Question

Eliminate the parameter in the given parametric equations. \(x=\ln t, \quad y=t^{2}-1\)

Short answer

The parameter is eliminated resulting in the equation \(y = e^{2x} - 1\).

Step-by-step solution

Express the Parameter from x in terms of t

We have the parametric equation for x as \(x = \ln t\). To express \(t\) in terms of \(x\), exponentiate both sides to get rid of the natural logarithm. This gives us: \(t = e^x\).

Substitute the Expression for t in the Equation for y

Now that we have \(t = e^x\), substitute it into the equation for \(y = t^2 - 1\). This results in the equation: \(y = (e^x)^2 - 1\).

Simplify the Expression for y in Terms of x

The equation \(y = (e^x)^2 - 1\) simplifies to \(y = e^{2x} - 1\). This is the expression of \(y\) purely in terms of \(x\), eliminating the parameter \(t\).

Key concepts

Natural Logarithm

The natural logarithm is often denoted by the symbol \( \ln \) and is the logarithm to the base \( e \), where \( e \approx 2.71828 \). This mathematical function is widely used in calculus and exponential growth calculations because \( e \) is the base of the natural exponential function. The natural logarithm of a number \( t \) is the power to which \( e \) must be raised to obtain \( t \).

For example, if \( t = e^2 \), then \( \ln t = 2 \). In the context of parametric equations, the natural logarithm can help express one variable in terms of another. In the given exercise, the parametric equation \( x = \ln t \) is solved for \( t \). Here, the natural logarithm is essential in setting up the equation so that it can be manipulated further.

Remember, the inverse of natural logarithm functions is exponential functions, making them very powerful in switching between the two forms when solving parametric equations.

Exponentiation

Exponentiation is the operation of raising one number (the base) to the power of another number (the exponent). In mathematical notation, \( a^b \) denotes \( a \) raised to the power \( b \). The number \( e \), the base of natural logarithms, is often used as the base in exponential functions, such as \( e^x \).

When dealing with logarithms and exponential functions, remember that they are inverse operations. For example, if you take the natural logarithm of \( e^x \), you simply get \( x \): \( \ln(e^x) = x \).

In the task of eliminating parameters from parametric equations, exponentiation is the key step when reversing the effects of a logarithm. Given \( x = \ln t \), exponentiation lets you solve for \( t \) as \( t = e^x \). This new expression for \( t \) allows further substitutions in parametric equations, making it a vital technique in simplifying complex equations.

Eliminating Parameters

Eliminating parameters in parametric equations involves expressing the dependent variable(s) entirely in terms of the independent variable. In parametric equations, you generally have equations that define \( x \) and \( y \) using another parameter, say \( t \). The goal is to find a direct relationship between \( x \) and \( y \) without the parameter.

For the equation \( x = \ln t \) and \( y = t^2 - 1 \), the process involves expressing \( t \) in terms of \( x \) and substituting this back into the equation for \( y \). By using previous steps of computing \( t = e^x \) from \( x = \ln t \), you substitute \( t \) in the second equation to get \( y = (e^x)^2 - 1 \).

The result is \( y = e^{2x} - 1 \), which is \( y \) expressed in terms of \( x \) with no \( t \) involved. This simplification makes it much easier to analyze the relationship between \( x \) and \( y \) directly, avoiding the complexities of dealing with an extra variable. Eliminating parameters is a useful technique for transforming parametric forms into more familiar algebraic forms.