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Math 240 Final Exam
Spring 2012
- Solve the equations for x. 2 x + 3 y + 2 z = 1 x + 0 y + 3 z = 2 2 x + 2 y + 3 z = 3 Hint:
If A =
(^) then A−^1 =
a) x = − 1 b) x = 0 c) x = 2 d) x = 5 e) x = 7 f) x = 11 g) none of the above
Answer: (f ). To find the solution vector, we need only apply the inverse:
x y z
This answer can easily be verified by substituting back into the original system.
- The lemniscate of Gerono is parametrized by the formulas
x(t) = cos t, y(t) = sin t cos t.
Compute the area of the right-hand lobe (corresponding to the range of parameters − π 2 ≤ t ≤ π 2 ). Hint: Use Green’s Theorem and the differential −y dx. Near the end you’ll likely need to use a u-substitution.
a)
b)
c)
d)
e)
f) 1 g) none of the above
Answer: (d). For this specific differential, we have P = −y and Q = 0, so Qx − Py = 1. Thus by Green’s Theorem we have (^) ∫ ∫
A
d A
∂A
−ydx,
that is, the area of A is equal to the line integral of −ydx along the boundary of A. Writing the line integral using the given parametrization, we have ∫
∂A
−ydx =
∫ π 2
− π 2
− sin t cos t(− sin t)dt =
∫ π 2
− π 2
sin^2 t cos tdt.
Substituting u = sin t gives (^) ∫ π 2 − π 2
sin^2 t cos tdt =
− 1
u^2 du =
- Calculate the outward flux of F~ across S if F~ (x, y, z) = 3xy^2 ~i + xez~j + z^3 ~k and S is the surface of the solid bounded by the cylinder y^2 + z^2 = 1 and the planes x = −1 and x = 2.
a) 0 b) −
π 4
c)
11 π 8
d) 3π e)
9 π 5
f)
9 π 2
g) none of the above
Answer: (f ). We use the divergence theorem. First we compute the divergence of F~ :
∇ · F~ =
∂x
(3xy^2 ) +
∂y
(xez^ ) +
∂z
(z^3 ) = 3(y^2 + z^2 ).
If C is the interior of the cylinder that is described in the problem and ∂C is the surface bounding it, then ∫ ∫
∂C
F^ ~ · ~ndS =
C
3(y^2 + z^2 )dV.
We can evaluate the integral in cylindrical coordinates (r, θ, x), where (r, θ) represent the polar coordinates in the yz-plane. ∫ ∫
∂C
F^ ~ · ~ndS =
C
3(y^2 + z^2 )dV =
− 1
∫ (^2) π
0
0
3 r^2 rdrdθdx =
9 π 2
- Compute the outward flux of ∇ × F~ through the surface of the ellipsoid 2x^2 + 2y^2 + z^2 = 8 lying above the plane z = 0, where F~ = (3x − y)~i + (x + 3y)~j + (1 + x^2 + y^2 + z^2 )~k.
a) 0 b) 2π c) 3π d) 8π e) 12π f) 16π g) none of the above
Answer: (d). We use Stokes’ theorem. If S is the top half of the ellipsoid, then its boundary will be a curve C obtained by setting z = 0. On the circle, 2x^2 + 2y^2 = 8, meaning that C is the circle of radius 2 centered at the origin and lying in the xy-plane. Since the outward flux points upwards in the z-direction, the correct orientation of C is counterclockwise. We conclude (^) ∫ ∫
S
(∇ × F~ ) · ~ndS =
C
(3x − y)dx + (x + 3y)dy + (1 + x^2 + y^2 + z^2 )dz.
We can parametrize C by ~r(t) = 〈2 cos t, 2 sin t, 0 〉 for t ∈ [0, 2 π]. We will get ∮
C
(3x − y)dx+(x + 3y)dy + (1 + x^2 + y^2 + z^2 )dz
∫ (^2) π
0
[
(6 cos t − 2 sin t)(−2 sin t) + (2 cos t + 6 sin t)(2 cos(t)) + (1 + 4 cos^2 t + 4 sin^2 t)(0)
]
dt.
