Parametric Equations: Eliminating Parameters
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x = 4t, y = 8t2, when 0≤t≤4
| Time, t | 0 | 1 | 2 | 3 | 4 |
| Distance, x x = 4t | 0 | 4 | 8 | 12 | 16 |
| Height, y y = 8t2 | 0 | 8 | 32 | 72 | 128 |

GUIDELINES FOR ELIMINATING THE PARAMETER:
1. Solve one parametric equation in terms of the parameter.2. Substitute the resulting expression for the parameter into the other parametric equation. 3. Simplify
Let's use this method on the pair of parametric equations above. Solve one of the parametric equations for the parameter, say x = 4t.x=4t→ x 4 =t
Substitute the resulting expression for the parameter into the other parametric equation and simplify.y=8 t 2 →y=8 ( x 4 ) 2 →y= 8 x 2 16 →y= x 2 2
The resulting rectangular equation, 1 2 x 2 , represents a parabola that opens up, has an axis of symmetry at x = 0 and a vertex of (0, 0).
x = t + 3
y = t - 49
Solve the first parametric equation for the parameter.x= t +3→x−3= t → ( x−3 ) 2 =t
Substitute the resulting expression for the parameter into the other parametric equation and simplify.y = t - 49
y = (x-3)2 - 49
y = (x2 - 6x + 9) - 49
y = x2 - 6x + 40
Based on the resulting rectangular equation, y = x2 - 6x + 40, the domain is all real numbers. However, the parametric equation, x= t +3, limits the domain of t to numbers where t > 0. Therefore the domain of the rectangular equation, y = x2 - 6x + 40, must be restricted to all numbers where x > 0. Let's try a couple of examples. Example 1: Eliminate the parameter and obtain the rectangular equation, state the domain and sketch the curve for the pair of parametric equations.x= t 2 −4 y=3 t 2 +5
| Step 1: Solve one of the parametric equations for the parameter | x = t2 - 4 Original x + 4 = t2 Add 4 ± x+4 =t Take square root |
| Step 2: Substitute the resulting expression for the parameter into the other parametric equation and simplify. | y=3 t 2 +5 Other equation y=3 ( ± x+4 ) 2 +5 Substitute for t y=3( x+4 )+5 Square y=3x+12+5 Multiply/Distribute y=3x+17 Add |
| Step 3: Determine the domain of the rectangular equation. | The domain for t in both parametric equations is all real numbers and the domain for x in the rectangular equation is all real numbers. Thus no adjustments are need to the domain of the rectangular equation. Domain: All real numbers |
| Step 4: Sketch the curve. | |
| Step 1: Solve one of the parametric equations for the parameter | x= 1 t+ 1 2 Original 1 x =t+ 1 2 Take inverse 1 x − 1 2 =t Subtract 1/2 |
| Step 2: Substitute the resulting expression for the parameter into the other parametric equation and simplify. | y= 1 t 2 +10 Other equation y= 1 ( 1 x − 1 2 ) 2 +10 Substitute for t y= 1 1 x 2 − 1 x + 1 4 +10 Square denom. y= x 2 −x+4+10 x by reciprocal y= x 2 −x+14 Add |
| Step 3: Determine the domain of the rectangular equation. | The domain for t in the first parametric equation, x= 1 t− 1 2 , is t> 1 2 . The domain for t in the second parametric equation, y= 1 t 2 +10 is t>0 . To meet both constraints, the domain of t must be t> 1 2 . Therefore the domain of the rectangular equation, x 2 −x+14 , is x> 1 2 Domain: x> 1 2 |
| Step 4: Sketch the curve. | |
| Related Links: Math algebra Parametric Equations: Eliminating Angle Parameters Finding Parametric Equations for a Graph Pre Calculus |
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