Calculus 10.3 Exercise
H. C. Tseng
†
December 26, 2005
Exercise (13). Find the distance between the points with polar coordinates (1, π6 ) and (3, 3π
4 ).
Solution. Since x = r cos θ, y = r sin θ, we have
√
π
3 1
π
(x1 , y1 ) = (1 cos , 1 sin ) = (
, ),
6
6
2 2√
√
3π
3π
3 2 3 2
(x2 , y2 ) = (3 cos
, 3 sin
) = (−
,
).
4
4
2
2
Hence, the distance
p
(x1 − x2 )2 + (y1 − y2 )2
s √
√
√
1 3 2 2
3 3 2 2
+
) +( −
)
= (
2
2
2
2
q
√
√
1
=
40 + 6 6 − 6 2.
2
d=
Exercise (27). For each of the described curves, decide if the curve would be more easily given by a
polar equation or a Cartesian equation. Then write an equation for a curve.
(a) A line through the origin that makes an angle of π/6 with the positive x-axis.
(b) A vertical line through the point (3, 3).
Solution.
(a) We have
θ = π/6
directly. Since tan θ =
y
x,
we obtain
y = tan
π
1
x = √ x.
6
3
(b) The easier description here is the Cartesian equation x = 3.
Exercise (31). Sketch the curve with the given polar equation
r = sin θ
Solution. By this polar equation, we have r2 = r sin θ by multiplying r both sides. Since r2 = x2 + y 2
and r sin θ = y, we have
2 2
1
1
2
x + y−
=
,
2
2
a circle of radius 21 centered at 0, 21 .
† Department of Applied Mathematics, National Chiao Tung University, 1001 Ta Hsueh Road, Hsinchu, Taiwan 300,
ROC.
E-mail: hctseng.am93g@nctu.edu.tw
Website: http://lcse.math.nctu.edu.tw/THChieh/
Figure 1: Ex31
Exercise (47). The figure shows the graph of r as a function of θ in Cartesian coordinates. Use it to
sketch the corresponding polar curve.
Solution. See Figure-Ex47. For θ = 0, π, 2π, r has its minimum value of about 12 . For θ = π2 , 3π
2 , r attains
its maximum value of 2. We see that the graph has a similar shape for 0 ≤ θ ≤ π and π ≤ θ ≤ 2π.
Figure 2: Ex47
Exercise (55). Find the slope of the tangent line to the given polar curve at the point specified by the
value of θ.
r = 2 sin θ,
θ = π/6.
Solution. For the polar equation r = 2 sin θ, we have
x = r cos θ = 2 sin θ cos θ = sin 2θ
y = r sin θ = 2 sin2 θ
and slope
dy/dθ
2 · 2 sin θ cos θ
sin 2θ
dy
=
=
=
= tan 2θ.
dx
dx/dθ
2 · cos 2θ
cos 2θ
When θ =
π
6,
the slope
dy
π √
= tan = 3.
dx
3
Exercise (65). Find the points on the given curve where the tangent line is horizontal or vertical.
r = cos 2θ.
Solution. For the polar equation r = cos 2θ, we have
x = r cos θ = cos 2θ cos θ
y = r sin θ = cos 2θ sin θ
and
dx
= −2 sin 2θ cos θ − cos 2θ sin θ
dθ
= −4 sin θ cos2 θ − (2 cos2 θ − 1) sin θ
= sin θ(1 − 6 cos2 θ)
dy
= −2 sin 2θ sin θ + cos 2θ cos θ
dθ
= −4 sin2 θ cos θ + cos3 θ − sin2 θ cos θ
= cos θ(cos2 θ − 5 sin2 θ)
= cos θ(1 − 6 sin2 θ)
To find the horizontal tangent line, we consider those θ such that dy
dθ = 0. Hence, cos θ = 0 or
−1 √1
. So the tangent line is
,
α,
π
−
α,
π
+
α,
2π
−
α
where
α
=
sin
sin θ = ± √16 , that is, θ = π2 , 3π
2
6
horizontal at
π
3π
2
2
2
2
(−1, ), (−1,
), ( , α), ( , π − α), ( , π + α), ( , 2π − α)
2
2
3
3
3
3
in polar coordinates.
√1
It is similar to find the vertical tangent line, For dx
dθ = 0, we have sin θ = 0 or cos θ = ± 6 , that is,
θ = 0, π, β, π − β, π + β, 2π − β where β = cos−1 √16 . So the tangent line is vertical at
2
2
2
2
(1, 0), (1, π), (− , β), (− , π − β), (− , π + β), (− , 2π − β)
3
3
3
3
> inwith(plots):
polar coordinates.
polarplot([cos(2*t),t,t=0..2*Pi],color=black);
>
Figure 3: Ex65