Introduction to Parametric Equations

Understanding curves defined parametrically

📐 Introduction to Parametric Equations

What Are Parametric Equations?

Instead of y=f(x)y = f(x), we define both xx and yy in terms of a third variable tt (the parameter):

x=f(t),y=g(t)x = f(t), \quad y = g(t)

💡 Key Idea: The parameter tt often represents time, and the equations trace out a path or curve!


Why Use Parametric Equations?

Advantages:

  1. Can describe curves that aren't functions

    • Example: A circle fails the vertical line test
    • But parametrically: x=costx = \cos t, y=sinty = \sin t works perfectly!
  2. Natural for motion problems

    • tt = time
    • Position: (x(t),y(t))(x(t), y(t))
    • Describes the path of a moving object
  3. More flexibility

    • Can go backwards, loop, cross itself
    • Can control speed along the curve

Example 1: A Simple Line

Consider: x=tx = t, y=2t+1y = 2t + 1 for tRt \in \mathbb{R}

Step 1: Make a table

| tt | x=tx = t | y=2t+1y = 2t + 1 | Point | |-----|---------|--------------|-------| | -2 | -2 | -3 | (-2, -3) | | -1 | -1 | -1 | (-1, -1) | | 0 | 0 | 1 | (0, 1) | | 1 | 1 | 3 | (1, 3) | | 2 | 2 | 5 | (2, 5) |


Step 2: Eliminate the parameter

From x=tx = t, we have t=xt = x

Substitute into y=2t+1y = 2t + 1: y=2x+1y = 2x + 1

This is a line with slope 2!

Direction: As tt increases, we move in the direction of increasing xx (left to right).


Example 2: A Circle

Consider: x=3costx = 3\cos t, y=3sinty = 3\sin t for 0t2π0 \leq t \leq 2\pi

Step 1: Eliminate the parameter

We know: cos2t+sin2t=1\cos^2 t + \sin^2 t = 1

From the equations: x3=cost,y3=sint\frac{x}{3} = \cos t, \quad \frac{y}{3} = \sin t

Square and add: (x3)2+(y3)2=cos2t+sin2t=1\left(\frac{x}{3}\right)^2 + \left(\frac{y}{3}\right)^2 = \cos^2 t + \sin^2 t = 1

x29+y29=1\frac{x^2}{9} + \frac{y^2}{9} = 1

x2+y2=9x^2 + y^2 = 9


This is a circle with center (0,0)(0, 0) and radius 3!

Direction: As tt goes from 0 to 2π2\pi:

  • At t=0t = 0: (3,0)(3, 0) (starting point)
  • At t=π/2t = \pi/2: (0,3)(0, 3) (top)
  • At t=πt = \pi: (3,0)(-3, 0) (left)
  • At t=3π/2t = 3\pi/2: (0,3)(0, -3) (bottom)

Counterclockwise motion!


Example 3: An Ellipse

Consider: x=4costx = 4\cos t, y=2sinty = 2\sin t for 0t2π0 \leq t \leq 2\pi

Eliminate the parameter:

x4=cost,y2=sint\frac{x}{4} = \cos t, \quad \frac{y}{2} = \sin t

(x4)2+(y2)2=1\left(\frac{x}{4}\right)^2 + \left(\frac{y}{2}\right)^2 = 1

x216+y24=1\frac{x^2}{16} + \frac{y^2}{4} = 1

This is an ellipse with semi-major axis 4 (horizontal) and semi-minor axis 2 (vertical)!


Direction of Motion

The orientation of the curve matters!

To find direction:

  1. Plug in a few values of tt
  2. See which way the point moves as tt increases
  3. Draw arrows on the curve

Example: For x=costx = \cos t, y=sinty = \sin t:

  • t=0π/2t = 0 \to \pi/2: Point moves from (1,0)(1,0) to (0,1)(0,1)
  • This is counterclockwise

Eliminating the Parameter

Common Techniques:

1. Solve for tt from one equation, substitute into other

Example: x=t+1x = t + 1, y=t2y = t^2

From first: t=x1t = x - 1

Substitute: y=(x1)2y = (x-1)^2


2. Use trigonometric identities

For x=acostx = a\cos t, y=bsinty = b\sin t:

