Parametric Curves and Vector Functions#

Parametric Curves in \(R^{2}\)#

If \(x=f(t)\) and \(y=g(t)\) over an interval \(I\) of \(t\)-values, then the set of points \((x, y)=(f(t), g(t))\) defined by these equations is a parametric curve.

  • \(t\) is the parameter, and its domain \(I\) is the parameter interval.

  • \(x=f(t)\) and \(y=g(t)\) are parametric equations for the curve.

  • If \(I\) is a closed interval, \(a \leq t \leq b\), then \((f(a), g(a))\) is the initial point and \((f(b), g(b))\) is the terminal point of the curve.

  • A parametrization of a curve comprises parametric equations together with a parameter interval.

  • Note that a given curve can be represented by many different sets of parametric equations.

Parametric Curves in \(R^{3}\)#

Similarly, a curve in \(\boldsymbol{R}^{3}\) can be described parametrically by

\[ x=f(t), \quad y=g(t), \quad z=h(t), \quad t \in I . \]

Vector-valued Functions#

A vector-valued function (or vector function) on a domain set \(D\) of real numbers is a rule that assigns a vector in space to each element in \(D\). The graph of a vector-valued function is the set of all terminal points of the output vectors with their initial points at the origin.

For example, consider a particle moving in \(\boldsymbol{R}^{3}\) on a space curve \(C\) with coordinates \((x, y, z)=\) \((f(t), g(t), h(t))\) at time \(t\).

  • \(x=f(t), y=g(t), z=h(t)\) are the parametric equations of the curve \(C\).

In this form, we visualize the curve as traced by a moving point \(c(t)\).

  • \(\overrightarrow{\boldsymbol{r}}(t)=\langle f(t), g(t), h(t)\rangle\) is the vector parameterization of \(C\).

In this form we visualize the curve as traced by the tip of a moving vector.

  • \(f(t), g(t), h(t)\) are called the component functions of \(\overrightarrow{\boldsymbol{r}}(\mathrm{t})\).

The domain of \(\overrightarrow{\boldsymbol{r}}(t)\) is the set of values of \(t\) for which \(\overrightarrow{\boldsymbol{r}}(t)\) is defined.

Limits & Continuity#

Consider a vector function \(\overrightarrow{\boldsymbol{r}}(t)=\langle f(t), g(t), h(t)\rangle\).

  • The limit of \(\overrightarrow{\boldsymbol{r}}(t)\) is found by taking the limits of the component functions (provided these limits exist):

\[ \lim _{t \rightarrow a} \vec{r}(t)=\left\langle\lim _{t \rightarrow a} f(t), \lim _{t \rightarrow a} g(t), \lim _{t \rightarrow a} h(t)\right\rangle \]

  • \(\overrightarrow{\boldsymbol{r}}(t)\) is continuous at \(a\) if \(\lim _{t \rightarrow a} \vec{r}(t)=\overrightarrow{\boldsymbol{r}}(a)\).

So \(\overrightarrow{\boldsymbol{r}}(t)\) is continuous at \(a\) if the component functions \(f(t), g(t), h(t)\) are all continuous at \(a\).

Derivative of a Vector-valued Function#

The derivative of a vector-valued function \(\overrightarrow{\boldsymbol{r}}(t)\) is defined to be:

\[ \frac{d \overrightarrow{\boldsymbol{r}}}{d t}=\overrightarrow{\boldsymbol{r}}^{\prime}(t)=\lim _{h \rightarrow 0} \frac{\overrightarrow{\boldsymbol{r}}(t+h)-\overrightarrow{\boldsymbol{r}}(t)}{h} \]

provided this limit exists.

