Parametric Equations

Understanding parametric equations and how to convert between parametric and rectangular forms

Parametric Equations

What Are Parametric Equations?

Parametric equations define both xx and yy in terms of a third variable called a parameter, usually tt.

Instead of y=f(x)y = f(x), we write: x=f(t)x = f(t) y=g(t)y = g(t)

Why Use Parametric Equations?

  1. Model motion: tt often represents time
  2. Describe curves that aren't functions (fail vertical line test)
  3. Separate horizontal and vertical motion
  4. Express direction and speed of motion

Example: Linear Motion

Consider: x=2t+1x = 2t + 1 y=3t−2y = 3t - 2

As tt increases, the point (x,y)(x, y) traces out a path.

| tt | x=2t+1x = 2t + 1 | y=3t−2y = 3t - 2 | Point (x,y)(x, y) | |-----|--------------|--------------|----------------| | 00 | 11 | −2-2 | (1,−2)(1, -2) | | 11 | 33 | 11 | (3,1)(3, 1) | | 22 | 55 | 44 | (5,4)(5, 4) |

Converting to Rectangular Form

To eliminate the parameter tt:

  1. Solve one equation for tt
  2. Substitute into the other equation
  3. Simplify to get yy in terms of xx (or vice versa)

Example

Given: x=2t+1x = 2t + 1 and y=3t−2y = 3t - 2

Step 1: Solve for tt from the first equation. x=2t+1x = 2t + 1 t=x−12t = \frac{x - 1}{2}

Step 2: Substitute into the second equation. y=3t−2=3(x−12)−2y = 3t - 2 = 3\left(\frac{x - 1}{2}\right) - 2

y=3x−32−2=3x−3−42=3x−72y = \frac{3x - 3}{2} - 2 = \frac{3x - 3 - 4}{2} = \frac{3x - 7}{2}

Rectangular form: y=32x−72y = \frac{3}{2}x - \frac{7}{2}

Parametric Equations for Common Curves

Circle (radius rr, center at origin)

x=rcos⁡(t)x = r\cos(t) y=rsin⁡(t)y = r\sin(t) 0≤t≤2π0 \leq t \leq 2\pi

Ellipse

x=acos⁡(t)x = a\cos(t) y=bsin⁡(t)y = b\sin(t)

Line Segment

x=x1+t(x2−x1)x = x_1 + t(x_2 - x_1) y=y1+t(y2−y1)y = y_1 + t(y_2 - y_1) 0≤t≤10 \leq t \leq 1

Direction and Orientation

The orientation shows the direction of motion as tt increases.

  • Plot points for increasing values of tt
  • Draw arrows to show direction of travel
  • Different parametrizations can trace the same curve with different orientations

Calculus with Parametric Equations

Derivative (slope of tangent line): dydx=dy/dtdx/dt=y′(t)x′(t)\frac{dy}{dx} = \frac{dy/dt}{dx/dt} = \frac{y'(t)}{x'(t)}

Note: This is NOT dydt\frac{dy}{dt}!

📚 Practice Problems

1Problem 1easy

❓ Question:

Eliminate the parameter to find the rectangular equation: x=t+3x = t + 3, y=t2−1y = t^2 - 1

💡 Show Solution

Solution:

Step 1: Solve one equation for tt.

From x=t+3x = t + 3: t=x−3t = x - 3

Step 2: Substitute into the other equation.

y=t2−1=(x−3)2−1y = t^2 - 1 = (x - 3)^2 - 1

Step 3: Expand and simplify.

y=x2−6x+9−1y = x^2 - 6x + 9 - 1 y=x2−6x+8y = x^2 - 6x + 8

Step 4: Identify the curve.

This is a parabola opening upward.

Answer: y=x2−6x+8y = x^2 - 6x + 8 (parabola)

2Problem 2medium

❓ Question:

Consider the parametric equations: x=2t−1x = 2t - 1 and y=t2+3y = t^2 + 3

a) Eliminate the parameter to find a Cartesian equation. b) Find the point on the curve when t=2t = 2. c) Sketch the direction of motion as tt increases.

💡 Show Solution

Solution:

Part (a): From the first equation: x=2t−1x = 2t - 1

Solve for tt: t=x+12t = \frac{x + 1}{2}

Substitute into the second equation:

y=t2+3=(x+12)2+3y = t^2 + 3 = \left(\frac{x + 1}{2}\right)^2 + 3

y=(x+1)24+3y = \frac{(x + 1)^2}{4} + 3

y=x2+2x+14+3y = \frac{x^2 + 2x + 1}{4} + 3

y=x2+2x+1+124=x2+2x+134y = \frac{x^2 + 2x + 1 + 12}{4} = \frac{x^2 + 2x + 13}{4}

Or: 4y=x2+2x+134y = x^2 + 2x + 13

Part (b): When t=2t = 2:

x=2(2)−1=3x = 2(2) - 1 = 3 y=22+3=7y = 2^2 + 3 = 7

Point: (3,7)(3, 7)

Part (c): As tt increases from negative to positive:

  • When t=−1t = -1: x=−3x = -3, y=4y = 4
  • When t=0t = 0: x=−1x = -1, y=3y = 3
  • When t=1t = 1: x=1x = 1, y=4y = 4
  • When t=2t = 2: x=3x = 3, y=7y = 7

The curve is a parabola opening upward, and motion is from left to right as tt increases.

