Review Session Questions
Michelle Bodnar
December 7, 2015
These are some of the questions I covered during my review session. Please
let me know right away if you spot any mistakes! I make no promises that this
document is error-free.
Problem 1
Find the decomposition a = a|| + a⊥ with respect to b, where a = h2, 5i and
b = h1, 1i.
Solution: Recall that a|| is the projection of a along b. It is given by the
formula
h2, 5ih1, 1i
2+5
a·b
b=
h1, 1i =
h1, 1i = h7/2, 7/2i.
a|| =
b·b
h1, 1ih1, 1i
1+1
To find a⊥ we just subtract:
a⊥ = a − a|| = h2, 5i − h7/2, 7/2i = h−3/2, 3/2i.
Problem 2
Determine whether the lines r1 (t) = h0, 1, 1i + th1, 1, 2i and r2 (s) = h2, 0, 3i +
sh1, 4, 4i intersect. If so, find the point of intersection.
Solution: To decide whether the lines intersect, we need to decide whether
there exist t and s such that r1 (t) = r2 (s), keeping in mind that s and t need not
be equal. Two vectors are equivalent if and only if they are equal componentwise,
so this gives us three equations:
t=2+s
1 + t = 4s
1 + 2t = 3 + 4s
Solving the system of linear equations gives t = 3 and s = 1. (Note: it’s
important to plug these back into the original equations to be sure you actually
achieve equality!) Doing so, we see that r1 (3) = r2 (1) = h3, 4, 7i, so the two
lines intersect at the point (3, 4, 7).
1
Problem 3
A UCSD student is running around campus, where the temperature at location
(x, y) is given by 3x2 y − y 3 . When the student is at (1, 2) she is anxious to move
in the direction in which the temperature drops the most rapidly.
a. Find a unit vector pointing in the direction she should move.
b. If she mistakenly moves toward the point (0, 1), what rate of change in temperature will she experience?
Solution: Recall that the gradient vector points in the direction of steepest
ascent, which means that the negative of the gradient points in the direction of
steepest descent, ie the direction in which the function drops most rapidly. We’ll
compute this, then normalize it to obtain a unit vector in the desired direction.
Let f (x, y) = 3x2 y − y 3 . Then
−∇f = −h6xy, 3x2 − 3y 2 i
so we have
−∇f (1, 2) = −h12, −9i = h−12, 9i.
Normalizing gives the final answer:
h−12, 9i
1
h−12, 9i
=√
=
h−12, 9i = h−4/5, 3/5i.
||h−12, 9i||
15
144 + 81
For part (b), we need to read carefully. The student isn’t moving in the
direction (0,1). She is moving in the direction of the point (0,1), so we need to
find the vector which points from (1,2) to (0,1) to use as our direction vector,
then compute the directional derivative in that direction. Subtracting, we see
that our direction vector is h−1, −1i. Thus, her rate of change is
√
1
1
∇f (1, 2) · h−1, −1i = √ h12, −9i · h−1, −1i = −3/ 2.
||h−1, −1i||
2
Problem 4
Find the global maximum and minimum of the function f (x, y) =
over the domain x2 + 3y 2 ≤ 4.
x3
3
2
+ x2 + 3y 2
Solution: The problem has two points. First we’ll look for critical points
in the plane, and pick on those which lie inside our domian. Next, we’ll look
for critical point on the boundary of the domain using the Lagrange multipliers
method. Then we’ll gather up all our critical points, evaluate the function at
all of them, and look for the global min and max.
2
To find critical points in the plane, we compute the partial derivatives of f
and set them equal to 0.
fx = x2 + x = 0
and
fy = 6y = 0.
The first equations tells us that x = 0 or −1 and the second equation tells
us that y = 0. Thus, the two critical points in the plane are (0, 0) and (−1, 0).
Both of them satisfy x2 + 3y 2 ≤ 4, so we care about both of them since they lie
in the domain.
The next step is to find critical points on the boundary. Remember, the
boundary is given by the equation x2 + 3y 2 = 4. Notice how the inequality is
gone!! Setting ∇f = λ∇g we have
hx2 + x, 6yi = λh2x, 6yi.
Our two Lagrange equations are
x2 + x = λ2x and 6y = λ6y.
Looking at the first Lagrange equation, we’d like to divide both sides by 2x.
However, we need to be careful,√because x could be 0. √
If x = 0, our constraint
√
equation tells us that y = ±2/ 3. The points (0, −2/ 3) and (0, 2/ 3) both
satisfy BOTH Lagrange equations for some value of λ, so they’re both critical
points we need to add to our list. (Note: If one of these points caused one or
more of the Lagrange equations to be false, we would throw it away.)
Now that we’ve handled the x = 0 case, we’ll assume that x 6= 0. Dividing
both sides by 2x gives λ = x+1
2 . Now we look to the second Lagrange equations.
