Download 1. /8 Answer the following short questions. Each question is

1. Answer the following short questions. Each question is independent of the others. No justification is
needed. One point each for part.
(a) Let Π be the plane in R3 whose equation is given by x + 2y + 3z = 1. Write down a three-variable
function g(x, y, z) such that Π is a level set of g.
ANSWER:
g(x, y, z) = x + 2y + 3z
REMARK:
A level set of g is represented by an equation of the form g(x, y, z) = c where c is a
constant. The given equation is x + 2y + 3z = 1, so by letting g(x, y, z) = x + 2y + 3z and c = 1, then the
plane equation becomes g(x, y, z) = 1 which is a level set of g.
(b) Let Π be the plane in R3 whose equation is given by x + 2y + 3z = 1. Write down a two-variable
function f (x, y) such that the plane Π is the graph of this function f .
ANSWER:
f (x, y) = 31 (1 − x − 2y)
REMARK:
A graph of f is represented by an equation z = f (x, y). Therefore, in order to find
such an f , we need to rearrange the given equation x + 2y + 3z = 1 so that z becomes the subject
of the equation. By a simple rearrangement, we get 3z = 1 − x − 2y and so z = 13 (1 − x − 2y). Let
f (x, y) = 13 (1 − x − 2y) then the plane is represented by the graph z = f (x, y).
(c) Consider the parametric curve r(t) = (cos 2t) i + (sin 2t) j. Write down an arc-length parametrization
r(s) of this curve.
ANSWER:
r(s) = (cos s) i + (sin s) j
REMARK:
By a simple computation, we get |r0 (t)| = 2, and so s = 2t or equivalently t = 2s .
Replace all t’s in r(t) by 2s , we get an arc-length parametrization r(s) = (cos 2( 2s )) i + (sin 2( 2s )) j.
(d) Let f (x, y) be a two-variable function such that ∇f (0, 0) = 3i + 4j. Write down a unit vector u such
that Du f (0, 0) is maximized.
ANSWER:
u = 35 i + 45 j
REMARK:
In class we proved that Du f (0, 0) = ∇f (0, 0) · u. By the definition of dot product,
∇f (0, 0) · u = |∇f (0, 0)| |u| cos θ where θ is the angle between ∇f (0, 0) and u. Since ∇f (0, 0) is already
given and u is unit, in order to maximize ∇f (0, 0) · u we need θ = 0. Therefore, the unit vector u
needs to be in the same direction as ∇f (0, 0) = 3i + 4j, and hence:
u=
3i + 4j
∇f (0, 0)
=
.
5
|∇f (0, 0)|
(e) Let f (x, y) be a two-variable function such that (x0 , y0 ) is its critical point. Suppose we know:
2
fxx fyy − fxy
(x0 , y0 ) > 0 and fyy (x0 , y0 ) < 0
What is the nature of the critical point (x0 , y0 )? Circle the correct answer:
saddle
not enough data
local minimum
local maximum
2
2
REMARK:
At (x0 , y0 ), it is given that fxx fyy > fxy . Since fxy must be non-negative, the product
fxx fyy must be positive. Therefore, fxx (x0 , y0 ) and fyy (x0 , y0 ) must be of the same sign, i.e. both
negative. By the Second Derivative Test, it is a local maximum. See 02:11 of Lecture 11 Video.
(f) Let f (x, y) be a two-variable function such that (0, 0) is its critical point. Suppose the Taylor’s series
of f about (0, 0) is given by:
f (x, y) = f (0, 0) − x2 + xy − y 2 + higher order terms
What is the nature of the critical point (0, 0)? Circle the correct answer:
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local minimum
local maximum
saddle
not enough data
The nature of a critical point is determined by the local behavior around the critical
REMARK:
point (0, 0), so we can consider only points (x, y) near (0, 0). The higher order terms is then
much smaller than the quadratic terms x2 , xy and y 2 , and so can be neglected. Near (0, 0), the
function f (x, y) is approximately equal to f (0, 0) − x2 + xy − y 2 . The quadratic function −x2 + xy − y 2
has discriminant ∆ = 12 − 4(−1)(−1) < 0 and the x2 -coefficient is −1. From elementary algebra,
−x2 + xy − y 2 ≤ 0 for any (x, y) and so f (x, y) ≤ f (0, 0) when (x, y) is close to (0, 0). In other words,
f (0, 0) is the maximum possible value when (x, y) is near (0, 0). It is a local maximum.