3Problem 3medium

❓ Question:

Find parametric equations for a circle with center (2,−3)(2, -3) and radius 55.

💡 Show Solution

Solution:

Step 1: Recall the parametric form for a circle centered at origin.

For a circle centered at (0,0)(0, 0) with radius rr: x=rcos⁡(t),y=rsin⁡(t)x = r\cos(t), \quad y = r\sin(t)

Step 2: Adjust for the center (h,k)=(2,−3)(h, k) = (2, -3).

To shift the center, add hh to xx and add kk to yy: x=h+rcos⁡(t)x = h + r\cos(t) y=k+rsin⁡(t)y = k + r\sin(t)

Step 3: Substitute h=2h = 2, k=−3k = -3, r=5r = 5.

x=2+5cos⁡(t)x = 2 + 5\cos(t) y=−3+5sin⁡(t)y = -3 + 5\sin(t)

Step 4: Specify the domain.

0≤t≤2π0 \leq t \leq 2\pi (for one complete revolution)

Answer: x=2+5cos⁡(t)x = 2 + 5\cos(t) y=−3+5sin⁡(t)y = -3 + 5\sin(t) 0≤t≤2π0 \leq t \leq 2\pi

4Problem 4hard

❓ Question:

A projectile is launched with parametric equations: x=40tx = 40t y=−16t2+60t+5y = -16t^2 + 60t + 5

where xx and yy are in feet and tt is in seconds.

a) Find the maximum height reached by the projectile. b) Find when and where the projectile hits the ground.

💡 Show Solution

Solution:

Part (a): Maximum height occurs at the vertex of the parabola y=−16t2+60t+5y = -16t^2 + 60t + 5.

For y=at2+bt+cy = at^2 + bt + c, vertex is at t=−b2at = -\frac{b}{2a}

t=−602(−16)=6032=1.875t = -\frac{60}{2(-16)} = \frac{60}{32} = 1.875 seconds

Maximum height:

y=−16(1.875)2+60(1.875)+5y = -16(1.875)^2 + 60(1.875) + 5 y=−16(3.516)+112.5+5y = -16(3.516) + 112.5 + 5 y=−56.25+112.5+5y = -56.25 + 112.5 + 5 y=61.25y = 61.25 feet

Part (b): The projectile hits the ground when y=0y = 0:

−16t2+60t+5=0-16t^2 + 60t + 5 = 0

Using the quadratic formula: t=−60±602−4(−16)(5)2(−16)t = \frac{-60 \pm \sqrt{60^2 - 4(-16)(5)}}{2(-16)}

t=−60±3600+320−32=−60±3920−32t = \frac{-60 \pm \sqrt{3600 + 320}}{-32} = \frac{-60 \pm \sqrt{3920}}{-32}

t=−60±62.61−32t = \frac{-60 \pm 62.61}{-32}

t=−60+62.61−32=−0.082t = \frac{-60 + 62.61}{-32} = -0.082 (reject, negative time)

t=−60−62.61−32=−122.61−32=3.832t = \frac{-60 - 62.61}{-32} = \frac{-122.61}{-32} = 3.832 seconds

Horizontal distance: x=40(3.832)=153.3x = 40(3.832) = 153.3 feet

The projectile hits the ground after 3.83 seconds at a distance of 153.3 feet.

5Problem 5medium

❓ Question:

Given x=2tx = 2t and y=t2+1y = t^2 + 1, find dydx\frac{dy}{dx} at t=3t = 3.

💡 Show Solution

Solution:

Step 1: Find dxdt\frac{dx}{dt} and dydt\frac{dy}{dt}.

dxdt=2\frac{dx}{dt} = 2

dydt=2t\frac{dy}{dt} = 2t

Step 2: Use the formula for the derivative.

dydx=dy/dtdx/dt=2t2=t\frac{dy}{dx} = \frac{dy/dt}{dx/dt} = \frac{2t}{2} = t

Step 3: Evaluate at t=3t = 3.

dydx∣t=3=3\frac{dy}{dx}\bigg|_{t=3} = 3

Answer: dydx=3\frac{dy}{dx} = 3 at t=3t = 3

This means the slope of the tangent line at the point where t=3t = 3 is 33.

🎴

Practice with Flashcards

Review key concepts with our flashcard system

📖

Browse All Topics

Explore other calculus topics