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MATH 51 MIDTERM 2 (MARCH 4, 2010)
Max Murphy Jonathan Campbell Jon Lee Eric Malm 11am 11am 10am 11am 1:15pm 2:15pm 1:15pm 1:15pm Xin Zhou Ken Chan (ACE) Jose Perea Frederick Fong 11am 1:15pm 11am 11am 1:15pm 1:15pm 1:15pm
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Instructions: Find your TA’s name in the table above, and circle the time that your TTh section meets. During the test, you may not use notes, books, or calculators. Read each question carefully, and show all your work. Each of the 10 problems is worth 10 points. You have 90 minutes to do all the problems.
Total
1(a). Find the inverse of the matrix A =
Solution:
To find the inverse of the matrix, we augment it with the identity matrix, and reduce it.
Add the first row from the third, and divide the second row by 2:
Now add the third row to the second, and also add double the third row to the first: (^)
Lastly, subtract the second row from the first:
This gives the inverse as the right side of the augmented matrix.
- Let T : R^2 → R^2 be the linear transformation defined by:
T
[
x y
]
[
x + y − 2 x + 4y
]
(a). Find the matrix A that represents the linear transformation T with respect to the standard basis S = {e 1 , e 2 }.
Solution: Recall e 1 = (1, 0)T^ and e 2 = (0, 1)T^. Plugging this into the formula we’re given we get
T
[
]
[
]
and T
[
]
[
]
so the desired matrix is
A =
[
]
(b). Consider the basis B = {v 1 , v 2 } given by: v 1 =
[
]
and v 2 =
[
]
Find the change of basis matrix C for the basis B. That is, find the matrix C such that v = C[v]B for all vectors v.
Solution: We use the formula
C =
v 1 v 2 | |
so we get
C =
[
]
(c). Find the matrix B that represents the linear transformation T with respect to the basis B.
Solution: The formula for finding T with respect to the basis B is
B = C−^1 AC.
So we first need to compute C−^1. We use the relation [ a b c d
]− 1
det
[
d −b −c a
]
which will give
C−^1 =
[
]
So finally we have
B = C−^1 AC =
[
][
][
]
and (carefully!) multiplying this out we get [ 3 4 0 2
]
3(b). Consider the matrix B =
Find an eigenvector of B with eigenvalue λ = 1.
Solution: To find an eigenvector of B with eigenvalue λ = 1, we find the null space of
I − B =
We do the row-reduced enchelon here.
(R 1 = R 2 + R 1 )
(^) (R 3 = R 3 + R 1 ) then (R 1 = (−1)R 1 )
(^) (R 2 = R 2 − 5 R 1 ) then (R 2 = (−1)R 2 )
(R 1 = R 1 − R 2 ).
Therefore x − 2 z = 0 y + z = 0.
Therefore N (I − B) = z
. One of the eigenvectors is
Remark: Many students end up getting
(^) as an eigenvector, which
is not possible.
4(a). Find the eigenvalues of the matrix A =
The eigenvalues are the roots of the characteristic polynomial PA, given by PA(x) = det(A − xI 3 ).
By expanding the determinant along the last row, one gets:
PA(x) = (3 − x)
(1 − x)^2 − 4
= (3 − x)^2 (−x − 1)
so the eigenvalues are 3 and −1.
4(b). Consider the quadratic form xT^ Ax, where A is the matrix in part (a).
Determine whether the quadratic form is positive definite, indefinite, or negative definite. If it is none of those, determine whether the quadratic form is positive semidefinite or negative semidefinite.
Solution: A has positive and negative eigenvalues, so the associated quadratic form is indefinite.
Comments:
- Question (b) has been graded regardless of the correctness of the answer of (a).
- Many students checked the (sign of the) trace and determi- nant in (b). THE ARGUMENT WORKS ONLY FOR 2-BY- MATRICES! There are 3-by-3 symmetric matrices with posi- tive trace and determinants, whose associated quadratic form is indefinite. For example, the diagonal matrix whose diagonal entries are − 1 , − 1 , 3. And the diagonal matrix with diagonal entries =-1 has determinant -1, yet the quadratic form is nega- tive definite...
