guide:8da4726d42

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Consider a function whose domain is a subset of the set of all real numbers and whose range is a subset of all vectors in the plane. If we denote this function by [math]\mbox'''{v'''}[/math], then its value at each number [math]t[/math] in the domain is the vector [math]\mbox'''{v'''}(t)[/math]. Every such vector-valued function [math]\mbox'''{v'''}[/math] of a real variable defines two real-valued coordinate functions [math]v_1[/math] and [math]v_2[/math] as follows: For every [math]t[/math] in the domain of [math]\mbox'''{v'''}[/math], the numbers [math]v_1(t)[/math] and [math]v_2(t)[/math] are the first and second coordinates of the vector [math]\mbox'''{v'''}(t)[/math], respectively. Hence, if the initial point of [math]\mbox'''{v'''}(t)[/math] is [math]P(t)[/math], then [math]v_1(t)[/math] and [math]v_2(t)[/math] are defined by the equation

[[math]] \begin{equation} \mbox'''{v'''}(t) = (v_1(t), v_2(t))_{P(t)}. \label{eq10.4.1} \end{equation} [[/math]]

Limits of vector-valued functions are defined in terms of limits of real-valued functions. Specifically, the limit of [math]\mbox'''{v'''}(t)[/math], as [math]t[/math] approaches [math]t_0[/math], will be denoted by [math]\lim_{t \rightarrow t_0} \mbox'''{v'''}(t)[/math] and is defined by

[[math]] \begin{equation} \lim_{t \rightarrow t_0} \mbox'''{v'''}(t) = (\lim_{t \rightarrow t_0} v_1(t), \lim_{t \rightarrow t_0} v_2(t)) _{\lim_{t \rightarrow t_0} P(t)} . \label{eq10.4.2} \end{equation} [[/math]]

[For the definition of [math]\lim_{t \rightarrow t_0} P(t)[/math], see page 542.] There is the possibility that all the vectors [math]\mbox'''{v'''}(t)[/math] have the same initial point [math]P_0[/math], i.e., that they all lie in the vector space [math]\mathcal{V}_{P_0}[/math]. If this happens, (2) reduces to the simpler equation

[[math]] \lim_{t \rightarrow t_0} \mbox'''{v'''}(t) = (\lim_{t \rightarrow t_0} v_1(t), \lim_{t \rightarrow t_0} v_2(t))_{P_0} . [[/math]]

Let [math]C[/math] be a curve in the plane defined by a parametrization [math]P: I \rightarrow R^2[/math]. If the coordinate functions of [math]P[/math] are denoted by [math]x[/math] and [math]y[/math], then [math]C[/math] is the set of all points

[[math]] P(t)= (x(t), y(t)) [[/math]]

such that [math]t[/math] is in the interval [math]I[/math]. A typical example is shown in Figure 14. Consider a number [math]t_0[/math] in [math]I[/math]. If [math]t[/math] is in [math]I[/math] and distinct from [math]t_0[/math], then the vector [math](P(t_0), P(t))[/math] represents the change in the value of [math]P[/math] from the point [math]P(t_0)[/math] to the point [math]P(t)[/math]. Thus for a change in the value of the parameter from [math]t_0[/math] to [math]t[/math], the scalar product

[[math]] \begin{equation} \frac{1}{t - t_0} (P(t_0), P(t)) \label{eq10.4.3} \end{equation} [[/math]]

File:Guide c5467 scanfig10 14.png

is the ratio of the corresponding change in the value of [math]P[/math] to the difference [math]t - t_0[/math]. Hence the vector (3) represents an average rate of change in position with respect to a change in the parameter. In analogy with the definition of the derivative of a real-valued function, we define the derived vector of [math]P[/math] at [math]t_0[/math], denoted by [math]\mbox'''{d'''}P(t_0)[/math], by the equation

[[math]] \mbox'''{d'''}P(t_0)= \lim_{t \rightarrow t_0} \frac{1}{t - t_0} (P(t_0),P(t)). [[/math]]

Since [math]P(t_0) = (x(t_0), y(t_0))[/math] and [math]P(t) = (x(t), y(t))[/math], the coordinate form of the vector [math](P(t_0), P(t))[/math] is given by

[[math]] (P(t_0), P(t)) = (x(t) - x(t_0), y(t) - y(t_0))_{P(t_0)}. [[/math]]