We’d like to solve for λ here as well, but again we need to worry about what
happens if y = 0. In this case, the constraint equation tells me that x = ±2.
Since the points (−2, 0) and (2, 0) both satisfy both Lagrange equations for some
value of λ, we add them to the list.
Finally, we consider the case where neither x nor y is 0. The second Lagrange
equations then implies that λ = 1. The first equation says λ = x+1
2 , so we must
have x = 1. When x = 1, the constraint equations implies that y = ±1, so
we add the critical points (1, −1) and (1, 1) to our list. Now that we have a
complete list of critical points, we evaluate the function at all of them:
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Critical Point f (x, y)
(0, 0)
0
(−1, 0)
1/6
√
(0, −2/√ 3)
4
(0, 2/ 3)
4
(−2, 0)
−2/3
14/3
(2, 0)
(1, −1)
23/6
23/6
(1, 1)
Scanning the list for the largest and smallest functions values, we see that
the global maximum is 14/3 and occurs at (2,0). The global minimum is -2/3
and occurs at (-2,0).
Problem 5
Let f (x, y) =
p
cos(x) + sin(y). Approximate f (.01, .04).
Solution: Since (.01, .04) is close to (0, 0), we’ll use a linear approximation
of the function at the origin. Recall that the linear approximation of f (x, y)
and (a, b) is given by
L(x, y) = f (a, b) + fx (a, b)(x − a) + fy (a, b)(y − b).
In our case, (a, b) = (0, 0). We’ll compute each piece separately then put them
all together.
f (0, 0) = 1
− sin(x)
fx = p
2 cos(x) + sin(y)
fx (0, 0) = 0
cos(y)
fy = p
2 cos(x) + sin(y)
fy (0, 0) = 1/2
Putting it all together,
L(x, y) = 1 + y/2.
Thus, a good approximation is
f (.01, .04) ≈ 1 + .04/2 = 1.02.
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Problem 6
Let W be the region bounded by z = 1−y 2 , y = x2 , z = 0, and y = 1. Calculate
the volume of W .
Solution: We know 0 ≤ z ≤ 1−y 2 , which tells us that y 2 ≤ 1, so −1 ≤ y ≤ 1.
However, we also know that y is bounded between x2 and 1, so x2 must be less
than 1. In other words, x2 ≤ y ≤ 1. This tells me that x2 ≤ 1, so −1 ≤ x ≤ 1.
Since z depends on y and y depends onR x,
R Rit makes sense to integrate in the
order dzdydx. Recalling that Vol(W ) =
1dV , we have
W
Z 1 Z 1 Z 1−y2
Vol(W ) =
dzdydx
x2
1
−1
1
Z
−1
1
Z
Z
=
Z
x2
1
−1
1
=
−1
1
Z
=
0
1 − y 2 dydx
=
Z
0
1−y2
z
dydx
x2
1
y − y 3 /3 2 dx
x
2/3 − x2 + x6 /3dx
−1
1
= (2/3)x − x3 /3 + x7 /21
−1
= 16/21.
Problem 7
Evaluate, or say why the limit does not exist.
xy + xy 2
(x,y)→(0,0) x2 + y 2
lim
Solution: We can approach this by switching to polar coordinates. Let
x = r cos θ and y = r sin θ. Then the limit becomes
r2 cos θ sin θ + r3 cos θ sin2 θ
= lim cos θ sin θ + r cos θ sin2 θ = cos θ sin θ.
r→0
r→0
r2
Since the limit depends on θ, the limit does not exist. To explicitly justify
the claim, note that the limit is 0 when θ = 0 and the limit is 1/2 when θ = π/4.
lim
Problem 8
Let u = h1, 0, 2i, v = h0, 1, −1i, and w = h0, 0, 3i. Compute the volume of the
parallelepiped spanned by these vectors. What is the angle between v and w?
5
Solution: Recall that the volume of the parallelepiped spanned by u, v, and
w is given by |u · (v × w)|. First we compute the cross product of v and w,
which is h3, 0, 0i. Then we take the dot product: h1, 0, 2i · h3, 0, 0i = 3. Thus,
the volume is 3 cubic units. To find the angle between v and w, it is tempting
to use the fact that ||v × w|| = ||v|| · ||w|| sin θ. However, keep in mind that
arcsin(x) is a multivalued function for x between 0 and 1. It is better to use the
cosine formula to avoid any potential errors
√
√ in deciding which angle to choose.
Recall that ||v|| · ||w|| cos θ = v · w. Thus, 2 · 3 · cos θ = −3, so cos θ = −1/√2.
Since θ = 3π/4 is the unique angle between 0 and π such that cos(θ) = −1/ 2,
3π/4 must be the angle between v and w.
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