CLP-4 Vector Calculus

Chapter 1 Curves 

We are now going to study vector-valued functions of one real variable. That is, we are going to study functions that assign to each real number t (typically in some interval) a vector1 r(t). For example

r(t) = ( x(t), y(t), z(t))

might be the position of a particle at time t. As t varies, r(t) sweeps out a curve.

While in some applications t will indeed be “time”, it does not have to be. It can be simply a parameter that is used to label the different points on the curve that r(t) sweeps out. We then say that r(t) provides a parameterization of the curve.

Example 1.0.1 Parametrization of x2 + y2 = a2. While we will often use t as the parameter in a parametrized curve r(t), there is no need to call it t. Sometimes it is natural to use a different name for the parameter. For example, consider the circle x2 + y2 = a2. It is natural to use the angle θ in the sketch below to label the point

That is,

r(θ) = ( a cos θ , a sin θ )        0 ≤ θ < 2π

is a parametrization of the circle x2 +y2 = a2. Just looking at the figure above, it is clear that, as θ runs from 0 to 2π, r(θ) traces out the full circle.

However beware that just knowing that r(t) lies on a specified curve does not guarantee that, as t varies, r(t) covers the entire curve. For example, as t runs over the whole real line, 2/π arctan(t) runs over the interval (−1, 1). For all t,

is well-defined and obeys x(t)2 + y(t)2 = a2. But this r(t) does not cover the entire circle because y(t) is always positive. 

Example 1.0.2 Parametrization of (x−h)2 +(y−k)2 = a2. We can tweak the parametrization of Example 1.0.1 to get a parametrization of the circle of radius a that is centred on (h, k). One way to do so is to redraw the sketch of Example 1.0.1 with the circle translated so that its centre is at (h, k).

We see from the sketch that

r(θ) = ( h+ a cos θ , k + a sin θ )      0 ≤ θ < 2π

is a parametrization of the circle (x− h)2 + (y − k)2 = a2. A second way to come up with this parametrization is to observe that we can turn the trig identity cos2 t+sin2 t = 1 into the equation (x−h)2+(y−k)2 = a2 of the circle by

• multiplying the trig identity by a2 to get (a cos t)2 + (a sin t)2 = a2 and then
• setting a cos t = x − h and a sin t = y − k , which turns (a cos t)2 + (a sin t)2 = a2 into (x− h)2 + (y − k)2 = a2.