(g) Let f (x, y) be a two-variable function such that (x0 , y0 ) is its critical point. Suppose we know:
fxx (x0 , y0 ) = 25, fxy (x0 , y0 ) = 5, fyy (x0 , y0 ) = 1.
What is the nature of the critical point (x0 , y0 )? Circle the correct answer:
local minimum
local maximum
saddle
not enough data
2
fxx fyy − fxy
REMARK:
Since
= 0 at (x0 , y0 ), the Second Derivative Test is inconclusive. There is
no other information about f given here, so we cannot conclude the nature of the critical point.
2. The spherical coordinates (ρ, θ, ϕ) in R3 are related to the rectangular coordinates (x, y, z) by the following
conversion rule:
x = ρ sin ϕ cos θ
y = ρ sin ϕ sin θ
z = ρ cos ϕ
Let f (x, y, z) be a three-variable function. By the above conversion rule, f can also be regarded as a
function of (ρ, θ, ϕ).
(a) Using the chain rule, show that:
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∂f
∂f
∂f
= −y
+x
∂θ
∂x
∂y
Solution: By the chain rule,
∂f
∂f ∂x ∂f ∂y ∂f ∂z
=
+
+
.
∂θ ∂x ∂θ ∂y ∂θ ∂z ∂θ
By direct computation, we get:
∂x
= −ρ sin ϕ sin θ = −y
∂θ
∂y
= ρ sin ϕ cos θ = x
∂θ
∂z
=0
∂θ
Therefore, by plugging these in to the chain rule above, we get:
∂f
∂f
∂f
= −y
+x
.
∂θ
∂x
∂y
REMARK:
Since f also depends on z, although z does not depend on θ. Therefore, you
∂f ∂z
should include the ∂z ∂θ
term in your work, and explain why it is zero.
2
(b) Suppose ∇f (x, y, z) is orthogonal to the vector −yi + xj for any (x, y, z) in R3 . Show that f does not
depend on θ.
/4
Solution: Given that
∇f (x, y, z) · (−yi + xj) = 0
(by orthogonality)
We get
!
∂f
∂f
∂f
i+
j+
k · (−yi + xj) = 0
∂x
∂y
∂z
∂f
∂f
∂f
−y
+x
+0·
=0
∂x
∂y
∂z
∂f
∂f
−y
+x
= 0.
∂x
∂y
Hence from part (a), we get
∂f
∂θ
= −y
∂f
∂x
+x
∂f
∂y
= 0, and so f is independent of θ.
TIPS:
Parts within the same problem are likely related to each other or under the same
theme. If you are stuck in (b), ALWAYS look at what was proved in (a). You can use the result
stated in (a) even though you cannot do it.
3. Consider the function f (x, y) = 1 − x2 − y 2 and the point P = (1, 0, 0).
(a) Verify that the point P is on the graph of f .
/2
Solution: We can calculate that
f (1, 0) = 1 − 12 − 02 = 0
and therefore P = (1, 0, 0) is on the graph of f .
(b) Find a normal vector n of the tangent plane to the graph of f at P . Justify your work.
/4
Solution: The graph of f is the level set g = 0 of the function
g(x, y, z) = z − f (x, y) = z − 1 + x2 + y 2 .
Since the gradient ∇g(1, 0, 0) is orthogonal to the level sets of g, it is therefore also orthogonal
to the graph of f . So we can calculate
∇g(x, y, z) = h2x, 2y, 1i
and plug in P to obtain
n = ∇g(1, 0, 0) = h2, 0, 1i .
