Calculus II (MAC2312-02)
Test 3 (2015/07/16)
Name (PRINT):
Please show your work. An answer with no work receives no credit.
You may use the back of a page if you need more space for a problem.
You may not use any calculators.
Page
Points Score
2
16
3
11
4
13
5
15
6
20
7
25
Total:
100
Page 1 of 7
Calculus II (MAC2312-02)
Page 2 of 7
1. Circle True of False.
(a) (3 points) True FALSE The polar coordinates (r, θ + 2nπ) and (r, θ + (2n + 1)π), where
n is any integer, represent the same point in the plane.
Solution: The polar coordinates (r, θ + 2nπ) and (−r, θ + (2n + 1)π) (note the negative
sign), where n is any integer, represent the same point in the plane.
(b) (3 points) TRUE False We get the entire graph of the curve r = sin(6θ/7) if we evaluate
r for all θ in [0, 14π].
Solution: We need to determine the minimum number n of complete rotations around
the circle before the curve starts repeating itself. That is, for any θ0 we want the polar
coordinates (r(θ0 ), θ0 ) and (r(θ0 + 2nπ), θ0 + 2nπ) to represent the same point. This
requires r(θ0 + 2nπ) = r(θ0 ), which means sin( 6(θ0 +2nπ)
) = sin( 6θ70 ). This requires that
7
6(2nπ)/7 be an even multiple of π, which happens first when n = 7, corresponding to
π/2
π
the interval [0, 2(7)π] = [0, 14π].
0
3π/2
(c) (5 points) True FALSE The set of all points in the plane that are equidistant from two
fixed points in that plane is an ellipse.
Solution: An ellipse is the set of all points in a plane the sum of whose distances from
two fixed points (foci) in that plane is a constant.
(d) (5 points) True
FALSE In polar coordinates ds, the element (differential) of arc length,
p
2
is given by (dr) + r(dθ)2 .
Solution: In polar coordinates ds, the element (differential) of arc length, is given by
p
(dr)2 + (rdθ)2 .
Calculus II (MAC2312-02)
Page 3 of 7
2. Consider the curve given parameterically by x = et − 3, y = e2t .
(a) (5 points) Eliminate the parameter to find a Cartesian definition of the curve.
Solution: Since y = e2t = (et )2 and x + 3 = et > 0, we have
y = (x + 3)2 ,
for x > −3.
Note that the restriction x > −3 for the non-parametric Cartesian curve is crucial to
make it equivalent to the parametric curve.
(b) (6 points) Sketch the graph of the curve and indicate with an arrow the direction in which
the graph is traced as the parameter increases.
As t increases, et = x + 3 increases, which means x and y = (x + 3)2 increase. So, as t
increases the graph is traced from left to right.
Calculus II (MAC2312-02)
Page 4 of 7
3. (13 points) Find the length of the curve x = 1 + 3t2 , y = 4 + 2t3 when 0 ≤ t ≤ 1.
Solution:
We have dx = 6tdt and dy = 6t2 dt.
p
(dx)2 + (dy)2 , which gives
1
Z
Z
ds =
L=
1
Z
p
2
2
2
(6t) + (6t ) dt = 6
0
0
Z
1
=6
√
t 1 + t2 dt
√
|t| 1 + t2 dt
(since t ≥ 0, |t| = t.)
1
2t(1 + t2 )1/2 dt
0
1
√
3
2 3/2 =
(1 + t ) = 2(2 2 − 1).
3/2
0
=3
1
0
0
Z
The element of arc length is ds :=
Calculus II (MAC2312-02)
Page 5 of 7
4. (15 points) Show that the cardioid r = 1 + sin θ has a vertical tangent at the point when
θ = 3π/2.
Hint: Find dy/dx using the chain rule. L’Hôspital’s rule may help with the final step.
π/2
π
Solution:
0
3π/2
dy
We need to show that limθ→ 3π ±
is ∞ or −∞. Since x = r cos θ and y = r sin θ, using the
2
dx
product rule, we have
dr
dx
=
cos θ − r sin θ
dt
dt
and
dy
dr
=
sin θ + r cos θ,
dt
dt
dy
dy dx
dr
= cos θ. From the chain rule,
=
, i.e.,
where r = 1 + sin θ and
dt
dt
dx dt
dy
dy
cos θ sin θ − (1 + sin θ) cos θ
= dt =
dx
dx
cos θ cos θ − (1 + sin θ) sin θ
dt
cos θ
=
.
cos 2θ − sin θ
So,
dy
cos θ
lim +
= lim +
(use L’H, since this is “0/0.”)
