2.1 Space Curves

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In this section we create the describe curves in space.

A curve in space can be described using a vector-valued function: $\vec r(t) = \vector {x(t), y(t), z(t)}$

Example 1 Express the line through the point $(3, -4, 1)$ with direction vector $\vector {2, 5, -8}$ using a vector-valued function.
The vector equation of the line is $\vector {x, y, z} = \vector {3, -4, 1} + t\vector {2, 5, -8}$ Distributing the parameter $t$ and adding the vectors on the right hand side, we have $\vector {x, y, z} = \vector {3+2t,-4+ 5t, 1-8t}$ Renaming the vector on the left hand side as $\vec r(t)$ we have $\vec r(t) = \vector {3+2t,-4+ 5t, 1-8t}$ which is a vector-valued function.

In general, a line with vector equation $\vector {x, y, z} = \vector {x_0, y_0, z_0} + t\vector {a, b, c}$ can be expressed as a vector-valued function as follows: $\vec r(t) = \vector {x_0 + at, y_0 + bt, z_0 + ct}$

(Problem 1a) Express the line through the point $(6, 7, 2)$ with direction vector $\vector {9, -3, 5}$ as a vector-valued function.

Here is a video solution of problem 1a:

(Problem 1b) Identify the space curve given by the vector-valued function: $\vec r(t) = \vector {3 -t, -2 + 2t, 7t}$

Here is a video solution of problem 1b:

The domain of a space curve is the largest subset of $\R$ for which each of the component functions are defined.

Example 2 Find the domain of the space curve $\vec r(t) = \vector {\frac {1}{t}, \sqrt t, \ln (1-t)}$ The domain of $f(t) = 1/t$ is $t \neq 0$ ; the domain of $g(t) = \sqrt t$ is $t \geq 0$ and the domain of $h(t) = \ln (t)$ is $1-t > 0$ or $t<1$ .
The domain of $\vec r(t)$ must satisfy all three conditions: $t\neq 0, \quad t \geq 0, \quad \text {and} \quad t <1$ Thus the domain of $\vec r(t)$ is the interval $(0, 1)$ .

(Problem 2a) Find the domain of the space curve $\vec r(t) = \vector {\frac {1}{t^2 - 4}, \sqrt {2- t}, \ln (t)}$

(Problem 2b) Find the domain of the space curve $\vec r(t) = \vector {\tan (t), \cot (t), \frac {1}{1+t^2}}$

The orientation of a space curve is in the direction of increasing values of the parameter $t$ .

A line segment from the point $P(x_1, y_1, z_1)$ to the point $Q(x_2, y_2, z_2)$ is a space curve that can be written as \begin{align*} \vec r(t) &= \vec P + t \arrowvec{PQ}, (\vec P \text{is the position vector of the point} \;P)\\ &=\vector{x_1, y_1, z_1} + t\vector{x_2-x_1, y_2-y_1, z_2-z_1} \\ &= (1-t)\vector{x_1, y_1, z_1} + t\vector{x_2, y_2, z_2}, \end{align*}

where $\; 0 \leq t \leq 1$ .
If we write the vector from the origin to $P$ as $\vec P$ and similarly for $\vec Q$ , then the line segment has the elegant form $\vec r(t) = (1-t) \vec P + t \vec Q, \quad 0 \leq t \leq 1$

Example 3 Express the line segment from the point $P(1, 2, -1)$ to the point $Q(-2, -2, 1)$ as a vector-valued function. Include the domain. Sketch the line segment with orientation arrows.
The vector form of the line segment is \begin{align*} \vec r(t) &= (1-t)\vector{1, 2, -1} + t \vector{-2, -2, 1}\\ &= \vector{1-t, 2-2t, -1+t} + \vector{-2t, -2t, t}\\ &= \vector{1-3t, 2-4t, -1+2t} \end{align*}

and its domain is $0 \leq t \leq 1$ . The sketch is below.

(Problem 3) Express the line segment from the point $P(-1, -1, 2)$ to the point $Q(2, 3, -1)$ as a vector-valued function. Include the domain. Sketch the line segment with orientation arrows.

Here is a video solution of problem 3:

Other interesting space curves include the helix $\vec r(t) = \vector {\cos t, \sin t, t}$ and the twisted cubic $\vec r(t) = \vector {t, t^2, t^3}$

Plot both of these space curves using this online graph plotter

2024-09-27 14:00:49