REMARK:
Since we ask you to justify your work, we expect you to explain as least why
n = ∇g(1, 0, 0) – mention that ∇g(1, 0, 0) is orthogonal to the level surface g = 0, which is also
the graph of f . Do NOT study multivariable calculus (or any other sciences or engineering
courses) in a mechanical way! Understand and be able to explain what you are doing!
(c) Find the equation of the tangent plane to the graph of f at P .
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/2
Solution: The vector n = h2, 0, 1i from above is the normal vector to the tangent plane. Therefore the tangent plane is given by the equation
n · hx, y, zi = n · P
and by plugging in values for n and P we find
2x + z = 2
as the equation of the tangent plane of f (x, y) at P .
2
2
4. (a) Consider the function f (x, y) = e−(x +y ) .
i. Find all critical point(s) of f (x, y).
/2
Solution: The gradient of f (x, y) is given by
∇f (x, y) = h−2xf (x, y), −2yf (x, y)i .
Setting ∇f (x, y) = 0 we find that (x, y) = (0, 0) is the only critical point, since f (x, y) > 0 for
all (x, y) ∈ R2 .
ii. It can be computed that (you do not need to verify these):
fxx = (4x2 − 2)e−(x
fyx = 4xye−(x
/4
2 +y 2 )
2 +y 2 )
fyy = (4y 2 − 2)e−(x
2 +y 2 )
Determine the nature of each critical point of f found in (a)(i)
Solution: Using the given formulas we calculate
fxx (0, 0) = −2 < 0
as well as
2
fxx (0, 0)fyy (0, 0) − fxy (0, 0) = (−2)(−2) − 02 = 4 > 0 .
The second derivative test therefore tells us that f admits a local maximum at the critical
point (0, 0).
(b) Consider the function g(x, y) = x4 + 6x2 y 2 − y 4 . It can be shown that (0, 0) is a critical point of g and
2 = 0 at (x, y) = (0, 0) – you do not need to verify all these. Determine the nature of
that gxx gyy − gxy
the critical point (0, 0).
Solution: On the one hand, the restriction of f to the x-axis is the function x 7→ f (x, 0) = x4 ,
which admits a local minimum at x = 0. On the other hand the restriction of f to the y-axis is
the function y 7→ f (0, y) = −y 4 , which admits a local maximum at y = 0. So the function f (x, y)
is increasing when moving away from (0, 0) along x-axis and decreasing when moving away
from (0, 0) along the y-axis. Therefore f (x, y) must have a saddle point at (x, y) = (0, 0).
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/4
5. Consider the function f (x, y) = x2 + 2y 2 − 4x. Let D be the semi-disk region (shaded in the diagram
below) defined by:
x ≥ 0 and x2 + y 2 ≤ 9.
The boundary of D has two components: L is the line segment joining (0, 3) and (0, −3), and Γ is the
circular arc joining (0, 3) and (0, −3).
y
3
L
D
x
Γ
−3
(a) Using Lagrange’s Multiplier, find the minimum value of f (x, y) when (x, y) is restricted to Γ .
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Solution: The circular arc Γ is part of the level set x2 + y 2 = 9. Let g(x, y) = x2 + y 2 then g = 9,
under the condition that x ≥ 0, is the constraint.
[Attn: The inequality x2 + y 2 ≤ 9 describes the region, not the boundary arc!]
Set-up the Lagrange’s Multiplier system:
∂f
∂g
=λ
∂x
∂x
∂f
∂g
=λ
∂y
∂y
2x − 4 = 2λx
4y = 2λy
x2 + y 2 = 9
g(x, y) = 9
The second equation simplifies to 2y = λy. [However, you can’t simply cancel out the y here,
because it can be zero! We need to branch out into two cases!] – Case (a): y , 0; Case (b): y = 0.
