In this module we look at geometric properties of curves in R2, or R3,
f(t) = (x(t), y(t))
f(t) = (x(t), y(t), z(t))
Notice that these functions describe both a geographical curve -- the physical path followed by an object -- and the way in which the object travels along the path -- that is, at each time, t, the location f(t), of a traveling object. In this section we study the geometric properties of the physical or geographical curve.
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Consider the cardioid shown at the right and described by the function
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| We are interested in the length of this curve. We can approximate the curve by a polygonal path made up of straight line segments as shown in the picture at the right. By using a large number of short straight line segments we can obtain a very good estimate. |
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We develop formulas for estimating the arclength of a curve described by a function f(t), with a <= t <= b in either R2 or R3. Let n denote the number of straight line segments and use the following notation.
b - a
h = -----
n
t0 = a
t1 = a + h
t2 = a + 2 h
.
.
ti = a + i h
.
.
tn = a + n h = b
So that the polygonal path connects the points
The i-th segment of this path has length
So the total length of the polygonal path is
Use your CAS window to estimate the arclength of each of the following curves.
x(t) = cos t
y(t) = sin t
0 <= t <= 2 Pi
x(t) = R cos t
y(t) = R sin t
0 <= t <= 2 Pi where R is a constant.
x(t) = 2 cos t
y(t) = sin t
0 <= t <= 2 Pi
x(t) = 3 cos t
y(t) = sin t
0 <= t <= 2 Pi
x(t) = 4 cos t
y(t) = sin t
0 <= t <= 2 Pi
x(t) = cos 2 Pi t
y(t) = sin 2 Pi t
z(t) = t
0<= t <= 1
x(t) = cos 2 Pi t
y(t) = sin 2 Pi t
z(t) = 0.2 t
0<= t <= 1
x(t) = cos 2 Pi t
y(t) = sin 2 Pi t
z(t) = m t where m is a constant.
0<= t <= 1
The formula
enables us to estimate the length of an arc as precisely as might be needed for any particular purpose. We can find the exact value of the arc length by taking the limit.
Theorem:
If the function f(t) on the interval a <= t <= b describes a curve then the arclength of the curve is
Proof:
Notice, in particular, that if f(t) = (x(t), y(t)) describes a curve in R2 then its arclength is
and that if f(t) = (x(t), y(t), z(t)) describes a curve in R3 then its arclength is
Set up the integrals to find the exact arclength for each of the curves in the first set of exercises. These exercises are repeated below. If you can evaluate the integrals exactly, do so.
x(t) = cos t
y(t) = sin t
0 <= t <= 2 Pi
x(t) = R cos t
y(t) = R sin t
0 <= t <= 2 Pi where R is a constant.
x(t) = 2 cos t
y(t) = sin t
0 <= t <= 2 Pi
x(t) = 3 cos t
y(t) = sin t
0 <= t <= 2 Pi
x(t) = 4 cos t
y(t) = sin t
0 <= t <= 2 Pi
x(t) = cos 2 Pi t
y(t) = sin 2 Pi t
z(t) = t
0<= t <= 1
x(t) = cos 2 Pi t
y(t) = sin 2 Pi t
z(t) = 0.2 t
0<= t <= 1
x(t) = cos 2 Pi t
y(t) = sin 2 Pi t
z(t) = m t where m is a constant.
0<= t <= 1
The Tangent Vector
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The movie at the right shows an object traveling along a path. We have attached two
vectors to the object. The first vector, in blue, is called the tangent
vector. It is a unit vector that points in the direction the object is
traveling. The second vector, in red, is called the normal vector.
It is also a unit vector. It points to the traveler's left or to the
traveler's right -- in the direction that the object
is turning. The normal vector is perpendicular to the tangent vector.
The tangent vector can be zero -- when the object's speed is zero. The normal vector can also be zero -- when the object is going straight ahead. Notice that at one point in the movie below the normal vector disappears. This point is a point where the object is going straight for a brief moment. |
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T(t) = --------- f'(t)
||f'(t)||
Find the tangent vector for each of the following curves.
where R is a constant.
where R and m are constants.
where a, b, and c are constants.
The Normal vector
The normal vector, N(t), to a curve, f(t), is defined by
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N(t) = --------- T'(t)
||T'(t)||
The following theorem shows that the normal vector is a unit vector that is perpendicular to the curve -- that is, to its tangent vector. Notice that since the normal vector is a unit vector in the direction of T'(t) it is pointing in the direction that the tangent vector, T(t), is turning.
Theorem
The normal vector is a unit vector that is perpendicular to the curve -- that is, to its tangent vector.
Proof:
It is obvious that the normal vector is a unit vector. To show that it is perpendicular to the tangent vector notice that
Thus,
and
So, differentiating both sides of this equation, we see that
which proves that T(t) and T'(t) are perpendicular. Since N(t) is a scalar multiple of T'(t), N(t) is also perpendicular to T(t). This completes the proof.
Use your CAS window to draw the tangent vector and the normal vector to each of the curves below at the indicated points.
where R is a constant.
where R and m are constants.
where a, b, and c are constants.
Suppose that f(t) for a <= t <= b describes a curve in R2 or R3. Suppose that h(t) is any function whose domain and range are [a, b]. Notice that the curve f(h(t)) follows the same geographical path as the curve f(t) but that the two objects may arrive at the same points at different times. What can you say about the tangent vector and the normal vector to each of these two curves at the same geographical points? Be careful.
When a vector, F, for example, a vector representing force, acts on an object as it is traveling it is often important to find the component of that vector that is in the same direction the object is traveling. This component is called the tangential component and is given by
Similarly, it is often important to find the component of the vector F that is in the direction of the normal vector, N(t). This component is called the normal component and is given by
Curvature
Next we want to determine how sharply a curve f(t) is bending. The obvious way to do this is to measure how quickly the tangent vector T(t) is changing by computing T'(t) but unfortunately this does not work because it depends on both the shape of the curve and the speed with which the object is traveling along the curve. The following example illustrates this.
Use your CAS window to look at the curve
where R and c are constants. Notice that the geographical path followed by this object is a circle of radius R and the speed of the object is c.
Find ||T'(t)||. How is ||T'(c)|| affected by the geography of the curve? How is it affected by the speed with which the object travels along the curve?
As we see from the example above, the rate at which the tangent vector T(t) is changing depends on both the shape of the curve (how sharply it is bending) and the speed with which the object is traveling. This leads us to the following definition for the curvature of a curve.
In view of the example above we define the radius of curvature by
where a is a constant.
where R and m are constants.