MA 22S1
Assignment 2
Due 19-21 October 2015
Id:
22S1-f2015-2.m4,v 1.1 2015/10/12 21:13:33 john Exp john
1. For each of the following sets given in polar coordinates determine
whether the set is symmetric about the x-axis, y-axis or origin.
(a) r = csc θ
(b) r 2 sin(2θ) < 1
(c) r = sin θ + cos θ
(d) r = 1/(1 − 2 cos θ)
(e) sec2 θ = 2
Solution: Using the symmetry criteria from lecture or the notes, these
are
(a) symmetric about the y-axis
(b) symmetric about about the orgin
(c) not symmetric about any of them
(d) symmetric about the x-axis
(e) symmetric about all three
2. Find the area of the region in the plane where
√
r 2 sin(2θ) > 2 3, r 2 < 4.
Solution: In this region
sin(2θ) >
1
√
3
2
Id:
22S1-f2015-2.m4,v 1.1 2015/10/12 21:13:33 john Exp john 2
This is possible if and only if π/3 < 2θ < 2π/3, or if 2θ differs by an
integer multiple of 2π from a value in that interval. In other words,
π/6 < θ < π/3 or θ differs by an integer multiple of π from a value in
that interval. Within the range −π < θ < π the possible values are
π/6 < θ < π/3 and −5π/6 < θ < −2π/3. We can just calculate the
area of the first region and double the result, since the two regions are
related by a rotation through an angle of ±π. So
√ !
Z π/3
2 3
A=
4−
dθ
sin(2θ)
π/6
Z π/3 √
=
4 − 2 3 csc(2θ) dθ
π/6
iπ/3
h
√
= 4θ − 3 log(csc(2θ) − cot(2θ))
π/6
!
!
4π √
2
2
4π √
1
1
− 3 log √ − − √ −
+ 3 log √ − √
=
3
6
3
3
3
3
2π √
− 3 log 3.
=
3
The numerical value is 0.1915428006. You don’t need to calculate that,
but it is at least comforting to see that it’s positive.
3. (a) Find foci of the ellipse
y2
x2
+
= 1.
25 16
Solution: This ellipse is in standard form
x2 y 2
+ 2 =1
a2
b
with
a > b so the foci are on the x axis, at a distance of c =
√
2
a − b2 from the origin. In this case a = 5 and b = 4 so c = 3
and the foci are at (3, 0) and (−3, 0).
(b) Find foci of the ellipse
x2
y2
+
= 1.
16 25
Solution: This is the same ellipse, but rotated through an angle
of ±π/2 so the foci are at (0, 3) and (0, −3).