By the definition of the scalar product,

[[math]] \frac{1}{t - t_0} (P(t_0), P(t)) = \left( \frac{x(t) - x(t_0)}{t - t_0}, \frac{y(t) - y(t_0)}{t - t_0} \right)_{P(t_0)} , [[/math]]

and so

[[math]] \mbox'''{d'''}P(t_0) = \left( \lim_{t \rightarrow t_0} \frac{x(t) - x(t_0)}{t - t_0}, \lim_{t \rightarrow t_0} \frac{y(t) - y(t_0)}{t - t_0} \right)_{P(t_0)} . [[/math]]

Recall that the derivatives of the functions [math]x[/math] and [math]y[/math] at [math]t_0[/math] are by definition

[[math]] \begin{eqnarray*} x'(t_0) &=& \lim_{t \rightarrow t_0} \frac{x(t) - x(t_0)}{t - t_0}, \\ y'(t_0) &=& \lim_{t \rightarrow t_0} \frac{y(t) - y(t_0)}{t - t_0}, \end{eqnarray*} [[/math]]

provided these limits exist. It follows that

Theorem

The parametrization defined by [math]P(t) = (x(t), y(t) )[/math] is differentiable at [math]t_0[/math] if and only if the derived vector [math]\mbox'''{d'''}P(t_0)[/math] exists. If it does exist, then

[[math]] \mbox'''{d'''}P(t_0) = (x'(t_0), y'(t_0))_{P(t_0)} . [[/math]]

Example Consider the curve parametrized by

[[math]] P(t) = (x(t), y(t)) = (t^2 - 1, 2t + 1), \;\;\; - \infty \lt t \lt \infty. [[/math]]

Compute the derived vectors of [math]P[/math] at [math]t_0 = - 1[/math], at [math]t_0 = 0[/math], and at [math]t_0 = 1[/math]. Draw the curve and the three derived vectors in the [math]xy[/math]-plane. As a result of (4.1), we have

[[math]] \mbox'''{d'''}P(t_0) = (x'(t_0),y'(t_0))_{P(t_0)} = (2t_0, 2)_{P(t_0)} . [[/math]]

Hence

[[math]] \begin{array}{ccl} \mbox'''{d'''}P(-1) = (- 2, 2)_{P(-1)}&\;\;\;\mathrm{and}&\;\;\; P(-1) = (0, -1), \\ \mbox'''{d'''}P(0) = (0, 2)_{P(0)} &\;\;\;\mathrm{and}&\;\;\; P(0) = (-1, 1), \\ \mbox'''{d'''}P(1) = (2, 2)_{P(1)} &\;\;\;\mathrm{and}&\;\;\; P(1) = (0, 3) . \end{array} [[/math]]

The terminal points of the three derived vectors are, respectively,

[[math]] \begin{eqnarray*} (0 - 2, -1 + 2) &=& (-2, 1), \\ (-1 + 0, 1 + 2) &=& (-1, 3), \\ (0 + 2, 3 + 2) &=& (2, 5) . \end{eqnarray*} [[/math]]

The parametrized curve is a parabola, as can be seen by setting

[[math]] \left \{ \begin{array}{l} x = t^2 - 1, \\ y = 2t + 1. \end{array} \right . [[/math]]

Solving the second equation for [math]t[/math], we get [math]t = \frac{y -1}{2}[/math], and substituting this value in the first, we obtain [math]x = \frac{(y - 1)^2}{4} - 1[/math], or, equivalently,

[[math]] 4(x + 1) = (y - 1)^2. [[/math]]

The latter is an equation of a parabola with vertex (-1, 1). If [math]x = 0[/math], then [math]4 = (y - 1)^2[/math], or, equivalently, [math]\pm 2 = y -1[/math], which implies that [math]y = -1[/math] or 3. The parametrized curve together with the three vectors is shown in Figure 15. Note that each of these vectors is tangent to the parabola.

If a parametrization [math]P: I \rightarrow R^2[/math] is differentiable at [math]t_0[/math], then we define a tangent vector to the resulting parametrized curve at [math]t_0[/math] to be any scalar multiple of the derived vector [math]\mbox'''{d'''}P(t_0)[/math]. In particular, the derived vector itself is a tangent vector. The set of all tangent vectors at [math]t_0[/math] is a subset of [math]\mathcal{V}_{P(t_0)}[/math], since every scalar multiple of [math]\mbox'''{d'''}P(t_0)[/math] has initial point [math]P(t_0)[/math]. Moreover,

File:Guide c5467 scanfig10 15.png

Theorem

The set of all tangent vectors to the parametrized curve [math]P(t)[/math] at [math]t_0[/math] is a vector space.

