Homework 1 Solutions
Math 352, Fall 2011
Problem 1.
(a) Let p be the point of contact between the bar and the circle, let q be the endpoint
of the bar. We may assume that the point p rotates around the unit circle at unit
speed:
p = (cos t, sin t).
The vector from p to q points in the direction of (sin t, − cos t). Assuming p
rotates at unit speed, the magnitude of this vector will increase at unit speed, so
q − p = t(sin t, −cos t) = (t sin t, −t cos t).
We conclude that
q = p + (q − p) = (cos t + t sin t, sin t − t cos t)
(b) The portion from (1, 0) to (1, −4π) corresponds to 0 ≤ t ≤ 4π, so the length is
Z
4π
Z
kγ̇(t)k dt =
0
4π
p
γ̇1 (t)2 + γ̇2 (t)2 dt
0
Z
=
4π
Z
p
2
2
(t cos t) + (t sin t) dt =
4π
|t| dt = 8π 2
0
0
Problem 2.
(a) Let p be the point of contact between the square and the catenary, let q be the
midpoint of the top edge of the square, and let r be the center point of the square.
The point p moves along the catenary:
p = (t, cosh t).
The vector ṗ = (1, sinh t) is tangent to the catenary at t, with (sech t, tanh t)
being the corresponding tangent unit vector. The vector from p to q points in
the exact opposite direction, and its magnitude is the arclength of the catenary
from (0, 1) to p. We conclude that
q − p = (sinh t)(−sech t, −tanh t) = (−tanh t, −sinh t tanh t).
The vector from q to r points in the direction of the unit vector (tanh t, −sech t),
and has magnitude L/2:
r−q =
1
L
(tanh t, −sech t).
2
Adding these together gives
L
L
r = t − tanh t + tanh t, cosh t − sinh t tanh t − sech t
2
2
L
L
= t− 1−
tanh t, 1 −
sech t
2
2
(b) When L = 2 , the path moves along a horizontal line.
Problem 3.
The curve is a parabola. There are a few different ways of figuring this out:
Method 1: “Consecutive” Lines
The idea of this method is that the intersection of two “consecutive” line segments
will be a point on the curve. That is, if we find the intersection point of L(t) and
L(u), then the limit as u → t will be a point on the curve.
This method is simple to implement. First, observe that the line L(t) is defined
by the formula
y = (2t − 1)x + 2t(1 − t).
Using straightforward algebra, we can compute that the intersection point of L(t)
and L(u) is
(t + u − 1, 2tu − t − u + 1) .
Taking the limit as u → t gives us a parametrization of the curve:
γ(t) = (2t − 1, 2t2 − 2t + 1).
Method 2: Guess and Check
The curve looks a bit like a parabola. It goes through the points (−1, 0), (0, 1/2),
and (1, 1), so it is reasonable to guess that the solution is the parabola
y =
x2 + 1
.
2
To verify that this guess is correct, observe that the tangent line to this parabola
at the point x = a is given by the formula
y = ax +
This is the same as the line L
1 − a2
.
2
1+a
, and therefore this parabola is tangent to
2
each of the lines L(t).
2
Method 3: A Differential Equation
Suppose then that the curve is of the form y = f (x). Then the tangent line at
the point (x0 , y0 ) is given by the formula
y = f 0 (x0 ) x + y0 − x0 f 0 (x0 ) .
Now, each line L(t) is given by the formula
y = (2t − 1)x + 2t(1 − t).
These must be the same lines, which gives us the following two equations:
2t − 1 = f 0 (x0 )
and
2t(1 − t) = y0 − x0 f 0 (x0 ).
Eliminating t and simplifying yields the following equation:
f 0 (x0 )2 − 2x0 f 0 (x0 ) + 2f (x0 ) = 1.
That is, the desired curve is a solution to the differential equation
(y 0 )2 − 2xy 0 + 2y = 1,
with initial condition y(0) = 1/2. This can be solved by completing the square
and making a substitution, eventually yielding y = (x2 + 1)/2.
Method 4: Optimization (due to Ke)
Each line L(t) is given by the formula
y = (2t − 1)x + 2t(1 − t).
If we fix x and vary t, then the maximum value of y will correspond to a point
on the curve. To find the maximum, we compute:
∂y
= 2x + 2 − 4t.
∂t
Setting this equal to zero, we find that the maximum value of y occurs when
t = (x + 1)/2. Plugging this into the formula for y gives y = (x2 + 1)/2.
3