Parametric Curves, Position Vectors, and Lines (Notes from Transcript)

Parametric Curves and Position Vectors

A vector can be viewed as the position vector of a point in the plane. For a parametric curve, the point corresponding to a parameter value t is given by the pair (x(t),\ y(t)) , i.e., the position vector \mathbf{r}(t) = (x(t),y(t)) .
For every t in the domain, you get one point in the plane; letting t run over its domain typically yields infinitely many points, which trace out a parametric curve.
The pair of functions x(t) and y(t) are called the parametric equations of the curve.
Advantages and Interpretation of Parametric Equations:
- The parameter t often represents time, allowing for the description of motion along a path, including velocity and direction.
- Parametric equations can describe complex curves that cannot be represented by a single function y=f(x) or x=g(y) (e.g., circles, spirals, or curves that self-intersect).
- They explicitly show the direction of traversal or orientation of the curve as t increases.
Reading a parametric description: pick values of the parameter t (often integers near 0 for quick sketches) and plot the corresponding points
Example domain: t

{ -3,-2,-1,0,1,2,3 }.

For a given t, the point on the curve is (x(t),y(t)) .
Example from transcript:
- Given the parametric equations x(t) = 2t - 1, \quad y(t) = t^2 - 1 .
- Compute points for t = -3,-2,-1,0,1,2,3:
- t = -3: (x,y)=(-7,8)
- t = -2: (x,y)=(-5,3)
- t = -1: (x,y)=(-3,0)
- t = 0: (x,y)=(-1,-1)
- t = 1: (x,y)=(1,0)
- t = 2: (x,y)=(3,3)
- t = 3: (x,y)=(5,8)
Graphing approach from the points: plot these points and connect with a smooth curve to sketch the parametric curve.
Relation between x and y (eliminate t):
- Eliminating the parameter helps to visualize the underlying Cartesian shape of the curve, if one exists.
- From x(t) = 2t - 1 we get t = \frac{x+1}{2} .
- Substitute into y(t) = t^2 - 1 to obtain a Cartesian form:
- y = \left(\frac{x+1}{2}\right)^2 - 1 = \frac{x^2 + 2x - 3}{4} .
- Equivalent form: 4y = x^2 + 2x - 3 .
Ellipses via a standard parametric form and elimination to Cartesian form:
- A common ellipse parameterization is x = a\cos\theta, \quad y = b\sin\theta , which yields the standard ellipse equation
- \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1.
- Example with a=2, b=3 : x = 2\cos\theta, \quad y = 3\sin\theta and hence \frac{x^2}{4} + \frac{y^2}{9} = 1.
Sample angle values for the ellipse (to read points on the ellipse):
- (\theta = 0): (x,y) = (2,0)
- (\theta = \frac{\pi}{2}): (x,y) = (0,3)
- (\theta = \pi): (x,y) = (-2,0)
- (\theta = \tfrac{3\pi}{2}): (x,y) = (0,-3)
- (\theta = \frac{\pi}{4}): (x,y) = (\sqrt{2}, \tfrac{3\sqrt{2}}{2})
Notes on trigonometric values referenced in class problems:
- (\cos\pi = -1, \ \sin\pi = 0)
- (\cos\tfrac{\pi}{4} = \sin\tfrac{\pi}{4} = \tfrac{\sqrt{2}}{2})
- These values are used to plug into ellipse parametrizations and check points.
Vector equation of a line in the plane:
- A line through a known point with position vector \mathbf{r}0 = (x0,y_0) and directed by a vector \mathbf{v} = (a,b) is written as
- \mathbf{r} = \mathbf{r}_0 + t\mathbf{v}, \quad t \in \mathbb{R}.
- In components: x = x0 + at, \quad y = y0 + bt.
Example from transcript (line through A and B):
- Take A = (-3,4) and B = (2,8). Then
- Direction vector: \mathbf{v} = \overrightarrow{AB} = B - A = (5,4).
- Point A has position vector \mathbf{r}_0 = (-3,4).
- Line L through A and B: \mathbf{r}(t) = (-3,4) + t(5,4).
- Component form: x = -3 + 5t, \quad y = 4 + 4t.
Intersections of two lines (two-parameter solve):
- If two lines are given by
- \mathbf{r}1(t) = \mathbf{r}{10} + t\mathbf{v}1, \quad \mathbf{r}2(s) = \mathbf{r}{20} + s\mathbf{v}2,
- The intersection is found by solving the system obtained from equating the two vectors:
- \mathbf{r}1(t) = \mathbf{r}2(s)
- This yields two linear equations in the unknowns (t) and (s) which one must solve simultaneously.
- Important caution: do not set (t = s); that is not generally valid. You must solve for (t) and (s) and then substitute back to get the intersection coordinates. The parameters (t) and (s) represent distinct 'times' or positions along each respective line.
Transcript example for line intersection (form and results):
- One setup in class was
- \mathbf{r}1(t) = (-4,5) + t(2,1), \quad \mathbf{r}2(s) = (5,0) + s(?,?)
- Solving the resulting 2

2 system gave

t = \frac{11}{5}, \quad s = -\frac{8}{15}.
The intersection point computed from (\mathbf{r}_1) is
\mathbf{r}_1 \left(\tfrac{11}{5}\right) = \left(\frac{2}{5}, \frac{36}{5}\right).
This matches the point obtained from (\mathbf{r}_2\left(-\tfrac{8}{15}\right)).
Note: The pair ((t,s) = (\tfrac{11}{5}, -\tfrac{8}{15})) are the parameters that yield the intersection coordinates, not the coordinates themselves.
- Practical exam and homework pointers:
In WebAssign, you may be asked to report the direction of traversal, not just the picture. This refers to the orientation of the curve as the parameter increases.
A single geometric shape can be traced in opposite directions depending on the parameterization; the orientation matters for some problems.
- Final recap of notation:
Position vector: \mathbf{r}(t) = (x(t),y(t)) .
A fixed point on a line: \mathbf{r}0 with coordinates (x0,y_0) .
Direction vector of the line: \mathbf{v} = (a,b) .
Vector form of a line: \mathbf{r} = \mathbf{r}0 + t\mathbf{v} , with component form x = x0 + at, \quad y = y_0 + bt.
You can also obtain Cartesian equations by eliminating the parameter whenever convenient, as done for the parabola from the example above or for the ellipse from the ellipse parameterization. This conversion helps in recognizing familiar geometric shapes.
If anything feels unclear, revisit the derivations at home and practice turning parametric forms into Cartesian relations and back, and practice solving line-intersection problems by solving the two-equation system for the two parameters.