Show Proof

This result has nothing to do with any special properties of the derived vector, since the set of all scalar multiples of any vector [math]\mbox'''{u'''}[/math] is a vector space. This result is proved, if [math]\mbox'''{u'''}[/math] is nonzero, in Example 3 of Section 3. If [math]\mbox'''{u'''}[/math] is a zero vector, the result is even simpler: The set of all scalar multiples of a zero vector [math]\mbox'''{0'''}[/math] is the set having [math]\mbox'''{0'''}[/math] as its only member, and the six conditions for a vector space are trivially satisfied. This completes the argument.

■

Consider a parametrization defined by [math]P(t)= (x(f), y(t))[/math], which is differentiable at [math]t_0[/math] and for which the derived vector [math]\mbox'''{d'''}P(t_0)[/math] is nonzero. If we set [math]x'(t_0) = d_1[/math] and [math]y'(t_0) = d_2[/math], then

[[math]] \mbox'''{d'''}P(t_0) = (d_1, d_2)_{P(t_0)}, [[/math]]

where not both coordinates [math]d_1[/math] and [math]d_2[/math] are zero. The set of all tangent vectors at to is the set of all scalar multiples

[[math]] s\mbox'''{d'''}P(t_0) = (sd_1, sd_2)_{P(t_0)}, [[/math]]

where [math]s[/math] is any real number. If [math]P(t_0) = (a, b)[/math], then the terminal point of [math]s\mbox'''{d'''}P(t_0)[/math] is equal to

[[math]] (sd_1 + a, sd_2 + b). [[/math]]

Hence the set of all terminal points of tangent vectors at [math]t_0[/math] is the set of all points [math](x, y)[/math] such that

[[math]] \begin{equation} \left \{ \begin{array}{l} x = sd_1 + a,\\ y = sd_2 + b, \end{array} \right . \label{eq10.4.4} \end{equation} [[/math]]

where [math]s[/math] is any real number and [math]d_1[/math] and [math]d_2[/math] are not both zero. It is easy to verify that this set is a straight line (see Problem 4). We conclude that if the derived vector $\mbox{d}P(t_0)$ exists and is nonzero, then the set of all terminal points of the tangent vectors at $to$ to the curve parametrized by $P$ is a straight line. It is called the tangent line to the parametrized curve at [math]t_0[/math].

Example Consider the ellipse defined parametrically by

[[math]] P(t) = (x(t), y(t)) = (4 \cos t, 2 \sin t), [[/math]]

for every real number [math]t[/math]. Compute the derived vector at [math]t_0 = \frac{\pi}{6}[/math], and draw it and the ellipse in the [math]xy[/math]-plane. In addition, write an equation for the tangent line at [math]t_0 = \frac{\pi}{6}[/math], and draw the tangent line in the figure. The derived vector is easily computed:

[[math]] \begin{eqnarray*} \mbox'''{d'''}P(t_0) &=& (x'(t_0), y'(t_0))_{P(t_0)} = (- 4 \sin t_0, 2 \cos t_0)_{P(t_0)} \\ &=& \Big(- 4\sin\frac{\pi}{6}, \cos\frac{\pi}{6} \Big) = (-2, \sqrt{3})_{P(t_0)}, \end{eqnarray*} [[/math]]

where

[[math]] P(t_0) = \Big(4\cos\frac{\pi}{6}, 2 \sin\frac{\pi}{6} \Big) = (2\sqrt{3}, 1). [[/math]]

The terminal point of the derived vector is therefore equal to

[[math]] (2\sqrt{3} - 2, 1 + \sqrt 3). [[/math]]

The parametrization [math]P[/math] can also be written in terms of the equations

[[math]] \left \{ \begin{array}{ll} x = 4 \cos t, & \\ y = 2 \sin t, \;\;\; &-\infty \lt t \lt \infty, \end{array} \right . [[/math]]

from which it follows that

[[math]] \frac{x^2}{4^2} + \frac{y^2}{2^2} = \cos^{2} t + \sin^{2} t = 1. [[/math]]

Hence every point [math](x, y)[/math] on the parametrized curve satisfies the equation

[[math]] \begin{equation} \frac{x^2}{4^2} + \frac{y^2}{2^2} = 1 . \label{eq10.4.5} \end{equation} [[/math]]

Conversely, it can be shown (as in Example 2, page 544) that any ordered pair [math](x, y)[/math] which satisfies (5) also lies on the parametrized curve. We recognize (5) as an equation of the ellipse shown in Figure 16. The derived vector [math]\mbox'''{d'''}P(t_0)[/math] and the tangent line at [math]\frac{\pi}{6}[/math] are also shown in the figure.