Component-wise Derivative#

Derivatives can be computed component-wise. For \(\overrightarrow{\boldsymbol{r}}(t)=\langle x(t), y(t), z(t)\rangle\), we have

\[ \overrightarrow{\boldsymbol{r}}^{\prime}(t)=\lim _{h \rightarrow 0} \frac{\overrightarrow{\boldsymbol{r}}(t+h)-\overrightarrow{\boldsymbol{r}}(t)}{h}=\lim _{h \rightarrow 0}\left\langle\frac{x(t+h)-x(t)}{h}, \frac{y(t+h)-y(t)}{h}, \frac{z(t+h)-z(t)}{h}\right\rangle . \]

Hence:

  • a vector function \(\overrightarrow{\boldsymbol{r}}(t)=x(t) \mathbf{i}+y(t) \mathbf{j}+z(t) \mathbf{k}=\langle x(t), y(t), z(t)\rangle\) is differentiable if and only if its component functions are differentiable

  • in that case, \(\overrightarrow{\boldsymbol{r}}^{\prime}(t)=x^{\prime}(t) \mathbf{i}+y^{\prime}(t) \mathbf{j}+z^{\prime}(t) \mathbf{k}=\left\langle x^{\prime}(t), y^{\prime}(t), z^{\prime}(t)\right\rangle\).

Rules of Differentiation#

All the usual rules of differentiation also apply to vector functions. For example, for vector functions \(\overrightarrow{\boldsymbol{u}}, \overrightarrow{\boldsymbol{v}}\), a scalar function \(f\) and constant \(c\) we have

  • \(\frac{d}{d t}[\mathbf{u}(t)+\mathbf{v}(t)]=\mathbf{u}^{\prime}(t)+\mathbf{v}^{\prime}(t)\)

  • \(\frac{d}{d t}[c \mathbf{u}(t)]=c \mathbf{u}^{\prime}(t)\)

  • \(\frac{d}{d t}[f(t) \mathbf{u}(t)]=f^{\prime}(t) \mathbf{u}(t)+f(t) \mathbf{u}^{\prime}(t)\)

  • \(\frac{d}{d t}[\mathbf{u}(t) \cdot \mathbf{v}(t)]=\mathbf{u}^{\prime}(t) \cdot \mathbf{v}(t)+\mathbf{u}(t) \cdot \mathbf{v}^{\prime}(t)\)

  • \(\frac{d}{d t}[\mathbf{u}(t) \times \mathbf{v}(t)]=\mathbf{u}^{\prime}(t) \times \mathbf{v}(t)+\mathbf{u}(t) \times \mathbf{v}^{\prime}(t)\)

  • \(\frac{d}{d t}[\mathbf{u}(f(t))]=f^{\prime}(t) \mathbf{u}^{\prime}(f(t))\)

The curve traced by \(\overrightarrow{\boldsymbol{r}}(t)\) is smooth if \(\frac{d \vec{r}}{d t}\) is continuous and never \(\overrightarrow{\mathbf{0}}\), that is, if the component functions \(x, y\), and \(z\) have continuous first derivatives that are not simultaneously 0 .

Interpretations of \(\vec{r}^{\prime}(t)\)#

  • If \(\overrightarrow{\boldsymbol{r}}(t)\) represents the position of a particle at time \(t\) then \(\overrightarrow{\boldsymbol{r}}^{\prime}(t)\) is its velocity, \(\left|\overrightarrow{\boldsymbol{r}}^{\prime}(t)\right|\) is its speed and \(\overrightarrow{\boldsymbol{r}}^{\prime \prime}(t)\) is its acceleration.

  • For a space curve \(\overrightarrow{\boldsymbol{r}}(t)\), the vector \(\overrightarrow{\boldsymbol{r}}^{\prime}\left(t_{0}\right)\) points in the direction of the tangent to the curve at the point \(t=t_{0}\). It is therefore called the tangent vector (provided it exists and is non-zero). The tangent line to \(\overrightarrow{\boldsymbol{r}}(t)\) at \(t=t_{0}\) thus has vector equation \(\overrightarrow{\boldsymbol{L}}(t)=\overrightarrow{\boldsymbol{r}}\left(t_{0}\right)+t \overrightarrow{\boldsymbol{r}}^{\prime}\left(t_{0}\right)\).

Integrating a vector-valued function#

If \(\overrightarrow{\boldsymbol{r}}(t)=x(t) \mathbf{i}+y(t) \mathbf{j}+z(t) \mathbf{k}\) then

\[ \int_{a}^{b} \overrightarrow{\boldsymbol{r}}(t) d t=\left(\int_{a}^{b} x(t) d t\right) \mathbf{i}+\left(\int_{a}^{b} y(t) d t\right) \mathbf{j}+\left(\int_{a}^{b} z(t) d t\right) \mathbf{k} \]

Note that an indefinite integral will yield a vector constant of integration, \(\vec{C}\).