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FINAL EXAM SOLUTIONS
Math 51, Spring 2003.
You have 3 hours.
No notes, no books, no calculators. YOU MUST SHOW ALL WORK AND EXPLAIN ALL REASONING TO RECEIVE CREDIT Good luck!
Name
ID number
- (/40 points)
- (/40 points)
- (/40 points)
- (/40 points)
- (/40 points)
Bonus (/20 points)
Total (/200 points)
“On my honor, I have neither given nor received any aid on this examination. I have furthermore abided by all other aspects of the honor code with respect to this examination.”
Signature:
Circle your TA’s name:
Byoung-du Kim (2 and 6)
Ted Hwa (3 and 7)
Jacob Shapiro (4 and 8)
Ryan Vinroot (A02)
Michel Grueneberg (A03)
Circle your section meeting time:
11:00am 1:15pm 7pm
- Consider the function f : R^2 → R^2 given by
f
([
x y
])
xy (x^2 +y^2 )^2 if
[
x y
]
[
]
0 if
[
x y
]
[
]
(a) Do ∂f ∂x and ∂f ∂y exist at the origin? If yes, compute them; if not, explain why.
Solution: We can compute these partial derivatives directly from the definition:
∂f ∂x
= lim h→ 0
f
([
0 + h 0
])
− f
([
])
h
= lim h→ 0
h
= 0
∂f ∂y
= lim h→ 0
f
([
0 + h
])
− f
([
])
h
= lim h→ 0
h
= 0
- (a) Suppose that a function f is differentiable at a given point −→a. Use the following theorem to derive the formula for the Jacobian matrix in terms of the partial deriva- tives of the components of f.
Theorem: If f is differentiable at −→a , then for any vector −→v ,
D−→v f (−→a ) = Df,−→a (−→v )
Solution: The Jacobian matrix is defined as the matrix Jf,−→a such that
Jf,−→a −→v = Df,−→a (−→v )
We know from linear algebra that the columns of a matrix are simply the images of the standard basis vectors by the corresponding linear transformation. So we have
ith^ column of Jf,−→a = Df,−→a (−→e (^) i)
The theorem above then tells us that this can be computed as the corresponding vector derivative; but of course since the vector in question is a unit vector, this can be interpreted as a directional derivative.
ith^ column of Jf,−→a = D−→e (^) i f (−→a )
Of course the limit defining this particular directional derivative is identical to that defining the corresponding partial derivative. So we get
ith^ column of Jf,−→a =
∂f ∂xi
Therefore we have
Jf,−→a =
∂f ∂x 1 · · ·^
∂f ∂xn | |
or we can expand the individual components of each column to get
Jf,−→a =
∂f 1 ∂x 1
∂f 1 ∂x 2 · · ·^
∂f 1 ∂f^ ∂xn 2 ∂x 1 · · ·^
∂f 2 ∂xn .. .
∂fm ∂x 1
∂fm ∂x 2 · · ·^
∂fm ∂xn
(b) Suppose that f is differentiable at the origin, and that f (
- Suppose further that
D^2 41 1
3 5
f (
D^2
3 5
f (
Use the Jacobian matrix to estimate the value of f
([
])
Solution: Since f is differentiable at the origin, we know that the derivative trans- formation exists; and the above vector derivatives allow us to conclude that
Df,−→ 0
([
])
(^) Df,−→ 0
([
])
We can then conclude the columns of the Jacobian matrix by looking at the appro- priate linear combinations of these vectors.
Df,−→ 0 (−→e 1 ) =
Df,−→ 0
([
])
([
]))
Df,−→ 0 (−→e 2 ) =
Df,−→ 0
([
])
− Df,−→ 0
([
]))
The Jacobian matrix is thus
Jf,−→ 0 =
The desired estimate is then
f
([
])
≈ f (
0 ) + Jf,−→ 0
[
]
[
]
Combining these two facts, we see that we can compute the desired partial derivative as the dot product of the first row of the left matrix with the second column of the right matrix. This gives us
∂y ∂u
= − sec^2 (3x − w)
t + 2u + v
t + u + v + 1
− sec^2 (3x − w) √ t + 2u + v
3 sec^2 (3x − w) 2
t + u + v + 1
− sec^2 (
t + u + v + 1 −
t + 2u + v) √ t + 2u + v
3 sec^2 (
t + u + v + 1 −
t + 2u + v) 2
t + u + v + 1
(c) Compute ∂z ∂n at the point
[
m n
]
[
]
(Hint: Express the equations above as a composition of four multivariable functions; then use the chain rule. It is also encouraged that you fully compute and evaluate the FIRST matrix (the one that acts first) before moving on to the others. Make sure to evaluate at the origin BEFORE you multiply.)
