Problem 23
Question
\(x=e^{t}, \quad y=e^{-2 t}, \quad t\) in \(\mathbb{R}\)
Step-by-Step Solution
Verified Answer
The relationship between \(x\) and \(y\) is \(y = x^{-2}\).
1Step 1: Understanding the Exercise
We need to convert the parametric equations into a single equation that relates the variables \(x\) and \(y\). The given equations are \(x = e^t\) and \(y = e^{-2t}\).
2Step 1: Express \(t\) in terms of \(x\)
From the first equation \(x = e^t\), we take the natural logarithm of both sides to express \(t\) in terms of \(x\). This gives us \(t = \ln(x)\).
3Step 2: Substitute \(t\) into \(y\)'s Expression
Now, substitute \(t = \ln(x)\) into the equation for \(y\), which is \(y = e^{-2t}\). This gives us \(y = e^{-2 \ln(x)}\).
4Step 3: Simplify the Expression for \(y\)
Use the power rule \(e^{a \cdot b} = (e^a)^b\) to simplify \(y = e^{-2 \ln(x)}\). This is equivalent to \(y = (e^{\ln(x)})^{-2}\). Since \(e^{\ln(x)} = x\), we have \(y = x^{-2}\).
5Step 5: Conclusion
The parametric equations can be represented in the simplified form \(y = x^{-2}\), which is an equation relating \(x\) and \(y\) directly.
Key Concepts
Convert Parametric to CartesianNatural LogarithmExponential Functions
Convert Parametric to Cartesian
Parametric equations are used to express the coordinates of the points making up a geometric object, such as a curve, by using one or more parameters. To convert a set of parametric equations to a Cartesian equation, which relates the variables directly, you follow a straightforward process of substitution.
Given the parametric form:
Understanding and implementing these steps not only aids in converting parametric forms but also strengthens your grasp of solving problems that involve multiple variables and parameters.
Given the parametric form:
- \(x = e^t\)
- \(y = e^{-2t}\)
Understanding and implementing these steps not only aids in converting parametric forms but also strengthens your grasp of solving problems that involve multiple variables and parameters.
Natural Logarithm
The natural logarithm, denoted as \(\ln\), is a useful mathematical function to understand due to its many applications in various fields. It is the inverse function of the exponential function. For a positive number \(x\), the natural logarithm \(\ln(x)\) is defined as the power to which the base \(e\) (approximately 2.718) must be raised to produce that number \(x\).
This function is particularly helpful in converting expressions involving exponential terms when dealing with equations. For instance, when given \(x = e^t\), applying the natural logarithm to both sides allows us to isolate \(t\), resulting in \(t = \ln(x)\). This transformation is indispensable when shifting between exponential and logarithmic forms, streamlining the process of mixing and rearranging exponential equations.
Natural logarithms not only simplify otherwise complex expressions but also enable easy manipulation and solution of equations in calculus and algebra.
This function is particularly helpful in converting expressions involving exponential terms when dealing with equations. For instance, when given \(x = e^t\), applying the natural logarithm to both sides allows us to isolate \(t\), resulting in \(t = \ln(x)\). This transformation is indispensable when shifting between exponential and logarithmic forms, streamlining the process of mixing and rearranging exponential equations.
Natural logarithms not only simplify otherwise complex expressions but also enable easy manipulation and solution of equations in calculus and algebra.
Exponential Functions
Exponential functions are mathematical expressions where a constant base is raised to a variable exponent, typically represented as \(f(x) = a^x\). The base \(a\) is a positive constant, the most common being the natural base \(e\). These functions model a variety of real-life scenarios from population growth to radioactive decay and are characterized by their rapid increase or decrease.
In the context of the given exercise, the equations \(x = e^t\) and \(y = e^{-2t}\) involve exponential functions. These equations demonstrate the properties of exponential functions, such as how changing the exponent affects growth or decay, a concept especially visible in \(y = e^{-2t}\), which simplifies to \(y = x^{-2}\) in its Cartesian form.
A fundamental property used in converting these parametric equations is \(e^{a \cdot b} = (e^a)^b\), which allows simplification of complex expressions. Exponential functions and their manipulation are crucial for many advanced concepts in mathematics, engineering, and sciences, showing how the initial growth or decay behavior of these functions can be reinterpreted in context-specific simplified forms.
In the context of the given exercise, the equations \(x = e^t\) and \(y = e^{-2t}\) involve exponential functions. These equations demonstrate the properties of exponential functions, such as how changing the exponent affects growth or decay, a concept especially visible in \(y = e^{-2t}\), which simplifies to \(y = x^{-2}\) in its Cartesian form.
A fundamental property used in converting these parametric equations is \(e^{a \cdot b} = (e^a)^b\), which allows simplification of complex expressions. Exponential functions and their manipulation are crucial for many advanced concepts in mathematics, engineering, and sciences, showing how the initial growth or decay behavior of these functions can be reinterpreted in context-specific simplified forms.
Other exercises in this chapter
Problem 22
Exer. 19-30: Find an equation of the parabola that satisfies the given conditions. $$ \text { Focus } F(-3,-2), \quad \text { directrix } y=1 $$
View solution Problem 22
Exer. 19-30: Find an equation for the ellipse that has its center at the origin and satisfies the given conditions. Foci \(F(\pm 3,0)\), minor axis of length 2
View solution Problem 23
Exer. 13-26: Find a polar equation that has the same graph as the equation in \(x\) and \(y\). $$ y^{2}-x^{2}=4 $$
View solution Problem 23
Find an equation for the hyperbola that has its center at the origin and satisfies the given conditions. Foci \(F(\pm 5,0)\), vertices \(V(\pm 3,0)\)
View solution