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To eliminate the parameter \( t \) from the equations \( x = e^{-3t} \) and \( y = e^{3t} \) and find a Cartesian equation, we can follow these steps:

1. Start with the given equations:
   \[
   x = e^{-3t} \quad \text{and} \quad y = e^{3t}
   \]

2. Take the natural logarithm of both sides of each equation to solve for \( t \):
   \[
   \ln(x) = -3t \quad \Rightarrow \quad t = -\frac{\ln(x)}{3}
   \]
   and
   \[
   \ln(y) = 3t \quad \Rightarrow \quad t = \frac{\ln(y)}{3}
   \]

3. Since both expressions equal \( t \), set them equal to each other:
   \[
   -\frac{\ln(x)}{3} = \frac{\ln(y)}{3}
   \]

4. Multiply both sides by 3 to eliminate the denominator:
   \[
   -\ln(x) = \ln(y)
   \]

5. Rewrite the equation by taking exponentials:
   \[
   \ln\left(\frac{1}{x}\right) = \ln(y) \quad \Rightarrow \quad \frac{1}{x} = y
   \]

Thus, the Cartesian equation of the curve is:
\[
y = \frac{1}{x}
\]

This equation describes the relationship between \( x \) and \( y \) on the curve after eliminating the parameter \( t \).

Would you like further details or have any questions?

Here are some related questions to explore:
1. How would the solution change if \( x = e^{2t} \) and \( y = e^{-2t} \)?
2. What is the domain of the function \( y = \frac{1}{x} \)?
3. How would you find the tangent line to this curve at a specific point?
4. How does changing the signs of the exponents in \( x \) and \( y \) affect the Cartesian equation?
5. Can you graph the equation \( y = \frac{1}{x} \) and describe its behavior?

**Tip:** Remember that logarithmic operations are helpful in converting exponential expressions, especially when eliminating parameters from parametric equations.

Eliminate the parameter to find a Cartesian equation of the curve given by x = e^(-3t) and y = e^(3t).

Learn how to eliminate the parameter in the parametric equations x = e^(-3t) and y = e^(3t) to find the Cartesian equation y = 1/x. This example demonstrates the application of logarithmic transformations to simplify exponential relationships.

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