Case (a): y , 0, then λ = 2 after canceling out the y from 2y = λy. Substitute λ = 2 into the first
equation, we get:
2x − 4 = 4x ⇒ x = −2
However, the arc Γ is on the side with positive x-coordinates. Therefore, we should reject this
solution (i.e. no solution in this case).
Case (b): y = 0, then from the constraint equation x2 + y 2 = 9, we get x2 = 9 and so x = 3
(again we only consider the points with x ≥ 0). Therefore, we get one boundary critical point
(x, y) = (3, 0)
Since Γ is not a closed curve, the minimum may take place at end-points of Γ as well. Evaluate
f at:
end-point
(0, −3) f (0, −3) = 18
end-point
(0, 3)
f (0, 3) = 18
boundary critical point
(3, 0)
f (3, 0) = −3
Evidently, the minimum value of f on Γ is −3 , which is attained at (3, 0).
(b) Find the minimum value of f (x, y) when (x, y) is restricted to L.
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/2
Solution: The constraint L is part of the y-axis, i.e. x = 0. [Although it works by using
Lagrange’s Multiplier by letting g(x, y) = x, it is easier to do it by substituting x = 0 into the
function f so that it becomes a single-variable problem.]
Since f (0, y) = 2y 2 ≥ 0 = f (0, 0) for any −3 ≤ y ≤ 3. Therefore, the minimum value of f restricted
on L is 0, attained at the point (0, 0).
(c) Determine the absolute minimum value of f (x, y) over the semi-disk region D.
/4
Solution:
COMMON MISTAKE:
Since D is a solid region, the minimum value of f over D may take
place in the interior or on the boundary. We have figured out the minimum value of f restricted
on the boundary already, but we need to see if there is any interior critical point!
To find interior critical points, we solve:
∂f
=0
∂x
∂f
=0
∂y
2x − 4 = 0
4y = 0
The only solution is (x, y) = (2, 0), which is in the region D, and so we should consider it as a
candidate point as well. Evaluate f at:
end-point
(0, −3) f (0, −3) = 18
end-point
(0, 3)
f (0, 3) = 18
minimum on Γ
(3, 0)
f (3, 0) = −3
minimum on L
(0, 0)
f (0, 0) = 0
interior critical point
(2, 0)
f (2, 0) = −4
Therefore, the absolute minimum value of f over the domain D is −4 , attained at (2, 0).
6. Determine whether each of the following statements is True or False. No justification is needed. Please
circle your answer.
(a)
True
False
Given any vector u in R3 , the dot product u · u must be non-negative.
REASON:
Since u · u = |u|2 ≥ 0.
(b)
True
False
Given any vectors u and v in R3 , we have u × v + v × u = 0.
REASON:
By the right-hand rule u × v and v × u are of opposite direction,
so u × v = −v × u.
(c)
True
False
Given any vectors u and v in R3 , the dot product u · v is equal to the angle
between u and v.
REASON:
The dot product is related to the angle by the formula u · v =
|u| |v| cos θ where θ is the angle between the two vectors.
(d)
True
False
Given any vectors u and v in R3 , we have u × v = |u| |v| sin θ where θ is the angle
between u and v.
REASON:
Be careful! The cross product u × v is a VECTOR and |u| |v| sin θ is
a SCALAR. They are different “animal” and so they can’t be equal to each other.
What we stated in class was about the magnitude: |u × v| = |u| |v| sin θ
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(e)
True
False
If r(t) is a parametric curve in R3 such that |r(t)| is a constant independent of t,
then r(t) = Ai + Bj + Ck for some constants A, B and C.
REASON:
There are many scenario that the magnitude |r(t)| is constant. For
instance r(t) = (cos t)i + (sin t)j + 0k is a quick counter-example. This curve has
magnitude |r(t)| = 1 for all time t, but the i and j components are not constants.
(f)
True
False
Let f (x, y) be a two-variable function. The graph of f (x, y) is a level set of the
function g(x, y, z) = z − f (x, y).
