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Math 51 - Winter 2011 - Final
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Nick Haber James Zhao Henry Adams 11:00 AM 10:00 AM 11:00 AM 1:15 PM 1:15 PM 1:15 PM Ralph Furmaniak Jeremy Miller Ha Pham 11:00 AM 11:00 AM 11:00 AM 1:15 PM 2:15 PM 1:15 PM Sukhada Fadnavis Max Murphy Jesse Gell-Redman 10:00 AM 11:00 AM 1:15 PM 1:15 PM 1:15 PM
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Instructions: Print your name and student ID number, select the time at which your section meets, and write your signature to in- dicate that you accept the Honor Code. There are 15 problems on the pages numbered from 1 to 16. Each problem is worth 10 points. In problems with multiple parts, the parts are worth an equal number of points unless otherwise noted. Please check that the version of the exam you have is complete, and correctly stapled. In order to receive full credit, please show all of your work and justify your answers. You may use any result from class, but if you cite a theorem be sure to verify the hypotheses are satisfied. You have 3 hours. This is a closed-book, closed-notes exam. No calculators or other electronic aids will be permitted. GOOD LUCK!
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1(a). Let f (x, y, z) = 3y^2 + 2y^3 − 3 x^2 + 6xy + z^2. Find the second order Taylor approximation of f at point (0, − 1 , 1).
Solution: We use Definition 11.3 from the DVC book, which says
T 2 (x) = f (a) + Df (a)(x − a) +
(x − a)T^ Hf (a)(x − a).
Here, x =
x y z
(^) and a =
We calculate f (a) = 2 and
Df (a) =
[
− 6 x + 6y 6 y + 6y^2 + 6x 2 z
]
(0,− 1 ,1) =^
[
]
and
Hf (a) =
6 6 + 12y 0 0 0 2
(0,− 1 ,1)
Hence
T 2 (x) = 2+
[
]
x y + 1 z − 1
+^1
[
x y + 1 z − 1
]
x y + 1 z − 1
It is totally fine (in fact preferred!) to not expand this expression any further.
1(b). Let S be the surface defined by 3y^2 + 2y^3 − 3 x^2 + 6xy + z^2 = 2. Find an equation for the tangent plane to S at (0, − 1 , 1).
Solution: Let F : R^3 → R be defined by F (x, y, z) = 3y^2 + 2y^3 − 3 x^2 + 6xy + z^2. Note that S = F −^1 (2) = F −^1 (F (0, − 1 , 1)). Hence Theorem 18 from the DVC book tells us that the tangent plane to S at (0, − 1 , 1) is given by the equation
0 = Fx(0, − 1 , 1)(x − 0) + Fy(0, − 1 , 1)(y + 1) + Fz (0, − 1 , 1)(z − 1).
As in part (a), we calculate that Fx(0, − 1 , 1) = −6, Fy(0, − 1 , 1) = 0 and Fz (0, − 1 , 1) = 2. Hence the equation for the tangent plane is
0 = −6(x − 0) + 0(y + 1) + 2(z − 1)
or 0 = − 6 x + 2(z − 1).
- Find all solutions to the following system of equations:
x 1 + 2x 2 + x 3 + x 4 = 7 x 1 + 2x 2 + 2x 3 − x 4 = 12 2 x 1 + 4x 2 + 6x 4 = 4.
Solution: Augmented matrix:
Reduced row echelon form:
There are infinitely many solutions verifying:
x 1 + 2x 2 + 3x 4 = 2 and x 3 − 2 x 4 = 5
That is: (^)
s
+^ t
,^ (s, t)^ ∈^ R
2
4(a). Let v 1 , v 2 , and v 3 be vectors in R^4. Prove that there is a nonzero vector x that is perpendicular to each of those vectors.
Solution: The problem asks us to find a non-zero x so that v 1 · x = v 2 · x = v 3 · x = 0. This is equivalent to finding a non-zero element in the null space of the matrix
A =
− vT 1 − − vT 2 − − vT 3 −
This is a 3×4 matrix (the vi are in R^4 ), so by the rank nullity theorem, null = 4 − rank. Since rank ≤ 3, null ≥ 1, i.e. there is at least one non-zero vector in the the null space of A.
4(b). Suppose that v 1 , v 2 , and v 3 are nonzero vectors that are orthog- onal to each other. Prove that {v 1 , v 2 , v 3 } is linearly independent.
Solution: Suppose that
c 1 v 1 + c 2 v 2 + c 3 v 3 = 0.
