1.3 Solutions
33
1.36 The radius of curvature
ρ =
1 + (dy/dx)
2
3/2
d 2 y/dx 2
(1)
x = v o t = 10 × 3 = 30 m
y =
1
2
gt
2
=
1
2
× 9.8 × 3
2
= 44.1 m.
∴ y =
1
2
g
x 2
v 2
0
v
2
0 =
9.8 × 30
10 2 = 2.94
(2)
d 2 y
dx 2 =
g
v 2
0
=
9.8
10 2 = 0.098
(3)
Using (2) and (3) in (1) we find ρ = 305 m.
1.37 Let P be the position of the boat at any time, Let AP = r , angle B ˆ
AP = θ ,
and let v be the magnitude of each velocity, Fig. 1.5:
dr
dt
= −v + v sin θ
and
r dθ
dt
= v cos θ
∴
1
r
dr
dθ
=
−1 + sin θ
cos θ
∴
dr
r
=
[− sec θ + tan θ ] dθ
∴ ln r = − ln tan
θ
2
+
π
4
− ln cos θ + ln C (a constant)
When θ = 0, r = a, so that C = a
∴ r =
a
tan
θ
2 +
π
4
cos θ
The denominator can be shown to be equal to 1 + sin θ :
∴ r =
a
1 + sin θ
This is the equation of a parabola with AB as semi-latus rectum.
1.38 Take the origin at O, Fig. 1.22. Draw the reference line OC parallel to AB, the
ground level. Let the ball hit the wall at a height H above C. Initially at O,
33
1.36 The radius of curvature
ρ =
1 + (dy/dx)
2
3/2
d 2 y/dx 2
(1)
x = v o t = 10 × 3 = 30 m
y =
1
2
gt
2
=
1
2
× 9.8 × 3
2
= 44.1 m.
∴ y =
1
2
g
x 2
v 2
0
v
2
0 =
9.8 × 30
10 2 = 2.94
(2)
d 2 y
dx 2 =
g
v 2
0
=
9.8
10 2 = 0.098
(3)
Using (2) and (3) in (1) we find ρ = 305 m.
1.37 Let P be the position of the boat at any time, Let AP = r , angle B ˆ
AP = θ ,
and let v be the magnitude of each velocity, Fig. 1.5:
dr
dt
= −v + v sin θ
and
r dθ
dt
= v cos θ
∴
1
r
dr
dθ
=
−1 + sin θ
cos θ
∴
dr
r
=
[− sec θ + tan θ ] dθ
∴ ln r = − ln tan
θ
2
+
π
4
− ln cos θ + ln C (a constant)
When θ = 0, r = a, so that C = a
∴ r =
a
tan
θ
2 +
π
4
cos θ
The denominator can be shown to be equal to 1 + sin θ :
∴ r =
a
1 + sin θ
This is the equation of a parabola with AB as semi-latus rectum.
1.38 Take the origin at O, Fig. 1.22. Draw the reference line OC parallel to AB, the
ground level. Let the ball hit the wall at a height H above C. Initially at O,