3π
3π
dx θ→ 2 cos 2θ − sin θ
θ→ 2
= lim +
− sin θ
−2 sin 2θ − cos θ
= lim +
sin θ
2 sin θ cos θ − cos θ
= lim +
sin θ
cos θ(2 sin θ − 1)
= lim +
1
= ∞,
cos θ
θ→ 3π
2
θ→ 3π
2
θ→ 3π
2
θ→ 3π
2
since cos θ ≥ 0 in the fourth quadrant and cos θ approaches zero from above (zero) as θ
approaches 3π
from the right. Similarly,
2
lim −
θ→ 3π
2
dy
1
= lim −
= −∞.
3π
dx θ→ 2 cos θ
Calculus II (MAC2312-02)
Page 6 of 7
5. (10 points) Find a Cartesian equation for the polar curve r = 2 cos θ and identify the curve.
Hint: Completing the square may help.
π/2
3π/4
π/4
π
0
−3π/4
−π/4
−π/2
Solution:
Recall that r2 = x2 + y 2 and x = r cos θ. Multiplying both sides of r = 2 cos θ by r we get
r2
x2 + y 2
x2 − 2x + y 2
x2 − 2x + 1 + y 2
(x − 1)2 + y 2
= 2r cos θ,
= 2x,
= 0,
= 1,
= 1,
which means this curve is a circle.
6. (10 points) Find the area enclosed by one loop of the four-petal rose r = cos 2θ.
π/2
3π/4
π/4
π
0
−3π/4
Solution:
−π/4
−π/2
Z
π/4
A=
−π/4
Z
=
0
π/4
1 2
1
r dθ =
2
2
Z
π/4
2
Z
cos 2θdθ =
−π/4
π/4
cos2 2θdθ
0
π/4
1 + cos 4θ
1
1
dθ = (θ + sin 4θ) = π/8
2
2
4
0
Calculus II (MAC2312-02)
Page 7 of 7
7. (10 points) Find all the points of intersection of the polar curves r = sin θ and r = sin 2θ.
π/2
3π/4
π/4
π
0
−3π/4
−π/4
−π/2
Solution:
Recall that a “collision point” is a point that is reached by two (or more) parametric curves
(including polar curves) at the same value of the parameter(s). Every intersection point is a
collision point, but not all intersections happen at collisions. Here, the collision points can
be found by looking for those values of θ that give the same r:
sin 2θ = sin θ,
2 sin θ cos θ = sin θ,
sin θ(2 cos θ − 1) = 0,
So, either sin θ = 0, which gives θ = kπ √and r = 0, or cos θ = 1/2, which gives θ =
π/3 + 2kπ and r√= sin(π/3) = sin(2π/3) = 3/2, or θ = −π/3 + 2kπ and r = sin(−π/3) =
sin(−2π/3) = − 3/2, where k is any integer. As seen from the graphs, these collision points
are the only
curves. In other words, the intersection points are
√
√ intersection points of these
(0, kπ), ( 3/2, π/3 + 2kπ), and (− 3/2, −π/3 + 2kπ).
8. (15 points) A conic is given by the equation r =
directrix, and sketch the conic.
10
. Identify the conic, locate the
3 − 2 cos θ
Hint: r = ed/(1 − e cos θ).
π/2
3π/4
π/4
π
0
−3π/4
Solution:
−π/4
−π/2
10/3
Since r = 1−(2/3)
cos θ , we have e = 2/3 and ed = (2/3)d = 10/3, i.e., d = 5. Because e < 1, this is
an ellipse. At θ = 0 we have r = 10/(3 − 2) = 10. At θ = π we have r = 10/(3 + 2) = 2. So, (10, 0)
and (2, π) are the polar coordinates of the major vertices of this ellipse. Recall that the pole is at
one of the foci of the ellipse. (For comparison, in Cartesian coordinates, when there are no shifts
the origin is at the center of an ellipse.) So, the directrix is x = −d = −5, shown as a red, dashed
line.