Math 222
Quiz 6 (Take Home) Solutions
Problem 1
(a) (1 point). Use the discriminant test to decide whether the equation −2x2 − xy +
y 2 + 7x − 8y + 15 = 0 represents an ellipse, a parabola, or a hyperbola.
Solution: A = −2, −B = 1 and C = 1 so the discriminant is
B 2 − 4AC = 1 − (4)(−2)(1) = 9 > 0
This implies that the equation represents a hyperbola.
(b) (2 points). Show that the graph of −2x2 − xy + y 2 + 7x − 8y + 15 = 0 is actually
the two lines y = 2x + 3 and y = −x + 5.
Solution: y = 2x + 3 is equivalent to y − 2x − 3 = 0, and y = −x + 5 is equivalent to
y + x − 5 = 0.
(y − 2x − 3)(y + x − 5) = y 2 + xy − 5y − 2xy − 2x2 + 10x − 3y − 3x + 15 =
−2x2 − xy + y 2 + 7x − 8y + 15
So − 2x2 − xy + y 2 + 7x − 8y + 15 = 0 factors into (y − 2x − 3)(y + x − 5) = 0
Therefore the graph of −2x2 − xy + y 2 + 7x − 8y + 15 = 0 is the two lines y = 2x + 3
and y = −x + 5.
Problem 2 (2 points). Evaluate the following integral.
Z π
√
1 − cos x dx
0
Solution: We use the identity sin2 θ = 1−cos(2θ)
. This allows us to replace 1 − cos x
2
2
with 2 sin (x/2). So
Z π
Z πs
√ Z π
√
x
2 x
1 − cos x dx =
2 sin
dx = 2
| sin
| dx
2
2
0
0
0
Since sin( x2 ) ≥ 0 on [0, π] we have
π
Z π
Z π √
√
√
x
x
1 − cos x dx =
sin
dx = −2 2 cos
=2 2
2
2 0
0
0
Problem 3 (2 points). Determine whether the following improper integral converges
or diverges.
Z ∞
cos2 x dx
x2 + 1
1
Hint: You only need to show whether or not the integral converges. Do not attempt to
actually compute it.
Solution: Since −1 ≤ cos x ≥ 1 for all real numbers, we have that 0 ≤ cos2 x ≤ 1 for
all real numbers, and therefore
0≤
cos2 x
1
≤ 2
2
x +1
x +1
This gives us that
Z
0≤
1
∞
cos2 x dx
≤
x2 + 1
Z
1
∞
dx
+1
x2
We can compute this new integral directly.
Z ∞
Z b
dx
dx
= lim
=
2
2
x + 1 b→∞ 1 x + 1
1
b
lim arctan x = lim arctan b − arctan 1 =
b→∞
1
b→∞
π π
π
− = <∞
2
4
4
Z ∞
dx
So
converges.
2
x +1
1
Therefore, by the Direct Comparison Test
Z ∞
cos2 x dx
converges.
x2 + 1
1
Problem 4
(a) (1 point). Sketch a graph of
r = sin(2θ)
Solution:
Your graph should resemble the following. The rθ graph is on the left and the xy graph
(i.e. the final answer) is on the right.
(b) (2 points). Find the equation of the tangent line to r = sin(2θ) at θ =
π
4
Solution:
r = f (θ) = sin(2θ) so f 0 (θ) = 2 cos(2θ). We apply the formula
dy
=
dx
=
dy
dθ
dx
dθ
=
f 0 (θ) sin θ + f (θ) cos θ
f 0 (θ) cos θ − f (θ) sin θ
2 cos(2θ) sin θ + sin(2θ) cos θ
2 cos(2θ) cos θ − sin(2θ) sin θ
Evaluating at θ = π/4 gives us that the slope of the tangent line is
m=
2 cos(π/2) sin(π/4) + sin(π/2) cos(π/4)
2 cos(π/2) cos(π/4) − sin(π/2) sin(π/4)
√
=
0+1·
0−1·
2
2
√
2
2
= −1
When θ = π/4, r = sin(π/2) = 1 so x = cos(π/4) =
Therefore, the equation of the tangent line is
√
√
2/2 and y = sin(π/4) =
√ !
2
2
y−
= −1 x −
2
2
√
2/2.