REASON:
The graph of f is the equation z = f (x, y), which is z − f (x, y) = 0
by rearrangement. It is the level set g = 0.
(g)
True
False
Let f (x, y, z) = ax + by + cz + d for fixed real numbers a, b, c, d, then the level sets
of f are parallel to each other.
REASON:
The level sets of f are ax + by + cz + d = D where D is a constant.
No matter what D-value we pick, the level set is a plane with normal vector
ai + bj + ck. Therefore, these level sets are parallel to each other.
(h)
True
False
(i)
True
False
Given any parametric curve r(t), the velocity r0 (t) and the acceleration r00 (t) must
be orthogonal to each other at all time t.
It is only true for motion with constant |r0 (t)|. A simple counterREASON:
example is just take r(t) = t 5 i, then r0 (t) and r00 (t) are in fact parallel to each
other.
Given a two-variable function f (x, y), the directional derivative of f at (0, 0) in
the direction parallel to i − 3j can be computed by the formula
∇f (0, 0) · (i − 3j)
REASON:
This formula only works for unit vectors u. The vector i − 3j is
not unit, so the above directional derivative should be instead:
i − 3j
∇f (0, 0) · √ .
10
(j)
(k)
(l)
True
True
True
False
False
False
Given a two-variable function f (x, y) and a point (x0 , y0 ) in the xy-plane,
the gradient ∇f (x0 , y0 ) is always orthogonal to the graph of f at the point
(x0 , y0 , f (x0 , y0 )).
REASON:
∇f (x0 , y0 ) is orthogonal to the LEVEL SET of f (i.e. f = c), not
the GRAPH of f (i.e. z = f (x, y)). The vector that is orthogonal to the graph of f
is the ∇g where g(x, y, z) = z − f (x, y). Since ∇g is orthogonal to the level set of
g = 0 which is the same as the graph of f .
A critical point (x0 , y0 ) of a two-variable function f (x, y) is a point at which the
tangent plane to its graph is orthogonal to k.
REASON:
At the critical point of f , the tangent plane is horizontal and
therefore having k as a normal vector to the tangent plane.
Given a two-variable function f (x, y) with (x0 , y0 ) as its local minimum point,
2 at (x , y ) must be non-zero.
then fxx fyy − fxy
0 0
REASON:
This is a tricky one – we admit that. The Second Derivative
2 > 0 and f
Test said that “If fxx fyy − fxy
xx > 0 at (x0 , y0 ), then (x0 , y0 ) is a local
minimum”. However, the converse, that is “If (x0 , y0 ) is a local minimum then
2 > 0 and f > 0 at (x , y )” is not true! An easy counter-example to
fxx fyy − fxy
xx
0 0
the given statement is f (x, y) = x4 + y 4 , then f has (0, 0) as its local minimum
2 = 0 at (0, 0).
point, but fxx fyy − fxy
7
7. Given a two-variable function h(x, y) such that for every scalar λ and point (x, y) in R2 , the following
equality holds:
h(λx, λy) = λ2 h(x, y)
Let u = u1 i + u2 j be a unit vector on the xy-plane.
(a) Find an arc-length parametrization r(s), with r(0) = 0, of the straight line on the xy-plane that passes
through (0, 0) and parallel to u.
/2
Solution:
WAY OF THINKING:
The problem asks for an arc-length parametrization of a certain path.
In class we learned how to find an arc-length parametrization from a given parametrization r(t),
so why not find any parametrization r(t) of the path first, then there is a standard procedure to
find the r(s).
To parametrize a straight-line, we need a point and a direction vector – they are both given in
the problem, namely: point (0, 0) and direction u. Therefore,
r(t) = h0 + tu1 , 0 + tu2 i = htu1 , tu2 i.
Then, r0 (t) = hu1 , u2 i = u, which is a unit vector! Therefore, this r(t) is already an arc-length
parametrization and so s = t:
r(s) = hsu1 , su2 i.