To show that v 1 , v 2 , v 3 are linearly independent, we must show that c 1 = c 2 = c 3 = 0. Keep in mind that the problem asks us to assume that the vectors are non-zero and mutually orthogonal. Taking the dot product of both sides of this equation with v 1 , we get
0 = v 1 · (c 1 v 1 + c 2 v 2 + c 3 v 3 ) = c 1 v 1 · v 1 + c 2 v 1 · v 2 + c 3 v 1 · v 3.
The second two terms on the right hand side are zero by orthogonality, so we get 0 = c 1 v 1 · v 1.
Many students reached this step and concluded that c 1 = 0 without justification. This is indeed true, but only because we are assuming that v 1 6 = 0, and therefore v 1 · v 1 , the square norm of v 1 , is not zero.
The same procedure, but this time using v 2 and v 3 , shows that c 2 and c 3 are also zero.
5(b). Suppose that
v 1 =
(^) v 2 =
(^) v 3 =
Find the matrix A for T with respect to the standard basis for R^3.
[Hint: You may use your answer to 5(a). However, it is easier to find A directly, without using the matrix B.]
Solution: We now want to find the matrix A for T relative to the standard basis {e 1 , e 2 , e 3 } are there are two ways we can do this. One is that we can relate A and B (from part a)) using the change of basis matrix but we prefer not to do this since we will need to calculate an inverse. Recall that the first column of A is given by T (e 1 ), second column by T (e 2 ) and third column by T (e 3 ). But note that v 1 = e 1 and so that T (e 1 ) = T (v 1 ) = 7v 3 = [7 7 21]T^.
Now v 2 = e 1 + e 2 and so we have that
e 2 = v 2 − e 1
and so that
T (e 2 ) = T (v 2 ) − T (e 1 ) = v 1 − 7 v 3 = [− 6 − 7 − 21]T^.
Note that v 3 = v 2 + 3e 3
and so
3 T (e 3 ) = T (v 3 ) − T (v 2 ) = 9v 2 − v 1 = [8 9 0]T^ ,
so
A =
- Find A−^1 , where A =
Solution: We need to row reduce the matrix
so the matrix on the left is in row reduced echelon form. The first step is to put the second row up top and the third row just below. The top row becomes the third row and so we have
To row reduce this we multiply the first row by −1 and add this to the third row and then multiply the second row by −1 and add this to the third row to get (^)
and so
A−^1 =
- Let f (x, y) be a scalar valued function of two variables describing the pressure at the point (x, y) on the (flat) Earth’s surface. Suppose that ∂f ∂x
∂f ∂y
8(a). Find the directional derivative of f at (− 1 , 2) in the direction
v =
[
]
Solution: The directional derivative at (− 1 , 2) in the direction of v is
∇f (− 1 , 2)
v ||v||
[
∂f ∂x (−^1 ,^ 2)^
∂f ∂y (−^1 ,^ 2)
][− 1
]
[
]
[
]
8(b). A dragon is flying along an isobar (i.e., a level set of f ) with speed 500. At time t = 0, the insect is at the point (− 1 , 2) and its x-coordinate is increasing. Find its velocity at time 0.
Solution: Let v be the dragonfly’s velocity. Since it is flying along a level set, it must be perpendicular to the gradient:
0 = v · ∇f (− 1 , 2) =
[
v 1 v 2
]
[
]
= −v 1 + 2v 2
Hence v 1 = 2v 2. We also know that the speed is 500, so:
500 = ||v|| =
v^21 + v^22
=
4 v^22 + v^22
=
5 v^22
100
v^22 = |v 2 |
Thus either v 2 = 100
5 and v 1 = 2v 2 = 200
5, or v 2 = − 100
5 and v 1 = 2v 2 = − 200
- The latter is impossible since we are told that the x-coordinate is increasing. Hence the velocity must be
v =
[
]
- Our dragon friend has now quit flying and is trying to build a box out of plywood. He has 12m^2 of plywood available, and is building a box in the shape of a rectangular prism without a top (so just needs to create four sides and a bottom). What is the greatest volume that this box can contain? (to be precise, suppose he has found a flat lid elsewhere).
Solution: We are trying to maximize the function V (x, y, z) = xyz, subject to the constraint S(x, y, z) = xy + 2xz + 2yz ≤ 12. Clearly the largest volume will use all of the plywood availiable, so we can assume S(x, y, z) = 12. Also, we can assume that x, y and z are all positive numbers, otherwise the equations have no physical interpretation. We use Lagrange multipliers.
