Notes on Parametric Curves and Vector Functions

Parametric Curves

  • A parametric curve represents x and y as functions of a parameter, usually t or θ, instead of a single equation y = f(x).
  • Given x = x(t) and y = y(t), the curve is traced by letting t vary.
  • Example: x = -3t + 1, y = 2t - 1.
    • Points on the curve (by tabulating t):
    • t = -1 → (x, y) = (4, -3)
    • t = 0 → (x, y) = (1, -1)
    • t = 1 → (x, y) = (-2, 1)
    • t = 2 → (x, y) = (-5, 3)
    • The plotted points lie on a straight line, so the parametric curve represents a line.
  • To convert to Cartesian form, eliminate the parameter t.
    • From x = -3t + 1: t = (1 - x)/3.
    • Substitute into y = 2t - 1:
      y=2(1−x3)−1=2−2x3−1=−2x−13.y = 2\left(\frac{1 - x}{3}\right) - 1 = \frac{2 - 2x}{3} - 1 = \frac{-2x - 1}{3}.
      So the Cartesian equation is:
      y=−23x−13.\displaystyle y = -\frac{2}{3}x - \frac{1}{3}.
  • Parameter ranges can be unbounded or bounded.
    • Strategies to eliminate the parameter:
    • Solve one equation for t and substitute into the other.
    • Use identities (e.g., trig identities such as $\sin^2\theta + \cos^2\theta = 1$) when x and y are given in terms of trig functions.
  • Problems: Eliminate the parameter and sketch the curve, noting the direction of traversal. 1) Curve: x=t+1,y=t3,−2≤t≤4.x = t + 1,\quad y = t^3,\quad -2 \le t \le 4.
    • Cartesian relation: y=(x−1)3.y = (x - 1)^3.
    • Domain of x: since $x = t + 1$, $x \in [-1, 5]$; corresponding y-range is $y \in [-8, 64]$.
    • Direction: as $t$ increases from -2 to 4, x increases from -1 to 5 (dx/dt = 1 > 0), so the curve is traversed from $(-1,-8)$ toward $(5,64)$.

Cartesian forms from parametric curves (circles and bounded/unbounded cases)

  • 2) x=3+sin⁡θ,y=2+cos⁡θ.x = 3 + \sin\theta,\quad y = 2 + \cos\theta.
    • Eliminate the parameter: (x−3)2+(y−2)2=sin⁡2θ+cos⁡2θ=1.(x-3)^2 + (y-2)^2 = \sin^2\theta + \cos^2\theta = 1.
    • This is a circle of radius 1 centered at (3,2)(3,2).
    • Note: this is described as an unbounded curve in the sense that the parameter $\theta$ is not bounded, but the locus itself is bounded (a circle). As $\theta$ runs over all real values, the full circle is traced.
    • Direction: with $x-3 = \sin\theta$ and $y-2 = \cos\theta$, increasing $\theta$ generally traces the circle in a clockwise orientation starting from $(3,3)$ when $\theta = 0$.
  • 3) x=3+sin⁡θ,y=2+cos⁡θ,0≤θ≤π.x = 3 + \sin\theta,\quad y = 2 + \cos\theta,\quad 0 \le \theta \le \pi.
    • This traces only a portion of the circle: the right semicircle (the eastern half).
    • The locus remains the same circle equation $(x-3)^2 + (y-2)^2 = 1$, but theta is restricted to half the circle.
    • Direction on $0\to\pi$: starts at $(3,3)$ when $\theta = 0$, moves toward $(4,2)$ at $\theta = \pi/2$, and ends at $(3,1)$ when $\theta = \pi$.

Vector Functions and lines in 2D

  • A vector function is written as $\mathbf{r}(t) = \langle x(t), y(t) \rangle$. Each value of t gives a point on a curve, and the set of points traced is the curve.
  • Example: $\mathbf{r}(t) = (4\cos t)\,\mathbf{i} + (\sin t)\,\mathbf{j}$.
    • This traces the ellipse with semi-axes 4 (along x) and 1 (along y):
      x216+y2=1.\frac{x^2}{16} + y^2 = 1.
  • Vector equation of a line and its relationship to parametric and Cartesian forms:
    • If $P0(x0, y0)$ is a point on the line and $\mathbf{v} = \langle a, b \rangle$ is a direction vector parallel to the line, then the vector equation is: r(t)=r</em>0+tv=⟨x<em>0,y</em>0⟩+t⟨a,b⟩.\mathbf{r}(t) = \mathbf{r}</em>0 + t \mathbf{v} = \langle x<em>0, y</em>0 \rangle + t \langle a, b \rangle.
    • The corresponding parametric equations are:
      x(t)=x<em>0+ta,y(t)=y</em>0+tb.x(t) = x<em>0 + t a, \quad y(t) = y</em>0 + t b.
    • The Cartesian equation of the line can be obtained by eliminating t (or by slope-intercept form).
  • 5) Find the parametric equation of the line through $(3,-2)$ parallel to $y = \tfrac{2}{3}x + 7$.
    • Slope is $m = \tfrac{2}{3}$, so a direction vector can be $\langle 3, 2 \rangle$.
    • Parametric form: r(t)=⟨3,−2⟩+t⟨3,2⟩.\mathbf{r}(t) = \langle 3, -2 \rangle + t \langle 3, 2 \rangle.
    • Component form: x(t)=3+3t,y(t)=−2+2t.x(t) = 3 + 3t, \quad y(t) = -2 + 2t.

