Math Problem Statement

Eliminate the parameter to find a Cartesian equation of the curve for 0 ≤ y ≤ 6.

Solution

To eliminate the parameter tt and express the relationship between xx and yy in Cartesian form, we’ll start with the given parametric equations:

x=t22x = t^2 - 2 y=t+3y = t + 3

  1. Solve for tt in terms of yy from the second equation: y=t+3t=y3y = t + 3 \Rightarrow t = y - 3

  2. Substitute t=y3t = y - 3 into the first equation to express xx in terms of yy: x=(y3)22x = (y - 3)^2 - 2

  3. Expand and simplify: x=(y26y+9)2x = (y^2 - 6y + 9) - 2 x=y26y+7x = y^2 - 6y + 7

So, the Cartesian equation of the curve is: x=y26y+7x = y^2 - 6y + 7

Range of yy:

Given that 3t3-3 \leq t \leq 3, substituting into y=t+3y = t + 3 gives: 0y60 \leq y \leq 6

Thus, the Cartesian equation is: x=y26y+7for0y6x = y^2 - 6y + 7 \quad \text{for} \quad 0 \leq y \leq 6

Would you like more details on this or have any further questions?


Related Questions

  1. How can we find the range of xx for this curve?
  2. What is the vertex of the resulting parabola in terms of xx and yy?
  3. How would the equation change if the parametric equations included a different transformation?
  4. Can we find the derivative dydx\frac{dy}{dx} of the curve given in Cartesian form?
  5. How do we plot the Cartesian form of this parametric equation?

Tip

When eliminating parameters, isolate the parameter in one equation and substitute it into the other to find the direct relationship.

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Math Problem Analysis

Mathematical Concepts

Parametric Equations
Cartesian Coordinates
Algebraic Manipulation

Formulas

x = t^2 - 2
y = t + 3

Theorems

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Suitable Grade Level

Grades 9-12