(xa)2+(yb)2=1\left(\frac{x}{a}\right)^2 + \left(\frac{y}{b}\right)^2 = 1


3. Use relationships between expressions

Example: x=t2x = t^2, y=t3y = t^3

Note: y=tt2=txy = t \cdot t^2 = t \cdot x

But t=±xt = \pm\sqrt{x}, so y=±xx=±x3/2y = \pm x\sqrt{x} = \pm x^{3/2}


Example 4: A Parabola

Consider: x=t2x = t^2, y=2ty = 2t for t0t \geq 0

Step 1: Solve for tt

From y=2ty = 2t: t=y2t = \frac{y}{2}


Step 2: Substitute

x=t2=(y2)2=y24x = t^2 = \left(\frac{y}{2}\right)^2 = \frac{y^2}{4}

y2=4xy^2 = 4x

This is a parabola opening to the right!

Restriction: Since t0t \geq 0, we have y=2t0y = 2t \geq 0, so only the upper half!


Domain Restrictions

Pay attention to the range of tt!

Example: x=costx = \cos t, y=sinty = \sin t for 0tπ0 \leq t \leq \pi

This traces only the upper semicircle (from (1,0)(1,0) to (1,0)(-1,0)), not the full circle!


Parametric vs Cartesian

Same Curve, Different Parametrizations

Example: The line y=xy = x can be represented as:

  1. x=t,y=tx = t, y = t
  2. x=2t,y=2tx = 2t, y = 2t
  3. x=t3,y=t3x = t^3, y = t^3

All trace the same line, but at different "speeds"!


Common Parametric Curves

Line through (x0,y0)(x_0, y_0) with slope mm:

x=x0+at,y=y0+btx = x_0 + at, \quad y = y_0 + bt

where ba=m\frac{b}{a} = m (slope)


Circle with center (h,k)(h, k) and radius rr:

x=h+rcost,y=k+rsintx = h + r\cos t, \quad y = k + r\sin t

for 0t2π0 \leq t \leq 2\pi


Ellipse with center (h,k)(h, k):

x=h+acost,y=k+bsintx = h + a\cos t, \quad y = k + b\sin t

where aa = horizontal radius, bb = vertical radius


Example 5: A Cycloid

The path traced by a point on the rim of a rolling wheel!

x=r(tsint),y=r(1cost)x = r(t - \sin t), \quad y = r(1 - \cos t)

where rr is the radius of the wheel.

Famous curve in physics - fastest descent path (brachistochrone)!


Finding Intersections

To find where a parametric curve intersects itself or another curve:

Step 1: Set x(t1)=x(t2)x(t_1) = x(t_2) and y(t1)=y(t2)y(t_1) = y(t_2)

Step 2: Solve for t1t_1 and t2t_2 (with t1t2t_1 \neq t_2)

Step 3: Those tt values give the intersection point(s)


⚠️ Common Mistakes

Mistake 1: Forgetting Domain Restrictions

When eliminating tt, the Cartesian equation might represent more than the parametric curve!

Always note restrictions on tt, xx, or yy.


Mistake 2: Losing Direction Information

The Cartesian equation doesn't tell you which way the curve is traced!

Always check a few points to determine orientation.


Mistake 3: Assuming One-to-One

A parametric curve can cross itself!

Different values of tt can give the same point (x,y)(x, y).


Applications of Parametric Equations

  1. Projectile motion:

    • x=(v0cosθ)tx = (v_0\cos\theta)t
    • y=(v0sinθ)t12gt2y = (v_0\sin\theta)t - \frac{1}{2}gt^2
  2. Planetary orbits: Elliptical paths

  3. Computer graphics: Bezier curves for design

  4. Engineering: Cam profiles, gear shapes


📝 Practice Strategy

  1. Make a table of values for small integer tt values
  2. Plot points and connect smoothly
  3. Draw arrows showing direction as tt increases
  4. Eliminate parameter to identify the curve type
  5. Check domain - does the parametric curve trace all or part of the Cartesian curve?
  6. Identify key features: starting point, ending point, orientation
  7. Look for symmetry or special structure

📚 Practice Problems

1Problem 1easy

Question:

Eliminate the parameter for x=1+2tx = 1 + 2t, y=3ty = 3 - t and sketch the curve, indicating direction.

