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# Curves in

Given a point and a vector there is always a unique point such that (q-p)=v; intuitively, q is at the point of the arrow v when its tail is at p. Then we can write q=p+v and the line segment from p (in the direction v) to q is given in parametric form by

or equivalently

and we say that the tangent vector, or velocity vector of this line at is v. Thus, we can write this as the derivative, which is the limit of differences:

because L is a linear function of t, since p and v are constant.

As we shall see, a curve may have a nonlinear dependence on its parameter and a velocity vector that varies in magnitude and direction, so curves are natural generalisations of line segments.

Tangent vector, speed and acceleration vector A curve in with parameter t satisfying is a continuous map

 (5)

Note that is a map and its image, (also called track or path), is a subset of we keep these concepts distinct. The curve starts at the point and ends at the point Sometimes, one or both endpoints of the curve are absent; so in general the domain of a curve may be an interval of any kind.

For our purposes, we shall suppose that our curves are differentiable, in the sense that the components, are real functions of t possessing derivatives of all orders--so no corners like those in the graph of |x|. The tangent vector or velocity of is the vector valued map which in components is given by

 (6)

and its speed is the absolute value of the velocity vector. The acceleration of is the vector given by the derivative of the velocity. Observe that the velocity and acceleration vectors are attached to the curve and change as the parameter moves the point of attachment.

We are particularly interested in regular curves which are differentiable and have nowhere zero velocity (they are always going somewhere, not stopped); then we make calculations easier if we choose the parameter set 0<s<L to make a unit vector for all s, and L is actually the total length of the curve.

The length of the curve (1) is defined as the integral of the speed over the domain [a,b]

 (7)

and it is independent of reparametrization. Unit speed curves are parametrized by arc length because then for all t. In general, it is difficult to calculate analytically the arc length of a given curve--because of the presence of the square root of sums of functions in the integrand; the same is true for other calculations for curves and surfaces but computer algebra software can help1.

Exercises on plane curves

1.
Show that a parametric equation for the line segment from (3,5) to is given by

Note that the line segment is the image of the map What is the tangent vector of
2.
Find another parametric equation for the line segment in the previous example, but having a tangent vector with half the magnitude of that used for
3.
Clearly, there is a well-defined continuous curve given by

 (8)

What happens to its tangent vector at t=0? Show that the curve has a well-defined length.
4.
Show that for all the matrix

when applied to the coordinates of a curve (viewed as a column vector) rotates the curve through angle in the plane, that is, round the z-axis. Find a suitable value that rotates the curve in the previous question through 30o; give an explicit equation for the rotated curve. Why does the rotation matrix leave the length of a curve unaltered? Can a reflection in the line y=x be a rotation?
5.
Find for the rotation matrix corresponding to the linear operator

and show that J gives an anticlockwise rotation of in
6.
Express the equation for the parabola y=3x2+2 in parametric form and find its velocity, speed and acceleration. Show that the same image can be obtained by another curve with velocity vector the negative of that of the first one.
7.
Verify that a curve defining the unit circle in with centre at the origin Ogiven by the set

has a parametric equation given by

Find its velocity, speed, acceleration and length. Show that the same image can be obtained by another curve with velocity vector the negative of that of the first one.
8.
Investigate the hyperbola

9.
Plot the limaçon (French name for slug--why?) given by

and show that the curve passes twice through the origin in different directions, which emphasises why we have to specify the parameter value and not the point on the curve when we require the velocity vector.
10.
Find a parametric equation for the equator of the sphere and for a perpendicular circle of longitude.

Implicitly defined plane curves We know that some curves are defined implicitly, like the unit circle,

 x2+y2-1=0 (9)

However, for f(x,y)=0 to define a parametrized curve near some point (x0,y0) where f is zero, it is sufficient for f to have at least one of its partial derivatives nonzero there.

Exercises on implicit curves

1.
Use the implicit function theorem to prove this assertion.
2.
Investigate the sets of zeros of the following function and a slightly perturbed version
 f(x,y) = x3+y3-3xy (10) f*(x,y) = x3+y3-3xy-0.01 (11)

for which Gray [6] gives graphs on pages 59 and 60.

Evolutes and involutes of plane curves The loci of centres of circles (called osculating circles), that are tangent to the plane curve in (1) and have radii equal to the radii of curvature at the points of tangency, is a new plane curve called the evolute of So, the evolute of a circle is its centre point. Explicitly,

 = (12) = (13)

The involute, starting at of the plane curve in (1) is the plane curve

 (14)

where s is the arc length function of

Exercises on evolutes and involutes

1.
The formula for evolute is independent of reparametrization of
2.
The evolute of an involute of is again

Next: Space Curves Up: Introducing Curves Previous: Group actions
Kit Dodson
2000-01-23