Homework 3 Solutions
Problem 1.
Let ~x(t) = (t − tanh t, sech t). Then
~x 0 (t) = (1 − sech2 t, − sech t tanh t) = (tanh2 t, −sech t tanh t) = tanh t (tanh t, −sech t)
so
s0 (t) = tanh t
and
T~ (t) = (tanh t, −sech t).
Then
~ (t).
T~ 0 (t) = (sech2 t, sech t tanh t) = sech t (sech t, tanh t) = (sech t) U
~ (t), it follows that
Since T~ (t) = s0 (t) κg (t) U
κg (t) =
sech t
sech t
=
= csch t
0
s (t)
tanh t
Problem 2.
We can parameterize this ellipse as ~x(t) = (a cos t, b sin t). At (a, 0), the curvature is
κg (0) =
−x00 (0) y 0 (0) + y 00 (0) x0 (0)
−(−a)(b) + (0)(0)
a
=
=
3/2
(02 + b2 )3/2
b2
x0 (0)2 + y 0 (0)2
and the curvature is the same at (−a, 0). At (0, b), the curvature is
κg (π/2) =
−x00 (π/2) y 0 (π/2) + y 00 (π/2) x0 (π/2)
−(0)(0) + (−b)(−a)
b
= 2
=
3/2
2
2
3/2
((−a) + 0 )
a
x0 (π/w)2 + y 0 (π/w)2
and the curvature at (0, −b) is the same.
The osculating circle
point
aradius of 1/κg (0) = b2 /a, and is therefore
at the
(a,2 0) has
2
2
b
a −b
centered at the point a − , 0 =
, 0 . Thus, the equation for the circle is
a
a
a2 − b 2
x−
a
2
+ y2 =
b4
a2
Problem 3.
By symmetry, the center of the circle must line on the y-axis, so the center is (0, b) for some
value of b. The distance from the center to the points (0, 1) and (h, cosh h) must be the
same, so
(b − 1)2 = h2 + (b − cosh h)2 .
sinh2 h + h2
sin2 h + h2
, so the center is at 0,
Solving for b gives b =
2 cosh h − 2
2 cosh h − 2
By L’Hôpital’s rule,
2 sinh h cosh h + 2h
2 cosh2 h + 2 sinh2 h + 2
sinh2 h + h2
= lim
= lim
= 2.
h→0
h→0
h→0 2 cosh h − 2
2 sinh h
2 cosh h
lim
Thus, the limit of the center as h → 0 is (0, 2) .
Since the radius of the osculating circle is equal to 1, it follows that the curvature of the
catenary at the point (0, 1) is equal to 1 .
Problem 4.
We have
Z
θ(s) =
Z
κg (s) ds =
(2s)−1/2 ds =
√
2s + C.
Since ~x(0) = (1, 0), we know that θ(0) = 0. It follows that C = 0, so θ(s) =
Since the curve is unit speed, we conclude that
√
√ ~x 0 (s) = cos θ(s), sin θ(s) = cos 2s, sin 2s
Then
√
2s .
√
√ ~x (s) ds =
cos 2s, sin 2s ds.
√
This integral is difficult. Substituting t = 2s gives
Z
Z
√
√
√
√
cos 2s ds =
t cos t dt = t sin t + cos t + C = 2s sin 2s + cos 2s + C
Z
~x(s) =
Z
and similarly
Z
0
√
√
√
√
sin 2s ds = − 2s cos 2s + sin 2s + C. Thus
~x(t) =
√
2s sin
√
2s + cos
√
√
√
√ ~
2s, − 2s cos 2s + sin 2s + C.
~ = (0, 0), so
Since ~x(0) = (1, 0), it follows that C
√
√
√
√
√
√ ~x(s) =
2s sin 2s + cos 2s, − 2s cos 2s + sin 2s
Substituting t =
√
2s gives a simpler parametrization:
t sin t + cos t, −t cos t + sin t for t > 0