5(b). If the determinant of B is 7, what are its eigenvalues? (Here B is the matrix from part (a).)
Solution: From the condition, we already know that 1 and 3 are two eigenvalues. Since the determinant is the multiple of all eigenvalues, we have that: 7 = det = 1 ∗ 3 ∗ λ 3.
Here λ 3 is the last eigenvalue, which is λ 3 = 7/3. So the set of eigen- values is { 1 , 3 , 7 / 3 }.
Grading Policy and Comments: If you write down that 1 and 3 are two eigenvalues, you will get 2 points. If you point out that the determinant is the multiple of all eigenvalues, you will get another 2 points. If you finally find the last eigenvalue λ 3 = 7/3, you will get the last 1 point. If you make obvious mistakes, for exmaple that someone wrote solution as { 1 , 7 }, or { 1 , 3(twice)}, we will cut your partial credit!
6(a).The position of a particle at time t is u(t) = (t, t^2 , t^3 ). Find the velocity of the particle at time t.
Solution (2 points): u′(t) = (1, 2 t, 3 t^2 ).
6(b). Find the acceleration of the particle at time t.
Solution (2 points): u′′(t) = (0, 2 , 6 t).
6(c). Find the speed of the particle at time t.
Solution (2 points): The speed is ‖u′(t)‖ =
1 + 4t^2 + 9t^4.
6(d). Find the tangent line to the path of the particle at the point (1, 1 , 1).
Solution (4 points): First we need to determine the time t when the particle is at the point (1, 1 , 1). The equation
u(t) = (t, t^2 , t^3 ) = (1, 1 , 1)
shows that this happens exactly when t = 1. Therefore the tangent vector at (1, 1 , 1) is u′(1) = (1, 2 , 3), and the required tangent line is
x y z
(^) + t
- The temperature at point (x, y) on the floor of a room is given by f (x, y) = xy^2.
7(a). A tweetle beetle crawls on the floor. At time t = 2, he is at the point (1, 3) and his velocity is (2, −1). Let u(t) be the temperature where the beetle is at time t. Find u′(2).
Let x(t) be the path of the tweetle beetle, then
u(t) = f (x(t)) = (f ◦ x)(t).
By Chain Rule, we have
u′(2) = Df (x(2)) · x′(2)
= Df (1, 3) ·
[
]
Df (1, 3) =
[
∂f ∂x
∂f ∂y
]
(x,y)=(1,3)
[
y^2 2 xy
]
(x,y)=(1,3) =^
[
]
Hence u′(2) =
[
]
[
]
Remark 1. Some students applied the chain rule incorrectly (for in- stance, wrote u′(t) = f ′(x, y)); and some gave an incorrect definition of the Jacobian matrix (for instance, wrote Df (x, y) = ∂f∂x + ∂f∂y )
Remark 2. Some students used directional derivatives to find out the rate of change. However, the directional derivative gives instead the spatial change of the temperature profile, i.e. the increase in temperature after moving along the direction for 1 unit length. Here what we look for is the rate of change against time, i.e. the change in temperature after moving along the path for 1 unit time.
Remark 3.[ Some assumed the beetle is moving along the path x(t) = 1 3
]
[
]
, i.e. a constant velocity. They may find the same
answer (fortunately), but it is not conceptually correct.
7(b). Another tweetle beetle is at the point (1, 3), where she finds it uncomfortably cold. In which direction should she start moving to warm up as quickly as possible?
The beetle should move in the direction u such that the directional derivative Duf is the maximum. As u is unit, we have
Duf = ∇f · u = ‖∇f ‖ cos θ,
where θ is the angle between ∇f and u.
Duf achieves its maximum when θ = 0.
Hence the beetle has to move in a direction parallel to ∇f.
∇f (1, 3) =
[∂f ∂x∂f ∂y
]
(x,y)=(1,3)
[
y^2 2 xy
]
(x,y)=(1,3)
[
]
Hence the beetle should start moving at the direction √^1117
[
]
in order
to warm up as quickly as possible.
Remark 4. The problem asked for the direction only but not the exact value of the directional derivative, so leaving the final answer as a unit
vector is optional. No point is deducted if students leave
[
]
or
[
]
as the final answer with correct reasoning.