Solution: Similarly to the solution to the previous problem, we write the given equations in terms of multivariable functions.
d : R^2 → R^3 d
([
m n
])
cos(2m − 3 n) cos(m + n) cos(m − n)
q r s
e : R^3 → R^3 e
q r s
q + r + s ln(q + r + s) ln(eq^ + er^ + es)
t u v
f : R^3 → R^2 f
t u v
[ √
√ t^ + 2u^ +^ v t + u + v + 1
]
[
w x
]
g : R^2 → R^2 g
([
w x
])
[
tan(3x − w) e^3 w+2x
]
[
y z
]
The desired partial derivative is one of the components of the Jacobian matrix for the composition of these four functions, which of course the chain rule allows us to write as the product of the Jacobians of the individual functions.
As suggested in the hint, we compute the Jacobian of d first.
J
d,
2 40 0
3 5
−2 sin(2m − 3 n) 3 sin(2m − 3 n)
− sin(m + n) − sin(m + n)
− sin(m − n) sin(m − n)
2 40 0
3 5
Since this first Jacobian matrix is identically zero, the product of the four Jacobians must be zero, independent of what the other Jacobians turn out to be! So, the Ja- cobian of the composition is identically zero, and thus the desired partial derivative must be zero.
(b) Suppose Bob tentatively decides to purchase what we will call “Combination A”, including components with d 1 = .10, d 2 = .05, d 3 = .02. He then decides he can afford to spend a few extra dollars, and thus considers two other alternatives: i. purchasing a more expensive amplifier that would reduce his d 1 to. ii. purchasing a more expensive pair of speakers that would reduce his d 3 to.
Using the gradient vector at Combination A, determine which of these two subse- quent options would be best for him; in other words, would make the most significant reduction to the total apparent distortion?
Solution: The gradient vector at the point representing Combination A is
∇f
4(.10)^3
4(.05)^3
4(.02)^3
We can use this to approximate the change in D caused by each of the two options:
i.
∆D ≈ ∇f
(^) · ∆d
ii.
∆D ≈ ∇f
(^) · ∆d
The greatest reduction happens with the first option; so, the extra money is in this case better spent on getting a better amplifier.
- Eventually Bob decides to make his decisions even more analytically. Of course he knows that the more money he spends on a given component, the lower its distortion level will be; along these lines, he collects enough data to determine precisely how the distortion of each component depends on the amount of money he spends on that component; in particular, he concludes that
d 1 = 8 e−p^1 /^8
4 = 8 e−p^1 /^4096 d 2 = 5 e−p^2 /^5 4 = 5 e−p^2 /^625 d 3 = 6 e−p^3 /^6
4 = 6 e−p^3 /^1296
where p 1 , p 2 , p 3 are the prices he pays for the amplifier, CD player, and speakers, respec- tively.
(a) Write the equations above as a single function from R^3 to R^3 , and then use the chain rule to find the gradient of D thought of as a function of p 1 , p 2 , and p 3.
Solution: Written as a single function h : R^3 → R^3 , we have
h
p 1 p 2 p 3
8 e−p^1 /^8 4
5 e−p^2 /^5 4
6 e−p^3 /^6
4
To determine the gradient of D thought of as a function of p 1 , p 2 , and p 3 , we are really asking for the gradient of the composition f ◦ h; we will determine that from the Jacobian, which we can compute with the chain rule.
Jf ◦h = Jf Jh
4 d^31 4 d^32 4 d^33
− 1 83 e
−p 1 / (^840 )
0 − 531 e−p^2 /^5 4 0
0 0 − 631 e−p^3 /^6
4
4(8e−p^1 /^8 4 )^3 − 831 e−p^1 /^8 4 4(5e−p^2 /^5 4 )^3 − 531 e−p^2 /^5 4 4(6e−p^3 /^6 4 )^3 − 631 e−p^3 /^6
− 4 e−^4 p^1 /^8 4 − 4 e−^4 p^2 /^5 4 − 4 e−^4 p^3 /^6
So we have that
∇(f ◦ h) =
− 4 e−^4 p^1 /^8 4
− 4 e−^4 p^2 /^5 4
− 4 e−^4 p^3 /^6
4
Bonus Question:
Using only Math 51 techniques, find a function f : R^2 → R, (or prove that such a function does not exist), with
∇f =
[
−y x
]
Solution: If this function were to exist, then we would have
∂f ∂x ∂f ∂y
[
−y x
]
This means that both first partial derivatives of f exist and are continuous, and in fact are also continuously differentiable. So, the function f itself must also be continuously differentiable since they are polynomials.
Given this, we conclude that the mixed partial derivatives of f must be equal. However, the same equation tells us that
∂ ∂y
∂f ∂x
∂y
(−y)
∂x
∂f ∂y
∂x
(x)
This is a contradiction. So, this function f cannot exist.