Perpendicular and intersecting lines

  • 6) Find the vector equation of the line through $(1,4)$ perpendicular to the line given by $\mathbf{r}(t) = (3 - 5t,\; 2 + 7t)$.
    • Direction of the given line: $\mathbf{v} = \langle -5, 7 \rangle$.
    • A line perpendicular to this has direction vector $\mathbf{w}$ with $\mathbf{v}\cdot\mathbf{w} = 0$. One choice: $\mathbf{w} = \langle 7, 5 \rangle$.
    • Perpendicular line: r(t)=⟨1,4⟩+t⟨7,5⟩.\mathbf{r}(t) = \langle 1, 4 \rangle + t \langle 7, 5 \rangle.
  • 7) Are the lines $\mathbf{r}1(t) = (-1 - 2t)\mathbf{i} + (2 + t)\mathbf{j}$ and $\mathbf{r}2(s) = (5 - 3s)\mathbf{i} + (3 + 6s)\mathbf{j}$ parallel, perpendicular, or neither?
    • Direction vectors: $\mathbf{v}1 = \langle -2, 1 \rangle$, $\mathbf{v}2 = \langle -3, 6 \rangle$.
    • Not parallel: no scalar $k$ with $\mathbf{v}1 = k\mathbf{v}2$ (check ratios: $-2/-3 \neq 1/6$).
    • Not perpendicular: $\mathbf{v}1 \cdot \mathbf{v}2 = (-2)(-3) + (1)(6) = 6 + 6 = 12 \neq 0$.
    • Intersection: solve for $t$ and $s$ with
      -x-coordinates: $-1 - 2t = 5 - 3s$,
      -y-coordinates: $2 + t = 3 + 6s$.
    • From y: $t = 1 + 6s$.
    • Substitute into x: $-1 - 2(1 + 6s) = 5 - 3s$ → $-3 - 12s = 5 - 3s$ → $-8 = 9s$ → $s = -\tfrac{8}{9}$.
    • Then $t = 1 + 6(-\tfrac{8}{9}) = 1 - \tfrac{48}{9} = -\tfrac{13}{3}$.
    • Intersection point: r1(−133)=(−1−2(−133),  2+(−133))=(233,  −73).\mathbf{r}_1\left(-\tfrac{13}{3}\right) = \left(-1 - 2\left(-\tfrac{13}{3}\right),\; 2 + \left(-\tfrac{13}{3}\right)\right) = \left(\tfrac{23}{3},\; -\tfrac{7}{3}\right).

Practice problems (additional problems for study and quick checks)

  • 1) Sketch the curve x=cos⁡t,  y=sec⁡t,  0≤t≤π/3.x = \cos t,\; y = \sec t,\; 0 \le t \le \pi/3.
    • Relationship: y=sec⁡t=1cos⁡t=1x.y = \sec t = \frac{1}{\cos t} = \frac{1}{x}.
    • Since $t \in [0, \frac{\pi}{3}]$, $\cos t \in [\tfrac{1}{2}, 1]$, so x∈[12,1],  y∈[1,2].x \in [\tfrac{1}{2}, 1],\; y \in [1, 2].
    • The curve is a portion of the rectangular hyperbola $xy = 1$ restricted to that x-range.
  • 2) Does the vector function $\mathbf{r}(t) = \langle t^2, t - 2 \rangle$ pass through the point $(49, 7)$ for $t \ge 0$?
    • If it passes, there would be a nonnegative $t$ with $t^2 = 49$ and $t - 2 = 7$.
    • From $t^2 = 49$ we get $t = 7$ (since $t \ge 0$).
    • Then $t - 2 = 5$, not $7$; so $(49,7)$ is not on the curve.
  • 3) Find the vector equation of a line that passes through $(2,5)$ and $(-1,7)$.
    • Direction vector: $\langle -1 - 2,\; 7 - 5 \rangle = \langle -3, 2 \rangle$.
    • Line: r(t)=⟨2,5⟩+t⟨−3,2⟩.\mathbf{r}(t) = \langle 2, 5 \rangle + t \langle -3, 2 \rangle.
    • Alternatively, x(t)=2−3t,y(t)=5+2t.x(t) = 2 - 3t, \quad y(t) = 5 + 2t.