More Exam Practice Problems
Math 352, Fall 2011
1. Let E be the ellipse 5x2 + 4xy + 8y 2 = 16.
(a) Find the lengths (i.e. diameters) of the major and minor axes of E.
(b) Find the equation of the line containing the major axis.
2. Let γ : R → R3 be a unit-speed curve. At the point γ(0), the tangent vector to this
curve is t = (1/9, 4/9, 8/9), and ṫ = (4, 7, −4) at this point.
(a) Compute the normal vector n and curvature κ for this curve at the point γ(0).
(b) Compute the binormal vector b at the point γ(0).
(c) Given that ṅ = (−41, 16, −13) at the point γ(0), compute the torsion τ for the
curve at this point.
3. Let T : R2 → R2 be reflection across the line x + 2y = 2. Find an orthogonal matrix
M and a vector b so that
T (x) = M x + b.
for all x ∈ R2 .
Solutions
1. (a) The given ellipse is defined by the equation
5 2 x
x y
= 16
2 8 y
The matrix has trace 13 and determinant 36, so the eigenvalues are 4 and 9.
Therefore, there exists an orthonormal system of coordinates x0 , y 0 under which
the ellipse takes the form
4(x0 )2 + 9(y 0 )2 = 16.
Then the vertices of the ellipse are at (x0 , y 0 ) = (±2, 0) and (x0 , y 0 ) = (0, ±4/3),
so the lengths of the axes are 4 and 8/3 .
(b) The major axis of the ellipse lies along the x0 -axis, i.e. the eigenspace for the
eigenvalue 4. To find this eigenspace, we solve
5 2 x
x
= 4
2 8 y
y
This gives the equations 5x + 2y = 4x and 2x + 8y = 4y, both of which reduce
to x + 2y = 0 .
2. (a) We know that ṫ = κn, where κ ≥ 0 and n is a unit vector. We conclude that
κ = ktk = k(4, 7, −4)k = 9 and n = ṫ/κ = (4/9, 7/9, −4/9) .
(b) We have
i
j
k
b = t × n = 1/9 4/9 8/9 = (−8/9, 4/9, −1/9)
4/9 7/9 −4/9
(c) We know that ṅ = −κt + τ b, and therefore
τ b = ṅ + κt = (−41, 16, −13) + (1, 4, 8) = (−40, 20, −5).
This vector is 45 times the binormal vector, so τ = 45 .
3. The following picture shows this reflection:
H0.8,1.6L
H0,1L
H0,0L
x+
2y
=2
H2,0L
As you can see, T (2, 0) = (2, 0) and T (0, 1) = (0, 1). Finding the image of (0, 0) is a
little bit tricky: we must first find the intersection of x + 2y = 2 with the perpendicular
line y = 2x, and then double the coordinates. The result is that T (0, 0) = (0.8, 1.6), so
b = (0.8, 1.6) . Then M (0, 1) = (0, 1) − b = (−0.8, −0.6) and M (1, 0) = 21 M (2, 0) =
1
(2,
0)
−
b
= (0.6, −0.8), so
2
0.6 −0.8
M=
−0.8 −0.6