∇V = (yz, xz, xy) and ∇S = (y + 2z, x + 2z, 2 x + 2y). Setting up the equation ∇V = λ∇S, we get three equations,
yz = λ(y + 2z) xz = λ(x + 2z) xy = λ(2x + 2y).
Solving each equation here for λ, we get one big equation yz y + 2z
xz x + 2z
xy 2 x + 2y
Notice that all our divisions are legal since our variables are positive. Cross multiplying fractions in the first equation gives yz(x + 2z) = xz(y + 2z), which reduces to 2yz^2 = 2xz^2. Since z 6 = 0, this implies x = y.
By cross multiplying our second equation, we get xz(2x + 2y) = xy(x + 2 z). By reducing, we get 2x^2 z = x^2 y, which implies 2z = y.
Now that we know x = y = 2z, we can plug everything back into our formula S(x, y, z) = 12, which says 4z^2 + 4z^2 + 4z^2 = 12. This tells us that z = 1, so x = y = 2. This gives a volume of 4, which must be the maximum, since there are no other critical points.
Some students started with the assumption that x = y, this is valid, because x and y are symmetric in the problem. Many students started with the assumption x = y = z, which does not work. The problem is not symmetrical this way, because of the open top.
10(c). Note that A
. Find all solutions to Ax =
Solution: Finally, for part (c), we are given an explicit solution to the
system of equations Ax =
, namely x 0 =
. Thus, the general
solution is given by the special solution x 0 plus an element of the null space. In part (b) we found a basis for the null space, so we conclude that a general solution is given by
+^ s
+^ t
with s, t ∈ R. That is an answer to part (c).
(A notable feature of many valid solutions to part (c) different from the above is that a lot of people ignored the given information of a special solution x 0 and instead “re-invented the wheel” by grinding through the usual manipulations to discover their own special solution. The
most popular one found was
. That is a valid and received full
marks, but it involves more work than is really necessary: as we saw above, all of the necessary work has already been done in part(b)!)
Typical errors for solutions to problem 10. A typical error when solving part (b) was to either lose some signs, or get the entries mixed up in the vertical direction. One should always double check that the proposed basis for the null space at least does satisfy the original homogeneous system of linear equations corresponding to the given matrix A! That will catch most computational errors.
Another typical error for solutions to part (b) was to not give a basis, but rather to say that the null space consists of vectors of the general parametric form in the displayed equation above. That is not a valid answer (led to a 1-point deduction) because a basis of the null space is not the same thing as the null space itself. It is important to understand
the distinction between these notions, even though passing between them is rather straightforward.
The typical error for solutions to part (c) was to “add” elements of
the null space to
(^) rather than to
, but it is meaningless to
“add” vectors with different numbers of entries (let alone to overlook the fact that one has to actually do the translation by a solution to the inhomogeneous system of equations). So that is a serious conceptual error.
- If t = x^2 + yz^2 and x = uve^2 s, y = u^2 − v^2 s, z = cos(uvs).
12(a). Find ∂t∂u (2, 1 , 0).
12(b). Find ∂t∂s (0, 1 , 5).
Solution. There are several ways to solve this. A widely-used method was brute force: explicitly plug the expressions for x, y, and z in terms of u, v, and s into the definition of t to get
t = u^2 v^2 e^4 s^ + (u^2 − v^2 s) cos(uvs)^2
(some erred here by not remembering the laws of exponents), and then explicitly differentiate this with respect to u and s separately. This leads to rather elaborate expressions (which are then evaluated at (2, 1 , 0) for (a) and at (0, 1 , 5) for (b)). Namely, one has
∂t ∂u
= 2uv^2 e^4 s^ + 2u cos(uvs)^2 − 2 vs(u^2 − v^2 s) cos(uvs) sin(uvs),
∂t ∂s
= 4u^2 v^2 e^4 s^ − v^2 cos(uvs)^2 − 2 uv(u^2 − v^2 ) cos(uvs) sin(uvs),
which respectively evaluate to 8 and −1 at the indicated points in parts (a) and (b). This is a very painful way to arrive at the answer.
