Problem 327
Question
Sketch the parametric curve and eliminate the parameter to find the Cartesian equation of the curve. $$ x=e^{t}, \quad y=1-e^{3 t}, \quad 0 \leq t \leq 1 $$
Step-by-Step Solution
Verified Answer
The Cartesian equation is \( y = 1 - x^3 \).
1Step 1: Analyzing Parametric Equations
We are given the parametric equations for the curve: \( x = e^t \) and \( y = 1 - e^{3t} \). Our task is to eliminate the parameter \( t \) and find a Cartesian equation relating \( x \) and \( y \).
2Step 2: Expressing t in Terms of x
From the equation \( x = e^t \), we can solve for \( t \) by taking the natural logarithm of both sides: \( t = \ln(x) \). This expression for \( t \) will be useful in eliminating the parameter.
3Step 3: Substituting t in the y-Equation
Substitute the expression for \( t \) into the equation for \( y \): \( y = 1 - e^{3t} = 1 - e^{3\ln(x)} \). To simplify, recall that \( e^{3\ln(x)} = (e^{\ln(x)})^3 = x^3 \). So, \( y = 1 - x^3 \).
4Step 4: Writing the Cartesian Equation
The Cartesian equation relating \( x \) and \( y \) is derived as \( y = 1 - x^3 \).
5Step 5: Sketching the Curve
Since we have the Cartesian equation \( y = 1 - x^3 \), and \( 0 \leq t \leq 1 \), which means \( 1 \leq x \leq e \), plot the curve for \( 1 \leq x \leq e \), taking note that as \( x \) increases from 1 to \( e \), \( y \) decreases from 0 to \( 1 - e^3 \). This gives an idea of the shape and direction of the curve.
Key Concepts
Eliminating the ParameterCartesian EquationCurve SketchingExponential Functions
Eliminating the Parameter
When working with parametric equations, one often needs to find a relationship between just the variables without the parameter. This is known as eliminating the parameter. In our example, we start with two parametric equations: \( x = e^t \) and \( y = 1 - e^{3t} \). The aim is to express one variable in terms of the other, eliminating \( t \).
To achieve this, we first solve for \( t \) in terms of \( x \) using the equation \( x = e^t \). By applying the natural logarithm, we find \( t = \ln(x) \). This step is crucial because it paves the way for substituting \( t \) in the equation for \( y \). This substitution then helps us eliminate the parameter.
To achieve this, we first solve for \( t \) in terms of \( x \) using the equation \( x = e^t \). By applying the natural logarithm, we find \( t = \ln(x) \). This step is crucial because it paves the way for substituting \( t \) in the equation for \( y \). This substitution then helps us eliminate the parameter.
Cartesian Equation
The Cartesian equation represents a relationship between \( x \) and \( y \) without involving the parameter. After eliminating the parameter \( t \) with \( t = \ln(x) \), we substitute this in the equation \( y = 1 - e^{3t} \). Doing so, we have:
\[ y = 1 - e^{3\ln(x)} \]
Relying on the property that \( e^{\ln(x)} = x \), we simplify \( e^{3\ln(x)} \) to \( x^3 \). Thus, we derive the Cartesian equation: \( y = 1 - x^3 \).
This equation shows us a direct relationship between \( x \) and \( y \) independent of \( t \), significantly simplifying the task of analyzing the curve.
\[ y = 1 - e^{3\ln(x)} \]
Relying on the property that \( e^{\ln(x)} = x \), we simplify \( e^{3\ln(x)} \) to \( x^3 \). Thus, we derive the Cartesian equation: \( y = 1 - x^3 \).
This equation shows us a direct relationship between \( x \) and \( y \) independent of \( t \), significantly simplifying the task of analyzing the curve.
Curve Sketching
Curve sketching involves drawing the graph of the Cartesian equation, which in our scenario is \( y = 1 - x^3 \). We must consider the range of \( t \), which varies from 0 to 1, affecting how \( x \) varies from 1 to \( e \).
As \( x \) progresses from 1 to \( e \), the value of \( y \) falls from 0 downwards to \( 1 - e^3 \). While sketching, it's beneficial to consider these endpoints to ensure the curve's correctness visually.
Additionally, checking the behavior as \( x \) approaches certain critical values, like intercepts or turning points, can give deeper insights into the curve's nature.
As \( x \) progresses from 1 to \( e \), the value of \( y \) falls from 0 downwards to \( 1 - e^3 \). While sketching, it's beneficial to consider these endpoints to ensure the curve's correctness visually.
Additionally, checking the behavior as \( x \) approaches certain critical values, like intercepts or turning points, can give deeper insights into the curve's nature.
Exponential Functions
Understanding exponential functions is pivotal in analyses like these since both parametric equations involve exponentials.
Exponential functions have the general form \( a^t \), and in our problem, it's presented as base \( e \), the natural exponential constant. These functions are notable for their rapid growth or decay, visible in both equations: \( x = e^t \) increases as \( t \) grows, while \( y = 1 - e^{3t} \) exhibits rapid decay due to the \( (-e^{3t}) \) component.
Analyzing how these functions alter with different ranges of \( t \) helps us comprehend the behavior of the underlying curve. Additionally, the properties such as transformations applied to these functions, like translations and reflections, play a significant role in shaping the curve effectively.
Exponential functions have the general form \( a^t \), and in our problem, it's presented as base \( e \), the natural exponential constant. These functions are notable for their rapid growth or decay, visible in both equations: \( x = e^t \) increases as \( t \) grows, while \( y = 1 - e^{3t} \) exhibits rapid decay due to the \( (-e^{3t}) \) component.
Analyzing how these functions alter with different ranges of \( t \) helps us comprehend the behavior of the underlying curve. Additionally, the properties such as transformations applied to these functions, like translations and reflections, play a significant role in shaping the curve effectively.
Other exercises in this chapter
Problem 325
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