- In part (a) and (b), find the indicated limit or else show that the limit does not exist.
9(a). lim(x,y)→(0,0)
xy x^2 + 2y^2
Solution: The limit does not exist. The are two ways to prove it. First put f (x, y) = (^) x 2 xy+2y 2.
(1) Let m be a real number, so f (t, mt) = (^) 1+2mm 2 , therefore limt→ 0 f (t, mt) = m 1+2m^2. If^ f^ had a limit^ ^ at 0, then one should have^^ =^
m 1+2m^2 , a contradiction because m is arbitrary (setting successively m = 0 , 1 gives = 0 and = 1/3). (2) Using the polar coordinates, one sets (x, y) = (r cos(θ), r sin(θ)), and one has g(r, θ) = f (r cos(θ), r sin(θ)) = sin(1+sinθ) cos( (^2) (θθ) ). Fixing θ, one looks at limr→ 0 g(r, θ), and one finds:
lim r→ 0 g(r, θ) =
sin(θ) cos(θ) 1 + sin^2 (θ)
By contradiction, if f had a limit at 0, one should have = sin(θ) cos(θ) 1+sin^2 (θ) , which cannot be, because the latter depends on^ θ (e.g. setting successively θ = 0, π/4 give ` = 0 = 1/3).
9(b). lim(x,y)→(0,0)
xy^2 x^2 + y^2
Solution: Here the limit exists, and equals 0. To prove it, one could start as in (a), and one finds that 0 is the only possible limit at 0. However, to prove it, one needs the squeeze theorem (because the ”line test”, or the polar coordinates only give a necessary condition). One possible way is to note that y^2 ≤ x^2 + y^2 for any (x, y) ∈ R^2 , so one has:
|f (x, y)| ≤ |x|
x^2 + y^2 x^2 + y^2
= |x|,
as if (x, y) → (0, 0), then so does |x|, so by the squeeze theorem lim(x,y)→(0,0) exists and is equal to 0. One can argue similarly with the polar coordinates, but one has to bound the expression involving θ
uniformly in θ (one gets r cos(θ) sin(θ), and bounds it by r.)
Comments:
- It was sufficient in (a) to use the line test only with two values for the slope, like m = 0, 1. It is crucial to understand that if the limit on t → 0 doesn’t depend on m, then it is the only possible limit, but to prove that it is the actual limit, one has to use the squeeze theorem a priori.
- When one uses the polar coordinates, one HAS to fix θ, then let r → 0. In (a), the limit in r does exist, and equals the ratio written above. The problem is not that the limit doesn’t depend on r (the limit of a function as r → 0 would never depend on r for sure), nor that the quotient doesn’t depends on r: the limit of a constant function equal to 2 say exists, and is equal to 2. The limit on r does exist, though the limit on (x, y) doesn’t.
- The dependence on θ should be more explicit, because if one ends up with sin
(^4) (θ)−cos (^4) (θ) sin^2 (θ)−cos^2 (θ) , it seems that this depends on^ θ, while it doesn’t! Morality: it is more safe to use the line-test, as the dependence on the slope is simpler to check.
- Two examples to look at: first f (x, y) = y^2 /x. This function has no limit at 0 (take (x, y) = (t, tm) for various values of m), yet a polar coordinates-test might lead to “limit=0”. This is why the squeeze theorem is necessary (and here it is not possible to bound uniformly the expression in θ of course!). Second example, let f be the function equal to 1 on the parabola of equation y = x^2 , and 0 otherwise. Here again, fix θ, and let r → 0. The expression one gets is 0, doesn’t depend on θ, and yet the limit doesn’t exist.
- To work rigourously with polar coordinates, one needs uniform estimates on θ – i.e. the squeeze theorem; exactly as one needs it without passing to the polar coordinates. It is not true that one can reduce a limit problem in 3 variables to a problem in 1 variable! 2 points for using it.
- In many exams, it was very unclear what limit doesn’t exist in 9(a). The limit on (x, y) doesn’t, while the limit on r does (and the latter has to be computed to be rigourous)! Especially, it seemed implicit often that a function which doesn’t depend on the variable(s) (i.e. a constant function) has no limit!