∫ (^2) π
0
[
sin^2 t + cos^2 t
]
dt = 8π.
- Solve the differential equation 9 x^2 y′′^ + 2y = 0 on the interval (0, ∞) subject to the initial conditions y(1) = 1 and y′(1) = 43. a) y = 2x (^23) − 3 x (^13) b) y = 3x (^23) − 2 x (^13) c) y = 3x (^32) − 3 x^3 d) y = 3x (^32) − 2 x^3 e) y = 2x^2 − 3 x f) y = 3x^2 − 2 x g) none of the above
Answer: (b). This is a Cauchy-Euler equation. The auxiliary equation is 9m(m − 1) + 2 = 0, which factors as (3m − 2)(3m − 1) = 0. Thus the general solution will have the form y = c 1 x
(^23)
(^13)
. Solving the system of equations
c 1 + c 2 = 2 3
c 2 +
c 2 =
gives c 1 = 3 and c 2 = −2.
- Let ~ω := 〈 1 , 2 , 3 〉, and let ~r(t) = 〈x(t), y(t), z(t)〉. Now consider the differential equation
d dt
~r = ~ω × ~r.
Select the answer which correctly expresses this system of equations in matrix notation when
X^ ~(t) =
x(t) y(t) z(t)
Do not solve the system.
a)
d dt
X~ =
(^) X~ b) d dt
X~ =
(^) X~ c) d dt
X~ =
X~
d) d dt
X~ =
(^) X~ e) d dt
X~ =
(^) X~ f) d dt
X~ =
X~
g) none of the above
Answer: (c). First we simply expand the formula for the cross product:
〈 1 , 2 , 3 〉 × 〈x, y, z〉 =
~i ~j ~k 1 2 3 x y z
y z
~i −
x z
~j +
x y
~k
= (2x − 3 y)~i + (3x − z)~j + (y − 2 x)~k.
We write this vector as a column vector and discover that
2 x − 3 y 3 x − z y − 2 x
x y z
d dt
X~ =
X.~
- Select the answer below which corresponds to the first few terms in a power series solution of the differential equation
x^2 y′′^ + (x^2 − x)y′^ + y = 0.
Will there be a second, linearly independent series solution for this equation? Explain your answer.
a) y = x
(^12)
x
(^32)
x
(^52)
x
(^72)
(^12)
x
(^32) −
x
(^52)
x
(^72) +· · · c) y = x
(^12)
(^32)
x
(^52) −
x
(^72)
d) y = x − x^2 +
x^3 −
x^4 + · · · e) y = x − x^2 +
x^3 −
x^4 + · · · f) y = x + x^2 +
x^3 +
x^4 + · · ·
g) none of the above
Answer: (e). The point x = 0 is a regular singular point of this ODE. We use the method of Frobenius to find a solution
y =
∑^ ∞
n=
cnxn+r^ xy′^ =
∑^ ∞
n=
(n + r)cnxn+r^ x^2 y′′^ =
∑^ ∞
n=
(n + r)(n + r − 1)cnxn+r
We substitute in and get
x^2 y′′^ + (x^2 − x)y′^ + y =
∑^ ∞
n=
(n + r)(n + r − 1)cnxn+r^ +
∑^ ∞
n=
(n + r)cnxn+r+1^ −
∑^ ∞
n=
(n + r)cnxn+r^ +
∑^ ∞
n=
cnxn+r^ = 0
∑^ ∞
n=
(n + r)(n + r − 1)cnxn+r^ +
∑^ ∞
n=
(n − 1 + r)cn− 1 xn+r^ −
∑^ ∞
n=
(n + r)cnxn+r^ +
∑^ ∞
n=
cnxn+r
= [r(r − 1)c 0 − rc 0 + c 0 ] +
∑^ ∞
n=
[(n + r)(n + r − 1)cn + (n − 1 + r)cn− 1 − (n + r)cn + cn] xn+r
We simplify and conclude that
(r − 1)^2 c 0 = 0 and (n + r − 1)^2 cn + (n + r − 1)cn− 1 = 0 when n ≥ 1.