💡 Show Solution

Step 1: Solve for tt from one equation

From x=1+2tx = 1 + 2t: 2t=x12t = x - 1 t=x12t = \frac{x-1}{2}


Step 2: Substitute into the other equation

y=3t=3x12y = 3 - t = 3 - \frac{x-1}{2}

y=3x2+12=72x2y = 3 - \frac{x}{2} + \frac{1}{2} = \frac{7}{2} - \frac{x}{2}

y=12x+72y = -\frac{1}{2}x + \frac{7}{2}

This is a line with slope 12-\frac{1}{2} and y-intercept 72\frac{7}{2}.


Step 3: Find direction

Make a table:

| tt | x=1+2tx = 1+2t | y=3ty = 3-t | |-----|------------|-----------| | 0 | 1 | 3 | | 1 | 3 | 2 | | 2 | 5 | 1 |

As tt increases, xx increases and yy decreases.

Direction: Left to right, downward along the line.

Answer: Line y=12x+72y = -\frac{1}{2}x + \frac{7}{2}, traced left to right as tt increases.

2Problem 2easy

Question:

A curve is defined by the parametric equations x=t22tx = t^2 - 2t and y=t+1y = t + 1.

a) Find the Cartesian equation by eliminating the parameter. b) Find the point on the curve when t=3t = 3.

💡 Show Solution

Solution:

Part (a): From y=t+1y = t + 1, we get t=y1t = y - 1.

Substitute into x=t22tx = t^2 - 2t:

x=(y1)22(y1)x = (y-1)^2 - 2(y-1)

=y22y+12y+2= y^2 - 2y + 1 - 2y + 2

=y24y+3= y^2 - 4y + 3

Or solving for yy: y24y+(3x)=0y^2 - 4y + (3 - x) = 0

Part (b): When t=3t = 3:

x=322(3)=96=3x = 3^2 - 2(3) = 9 - 6 = 3

y=3+1=4y = 3 + 1 = 4

Point: (3,4)(3, 4)

3Problem 3medium

Question:

For x=2costx = 2\cos t, y=3sinty = 3\sin t where 0t2π0 \leq t \leq 2\pi, eliminate the parameter and describe the curve.

💡 Show Solution

Step 1: Use the Pythagorean identity

From the equations: cost=x2,sint=y3\cos t = \frac{x}{2}, \quad \sin t = \frac{y}{3}


Step 2: Apply cos2t+sin2t=1\cos^2 t + \sin^2 t = 1

(x2)2+(y3)2=1\left(\frac{x}{2}\right)^2 + \left(\frac{y}{3}\right)^2 = 1

x24+y29=1\frac{x^2}{4} + \frac{y^2}{9} = 1


Step 3: Identify the curve

This is an ellipse with:

  • Center: (0,0)(0, 0)
  • Horizontal semi-axis: a=2a = 2
  • Vertical semi-axis: b=3b = 3

Step 4: Find direction

Check a few points:

  • t=0t = 0: (2,0)(2, 0)
  • t=π/2t = \pi/2: (0,3)(0, 3)
  • t=πt = \pi: (2,0)(-2, 0)
  • t=3π/2t = 3\pi/2: (0,3)(0, -3)

Direction: Counterclockwise, starting at (2,0)(2, 0).

Answer: Ellipse x24+y29=1\frac{x^2}{4} + \frac{y^2}{9} = 1, traced counterclockwise once.

4Problem 4hard

Question:

Eliminate the parameter for x=etx = e^t, y=e2t+1y = e^{2t} + 1 and identify any restrictions.

💡 Show Solution

Step 1: Express tt in terms of xx

From x=etx = e^t: t=lnxt = \ln x

(Note: This requires x>0x > 0)


Step 2: Substitute into yy equation

y=e2t+1=(et)2+1y = e^{2t} + 1 = (e^t)^2 + 1

Since et=xe^t = x: y=x2+1y = x^2 + 1


Step 3: Identify restrictions

Since x=etx = e^t and et>0e^t > 0 for all tt:

  • Domain: x>0x > 0
  • This is only the right half of the parabola y=x2+1y = x^2 + 1

Step 4: Additional restriction on yy

When x>0x > 0: x2>0x^2 > 0, so y=x2+1>1y = x^2 + 1 > 1

Therefore: y>1y > 1


Answer: Parabola y=x2+1y = x^2 + 1 for x>0x > 0 (equivalently, y>1y > 1).

The curve is only the right branch of the parabola.