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⁡2θ+cos⁡2θ=1\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∈[−1,5]x \in [-1, 5]; corresponding y-range is y∈[−8,64]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 θ\theta is not bounded, but the locus itself is bounded (a circle). As θ\theta runs over all real values, the full circle is traced.
    • Direction: with x−3=sin⁡θx-3 = \sin\theta and y−2=cos⁡θy-2 = \cos\theta, increasing θ\theta generally traces the circle in a clockwise orientation starting from $(3,3)$ when θ=0\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→π0\to\pi: starts at $(3,3)$ when θ=0\theta = 0, moves toward $(4,2)$ at θ=π/2\theta = \pi/2, and ends at $(3,1)$ when θ=π\theta = \pi.

Vector Functions and lines in 2D

  • A vector function is written as r(t)=⟨x(t),y(t)⟩\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: r(t)=(4cos⁡t) i+(sin⁡t) j\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 v=⟨a,b⟩\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=23x+7y = \tfrac{2}{3}x + 7.
    • Slope is m=23m = \tfrac{2}{3}, so a direction vector can be ⟨3,2⟩\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 r(t)=(3−5t,  2+7t)\mathbf{r}(t) = (3 - 5t,\; 2 + 7t).
    • Direction of the given line: v=⟨−5,7⟩\mathbf{v} = \langle -5, 7 \rangle.
    • A line perpendicular to this has direction vector w\mathbf{w} with v⋅w=0\mathbf{v}\cdot\mathbf{w} = 0. One choice: w=⟨7,5⟩\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 r<em>1(t)=(−1−2t)i+(2+t)j\mathbf{r}<em>1(t) = (-1 - 2t)\mathbf{i} + (2 + t)\mathbf{j} and r</em>2(s)=(5−3s)i+(3+6s)j\mathbf{r}</em>2(s) = (5 - 3s)\mathbf{i} + (3 + 6s)\mathbf{j} parallel, perpendicular, or neither?
    • Direction vectors: v<em>1=⟨−2,1⟩\mathbf{v}<em>1 = \langle -2, 1 \rangle, v</em>2=⟨−3,6⟩\mathbf{v}</em>2 = \langle -3, 6 \rangle.
    • Not parallel: no scalar $k$ with v<em>1=kv</em>2\mathbf{v}<em>1 = k\mathbf{v}</em>2 (check ratios: −2/−3≠1/6-2/-3 \neq 1/6).
    • Not perpendicular: v<em>1⋅v</em>2=(−2)(−3)+(1)(6)=6+6=12≠0\mathbf{v}<em>1 \cdot \mathbf{v}</em>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=−89s = -\tfrac{8}{9}.
    • Then t=1+6(−89)=1−489=−133t = 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∈[0,π3]t \in [0, \frac{\pi}{3}], cos⁡t∈[12,1]\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 r(t)=⟨t2,t−2⟩\mathbf{r}(t) = \langle t^2, t - 2 \rangle pass through the point $(49, 7)$ for t≥0t \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≥0t \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: ⟨−1−2,  7−5⟩=⟨−3,2⟩\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.