File:Guide c5467 scanfig10 16.png

If [math]s[/math] is an arbitrary real number, then the scalar product [math]s\mbox'''{d'''}P(t_0)[/math] in this example is the vector

[[math]] s\mbox'''{d'''}P(t_0) = s(- 2, \sqrt 3)_{P(t_0)} = (- 2s, \sqrt{3} s)_{P(t_0)}. [[/math]]

The terminal point of this vector, since [math]P(t_0) = (2\sqrt{3}, 1)[/math], is the point

[[math]] (-2s + 2\sqrt 3, \sqrt{3} s + 1). [[/math]]

Hence the tangent line at [math]\frac{\pi}{6}[/math] is parametrically defined by the equations

[[math]] \begin{equation} \left \{ \begin{array}{l} x = - 2s + 2 \sqrt 3 ,\\ y= \sqrt {3} s + 1, \;\;\; -\infty \lt s \lt \infty . \end{array} \right . \label{eq10.4.6} \end{equation} [[/math]]

Solving the first of these for [math]s[/math], we obtain [math]s = \frac{-x + 2\sqrt 3}{2}[/math]. Substitution in the second then yields

[[math]] \begin{eqnarray*} y &=& \sqrt 3 \Big(\frac{-x + 2\sqrt 3}{2} \Big) + 1,\\ y &=& - \frac{\sqrt 3}{2}x + 4. \mbox{ (7)} \end{eqnarray*} [[/math]]

Thus any point on the tangent line satisfies (7), and it is easy to verify that, for any [math]x[/math] and [math]y[/math] which satisfy (7), there is a unique s such that equations (6) hold. We conclude that (7) is an equation of the tangent line.

It is important to know that the ideas introduced in this section are consistent with related concepts developed earlier. For example, consider a differentiable parametrization defined by

[[math]] P(t) = (x(t), y(t)), \;\;\;\mathrm{for every}~t~\mathrm{in some interval}~I. [[/math]]

Suppose that, for some [math]t_0[/math] in [math]I[/math], there exists a differentiable function [math]f[/math] such that

[[math]] y(t) = f(x(t)), [[/math]]

for every [math]t[/math] in some subinterval of [math]I[/math] containing [math]t_0[/math] in its interior. This situation was described in Section I and was illustrated in Figure 3 (page 545). If such a function [math]f[/math] exists, we say that [math]y[/math] is a differentiable function of [math]x[/math] on the parametrized curve [math]P(t)[/math] in a neighborhood of [math]P(t_0)[/math]. Formulas (5) and (6), page 546, assert that, for every [math]t[/math] in the subinterval,

[[math]] \frac{dy}{dx} = f'(x(t)) = \frac{y'(t)}{x'(t)}, [[/math]]

provided [math]x'(t) \neq 0[/math]. Hence [math]\frac{y'(t)}{x'(t)}[/math] is the slope of the line tangent to the graph of [math]f[/math] at the point

[[math]] (x(t), f(x(t))) = (x(t), y(t)) = P(t) . [[/math]]

Moreover, in the vicinity of [math]P(t_0)[/math], the graph of [math]f[/math] is the curve parametrized by [math]P[/math]. At every [math]t[/math] in the subinterval, the derived vector of [math]P[/math] is equal to

[[math]] \mbox'''{d'''}P(t) = (x'(t), y (t))_{P(t)} . [[/math]]

This vector is, by definition, a tangent vector to the parametrized curve. Its initial point is [math]P(t) = (x(t), y(t))[/math] and its terminal point is

[[math]] Q(t) = (x(t) + x'(t), y(t) + y'(t)). [[/math]]

The slope of the line segment joining these two points is given by

[[math]] m(P(t), Q(t)) = \frac{(y(t) + y'(t)) - y(t)}{(x(t) + x'(t)) - x(t)} = \frac{y'(t)}{x'(t)}, [[/math]]

provided [math]x'(t) \neq 0[/math]. We conclude that the concept of tangency, as defined in terms of the derived vector to a parametrized curve, is consistent with the earlier notion, defined in terms of the derivative.

General references

Doyle, Peter G. (2008). "Crowell and Slesnick's Calculus with Analytic Geometry" (PDF). Retrieved Oct 29, 2024.