Question

For the following exercises, sketch the parametric curve and eliminate the parameter to find the Cartesian equation of the curve. $$ x=e^{t}, y=1-e^{3 t}, 0 \leq t \leq 1 $$

Short answer

The Cartesian equation of the curve is \(y = 1 - x^3\).

Step-by-step solution

Understanding Parametric Equations

The given problem provides parametric equations: \(x = e^{t}\) and \(y = 1 - e^{3t}\). Your task involves plotting this curve for a parameter \(t\) ranging from 0 to 1.

Setting Up the Parameter Range

The parameter \(t\) is in the interval \([0, 1]\). Substitute \(t\) values from this range into both equations, \(x = e^{t}\) and \(y = 1 - e^{3t}\), to generate points on the curve.

Generating Points from Parametric Equations

For \(t = 0\), \(x = e^{0} = 1\) and \(y = 1 - e^{0} = 0\).For \(t = 0.5\), \(x = e^{0.5} \approx 1.65\) and \(y = 1 - e^{1.5} \approx -3.48\).For \(t = 1\), \(x = e^{1} \approx 2.72\) and \(y = 1 - e^{3} \approx -19.09\).These points help in sketching the parametric curve.

Sketching the Curve

Plot the points \((1, 0)\), \((1.65, -3.48)\), and \((2.72, -19.09)\) on a graph. Sketch a smooth curve through these points showing the path of the curve from \(t = 0\) to \(t = 1\).

Eliminate the Parameter

To find the Cartesian equation, eliminate the parameter \(t\). Start by expressing \(t\) in terms of \(x\) from the equation \(x = e^{t}\), giving \(t = \ln(x)\). Substitute this into \(y = 1 - e^{3t}\) to get \(y = 1 - e^{3\ln(x)}\).

Simplify the Expression

Simplify \(1 - e^{3\ln(x)}\) using the property \(e^{\ln(a^b)} = a^b\), leading to \(y = 1 - x^3\). This is the Cartesian form of the given parametric equations.

Key concepts

Cartesian Equation

A Cartesian equation is a formula that relates the coordinates \((x, y)\) of points on a curve directly without involving a parameter like \(t\). In the context of converting a parametric equation into its Cartesian form, the parameter is eliminated to relate \(x\) and \(y\) directly. For our example, we have parametric equations \(x = e^t\) and \(y = 1 - e^{3t}\). By eliminating the parameter \(t\), we express \(t\) in terms of \(x\) as \(t = \ln(x)\). Substituting this into the equation for \(y\), we arrive at a direct relationship: \(y = 1 - x^3\). This equation is now in Cartesian form, describing the curve on a standard \(xy\)-plane without any reference to a parametric variable.

Curve Sketching

Curve sketching helps visualize the behavior of mathematical functions and parametric equations. By plotting key points, we can understand the shape and direction of the curve over a specified range. In our example, we evaluate the parametric equations at specific values of \(t\):

• \(t = 0\): \((x, y) = (1, 0)\)
• \(t = 0.5\): \((x, y) \approx (1.65, -3.48)\)
• \(t = 1\): \((x, y) \approx (2.72, -19.09)\)

By plotting these points on a graph and connecting them smoothly, we envisage the curve’s trajectory from \(t = 0\) to \(t = 1\). The curve’s shape provides insight into how \(x\) and \(y\) change as \(t\) varies.

Parameter Elimination

Parameter elimination involves removing the parameter from parametric equations to derive a relationship directly between \(x\) and \(y\). This process provides a Cartesian equation representing the same curve. For our equations \(x = e^t\) and \(y = 1 - e^{3t}\), the goal is to express \(y\) directly in terms of \(x\). First, solve for \(t\) in terms of \(x\): \(t = \ln(x)\). Next, substitute this expression into \(y = 1 - e^{3t}\) to transform \(y = 1 - e^{3\ln(x)}\). By utilizing the exponential property where \(e^{\ln(a^b)} = a^b\), the expression simplifies to \(y = 1 - x^3\), removing the parameter \(t\) and producing the Cartesian representation.

Exponential Functions

Exponential functions are characterized by a constant base raised to a variable exponent, typically written as \(e^x\) with \(e\) approximately equal to 2.718. These functions grow rapidly and are essential in describing processes like exponential growth and decay. In parametric equations, exponential functions allow complex curves to be expressed simply using parameters like \(t\). For instance, \(x = e^t\) captures exponential growth where \(x\) increases as \(t\) increases. Similarly, \(y = 1 - e^{3t}\) shows how curves like \(1 - x^3\) are derived from manipulating exponential terms. Understanding how to handle and integrate exponential functions is crucial in both calculus and algebra for analyzing various practical phenomena.