One can easily verify that it fulfills the requirement r(0) = h0 · u1 , 0 · u2 i = 0.
(b) Show that h(r(s)) = s2 h(u) where r(s) is the parametric equation of the straight line found in (a).
[Notation: For a vector v = v1 i + v2 j, the notation h(v) simply means h(v1 , v2 ).]
/3
Solution:
WAY OF THINKING:
Now you are asked to show h(r(s)) equals to something, so why not
plugging in r(s) found in (a) into h and see if there is a way out!
h(r(s)) = h(su1 i + su2 j)
= h(su1 , su2 )
= s2 h(u1 , u2 )
(using the given equality by plugging in λ = s, x = u1 and y = u2 )
2
= s h(u1 i + u2 j)
= s2 h(u).
(c) Using (b), find the directional derivative Du h(0, 0).
Solution:
REMARK:
This is probably the hardest problem in the test. The hint “using (b)” suggested
that it should have something to do with the straight-line through (0, 0) parallel to u, and
the value of h along this straight-line. If you paid attention to how directional derivative is
geometrically defined, you might probably realize that the intention of the problem is to figure
out Du h(0, 0) from its geometric definition.
The directional derivative Du h(0, 0) is defined to be the rate of change of h(x, y) when (x, y)
travels along the straight-line through (0, 0) in the direction of u. Along this straight-line,
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/3
hx, yi = r(s) where r(s) is the path found in (a). Therefore, the above rate of change is given by:
d
d
h(r(s)) = s2 h(u) = 2s h(u)
ds
ds
according to (b).
The origin (0, 0) corresponds to s = 0 of the path r(s) since r(0) = 0, and so the direction
derivative Du h(0, 0) is given by the above rate of change evaluated at s = 0:
Du h(0, 0) = 2s h(u)|s=0 = 0.
Number of students who succeeded in this part ≈ 10
8. Given a fixed point (x0 , y0 , z0 ) in R3 , we denote its position vector by p = x0 i + y0 j + z0 k. Let r(t) be the
path of a particle in R3 with mass m. The angular momentum about p of this particle is defined to be:
Lp (t) := (r(t) − p) × mr0 (t).
Suppose r(t) − p is parallel to r00 (t) at all time t.
(a) Show that Lp (t) is conserved, i.e. independent of t.
/4
Solution: [c.f. Homework 2]
By product rule:
dLp
d
d
= (r(t) − p) × mr0 (t) + (r(t) − p) × mr0 (t)
dt
dt
dt
= r0 (t) × mr0 (t) + (r(t) − p) × mr00 (t)
r0 (t) × mr0 (t) = 0 since r0 (t) and mr0 (t) are parallel – one is a constant multiple of another.
(r(t) − p) × r00 (t) = 0 too as r(t) − p and r00 (t) are given to be parallel.
Therefore, we have L0p = 0 and so Lp is a constant vector.
(b) Using (a), show that the path of the particle is contained in a single plane.
[Hint: consider the dot product Lp · (r(t) − p).]
Solution: [c.f. Lecture 3 (see YouTube video posted)]
Using (a), we know Lp is a constant vector, so we can let it to be Lp = Ai + Bj + Ck where A, B
and C are constants.
By the definition of cross product, Lp = (r(t) − p) × mr0 (t) must be orthogonal to both r(t) − p
and mr0 (t). Therefore, the following dot product equals zero:
Lp · (r(t) − p) = 0
Let r(t) = x(t)i + y(t)j + z(t)k. By rewriting the above in terms of components, we get:
(Ai + Bj + Ck) · hx(t) − x0 , y(t) − y0 , z(t) − z0 i = 0
A(x(t) − x0 ) + B(y(t) − y0 ) + C(z(t) − z0 ) = 0
In other words, the path (x(t), y(t), z(t)) satisfies the plane equation A(x−x0 )+B(y−y0 )+C(z−z0 ) =
0 at all time t. Therefore, the path of the particle is contained in the above plane.
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