The method which was intended for the solution was to use the Chain Rule (to ease the algebra burden):
∂t ∂u
∂t ∂x
∂x ∂u
∂t ∂y
∂y ∂u
∂t ∂z
∂z ∂u
and similarly for ∂t∂s by replacing u with s everywhere. In this approach, one has to keep a clear distinction between the point (u 0 , v 0 , s 0 ) at which we wish to evaluate the left side (namely, (2,1,0) for part (a) and (0,1,5) for part (b)) and the corresponding point in “(x, y, z) co- ordinates”:
(x(2, 1 , 0), y(2, 1 , 0), z(2, 1 , 0)) = (2, 4 , 1), (x(0, 1 , 5), y(0, 1 , 5), z(0, 1 , 5)) = (0, − 5 , 1).
To be precise, the “Chain Rule” expression for (^) ∂u∂t evaluated at (2, 1 , 0) simplifies to give
∂t ∂u
(2, 1 , 0) = (2x + 4z^2 )|(2, 4 ,1) = 8,
and proceeding similarly for the s-partial gives
∂t ∂s
(0, 1 , 5) = −z^2 |(0,− 5 ,1) = − 1.
There is a third method of solution which truly minimizes the odds of algebra errors by going back to the very definition of a partial derivative and avoiding the Chain Rule altogether: when computing a u-partial at the point (2, 1 , 0), it doesn’t matter whether we replace (v, s) with (1, 0) before or after the derivative calculation with respect to u. (Why not? Think about it.) So let’s do that evaluation before the differentiation: t(u, 1 , 0) = u^2 + u^2 · 1 = 2u^2. The u-derivative of this is 4u, and setting u = 2 then recovers the answer of 8 for part (a). Likewise, for part (b) we observe that t(0, 1 , s) = −s cos(0)^2 = −s, so its s-derivative is − 1 (a constant!). Evaluating at s = 5 recovers the answer −1 for part (b).
Typical errors for solutions to problem 12. In solutions to by the brute force method, apart from the many varieties of algebra errors the most typical error by far was to overlook the minus sign in front the v^2 in the middle term of the “brute force” expression for ∂t∂s (leading to an answer of 1 rather than −1 for part (b)). In the solutions via the Chain Rule method, the typical error was to forget that when working in terms of (x, y, z) we need to recompute the point we’re at (namely, (2, 4 , 1) for part (a), and (0, − 5 , 1) for part (b)). Also, a lot of people mistakenly evaluated the expression (^) ∂x∂t = 2x at the point (2, 4 , 1) to be 2 rather than 2 · 2 = 4.
- Suppose F : R^2 → R^3 is defined by F (x, y) =
ee xy esin^ x^ − y^2 ecos^ y^ + x^2
Find the Jacobian matrix (i.e, the total derivative matrix) DF (x, y).
Solution: Recall that the Jacobian of a function F : R^2 −→ R^3 is a 3 × 2 matrix given by the formula:
DF (x, y) =
∂f 1 ∂x
∂f 1 ∂y ∂f 2 ∂x
∂f 2 ∂y ∂f 3 ∂x
∂f 3 ∂y
where F (x, y) =
f 1 (x, y)
f 2 (x, y)
f 3 (x, y)
We proceed now to computing the relevant derivatives. To make the use of the chain rule more transparent, we let f (t) = et. It follows that
∂x
ee xy =
∂x
f (f (xy)) =
∂f ∂t
(f (xy)) ·
∂f ∂t
(xy) ·
∂x
(xy)
= ef^ (xy)^ · exy^ · y
= yee
xy (^) +xy
Following this procedure, we get the matrix of derivatives:
DF (x, y) =
yee xy (^) +xy xee xy (^) +xy
cos x · esin^ x^ − 2 y
2 x − sin y · ecos^ y
15(a). Find an eigenvector with eigenvalue λ = 1 for the matrix
A =
Solution: Since each row in this matrix adds up to 1, it follows that
(^) is an eigenvector of A with eigenvalue 1. Another way of going
about this, is to actually compute the eigenspace associated to λ 1. That is,
E 1 = N (I − A) = N
Putting this new matrix in RREF we get
and therefore
E 1 =
x y z
∣ x = y = z
15(b). Find an eigenvector with eigenvalue λ = 1 for the matrix A^2 + A − I, where I is the identity matrix and A is the matrix in part (a).
Solution: The thing to notice here is that if v is an eigenvector of A with eigenvalue 1, then
(A^2 − A + I)v = A(Av) − Av + Iv = Av − v + v = v
and therefore v is also an eigenvector of A^2 − A + I with eigenvalue 1. That is, any answer for part (a), will also be an answer for part (b).