The indicial roots are both r = 1, so the method of Frobenius says that only one such series solution will exist (the other will involve a logarithm). We have
c 1 = −
c 0 = −c 0 c 2 = −
c 1 = c 0 2
c 3 = −
c 2 = − c 0 6
In fact, we can see that there is a pattern: cn = (−1)
n n!. Our full series will be
y =
∑^ ∞
n=
(−1)n n!
xn+1.
We can even identify this as a Taylor series: y = xe−x.
- Let y be a function satisfying y(0) = y′(0) = y′′(0) = 0 which is a solution of the ODE
y′′′^ − 4 y′′^ + 4y′^ = 4.
Compute y(1). a) y(1) = − 5 b) y(1) = 4 c) y(1) = − 3 d) y(1) = 2 e) y(1) = − 1 f) y(1) = 0 g) none of the above Answer: (d). The complementary solution will have the form yc = c 1 e^2 x^ + c 2 xe^2 x^ + c 3. The method of undetermined coefficients says that we should find a particular solution in the form yp = Ax (we multiply by x because the com- plementary solution includes constants). That method gives a solution when A = 1. Thus the solution is of the form
- Find a solution to the initial value problem y′′^ − 2 xy′^ − 4 y = 0 subject to the initial conditions y(0) = 0 and y′(0) = 1 which takes the form of a power series centered at the origin. What is the coefficient in front of x^5 in the series? a) − 1 b) 0 c)
d) 1 e) 2 f) 6 g) none of the above
Answer: (c). The point x = 0 is ordinary, so we make a standard power series solution:
y =
∑^ ∞
n=
cnxn^ y′^ =
∑^ ∞
n=
ncnxn−^1 y′′^ =
∑^ ∞
n=
n(n − 1)cnxn−^2.
We substitute these into the equation and get
∑^ ∞
n=
n(n − 1)cnxn−^2 − 2 x
∑^ ∞
n=
ncnxn−^1 − 4
∑^ ∞
n=
cnxn^ = 0.
Next we can combine the last two sums: ∑^ ∞
n=
n(n − 1)cnxn−^2 +
∑^ ∞
n=
(− 2 n − 4)cnxn^ = 0.
After that we shift the index of summation in the first sum: ∑^ ∞
n=− 2
(n + 2)(n + 1)cn+2xn^ +
∑^ ∞
n=
(− 2 n − 4)cnxn^ = 0.
Because (n + 2)(n + 1) = 0 both when n = −2 and n = −1, we can drop these two terms from the first sum and combine again: ∑∞
n=
[(n + 2)(n + 1)cn+2 − 2(n + 2)cn] xn.
Our recurrence formula is found by setting these coefficients equal to zero:
(n + 2)(n + 1)cn+2 − 2(n + 2)cn = 0 ⇒ cn+2 = 2 cn n + 1
The initial conditions give c 0 = 0 and c 1 = 1. Computing the rest by means of the formula gives
c 2 =
2 c 0 1
= 0 c 3 =
2 c 1 2
= 1 c 4 = 0 c 5 =
2 c 3 4
- Circle “T” for true or “F” for false in the space provided to the left of the following statements. You DO NOT need to justify your answer for full credit.
( T F ) Every 2 × 2 diagonalizable matrix with repeated eigenvalue is a diagonal matrix.
True: A = P DP −^1 and D is a multiple of the identity, so A must also be.
( T F ) There is a vector field F~ such that ∇ × F~ = 〈x, y, z〉.
False: The divergence of the curl is always equal to zero, but the divergence of 〈x, y, z〉 equals 3.
( T F ) If det(A) = 0, then the system A X~ = 0 has infinitely many solutions.
True: det(A) = 0 means that A has an eigenvector E~ which has eigenvalue zero. Any multiple of E~ will solve the system A X~ = 0.
( T F ) If y 1 and y 2 are solutions to a non-homogeneous linear differential equation, then y 1 + y 2 is also a solution.
False: Try the differential equation y′^ = 1 and take y 1 = y 2 = x.
( T F ) If A and B are square matrixes such that AB^2 = I, then B is invertible.
True: det(AB^2 ) = (det A)(det B)(det B) = det I = 1. In particular, det A and det B must both be nonzero, meaning they’re both invertible.
