ρ cos φ = ρ sin φ.
When is this true? Only when φ = π/4. Thus, the equation of the cone is φ = π/4, which is
exactly the form of a cone in spherical coordinates, which is assuring.
We can now construct our triple integral. Note that ρ varies from the inside of the cone to the
outside of the sphere, meaning that its bounds are ρ = 0 and ρ = cos φ. The angle φ varies
from vertical, φ = 0, to the outside of the cone, φ = π/4. We also have a full rotation around,
meaning the range of θ is θ = 0 to θ = 2π.
We also need to convert f (x, y ) = xy into spherical coordinates, which becomes:
f (ρ sin φ cos θ, ρ sin φ sin θ) = ρ2 sin2 φ cos θ sin θ.
Lastly, we recall the volume element in spherical:
dV = ρ2 sin φdρ dφ dθ.
Piecing all this information, our triple integral becomes:
ZZZ
Z
2π
Z
π
4
Z
cos φ
f (x, y , z) dV =
V
0
0
ρ2 sin2 φ cos θ sin θ ρ2 sin φ dρ dφ dθ.
0
11/12
EXAM
SOLUTION
where f (x, y , z) = xy , and V is the regionp
below the sphere
x 2 + y 2 + z 2 = z and above the cone z = x 2 + y 2 .
x2 + y2 + z2 = z
v
to find th
sphere x
/10 pts
write the
The equa
Math 1321: Exam 2
April 4, 2016
Bonus Questions
9. (Politics) Of the remaining presidential candidates, only one provided the vocals for a folk album,
titled We Shall Overcome. Which candidate is this?
Solution:
https://www.youtube.com/watch?v=iV4L_OHGpyo
10. (Current Events) After crowdfunding a Nikola Tesla museum, which popular online comic artist
is set to receive the newly announced Tesla Model 3 early, in order to review it?
Solution: The Oatmeal, found here: http://theoatmeal.com/.
Here’s the comic about Tesla himself, worth checking out:
http://theoatmeal.com/comics/tesla
And here’s a comic about the car Tesla:
http://theoatmeal.com/comics/tesla_model_s
12/12
/0 pts