Problem 36 - Calculus Volume 3

Question

Convert the parametric equations of a curve into rectangular form. No sketch is necessary. State the domain of the rectangular form. $$ \begin{array}{l} x=\ln (5 t) \\ y=\ln \left(t^{2}\right) \text { where } 1 \leq t \leq e \end{array} $$

Step-by-Step Solution

Answer

The rectangular form is \( y = 2x - \ln(25) \) with domain \( \ln(5) \leq x \leq \ln(5) + 1 \).

Step 1: Express the parameter in terms of x

The given equation for x is \( x = \ln(5t) \). To isolate \( t \), rewrite the equation as an exponential function: \( 5t = e^x \). Therefore, \( t = \frac{e^x}{5} \).

Step 2: Substitute into the equation for y

Using the expression for \( t \) from Step 1, substitute it into the equation for \( y \): \( y = \ln(t^2) \). Replace \( t \) with \( \frac{e^x}{5} \), resulting in \( y = \ln\left(\left(\frac{e^x}{5}\right)^2\right) \).

Step 3: Simplify the expression for y

Simplify the expression from Step 2: \( y = \ln\left(\frac{e^{2x}}{25}\right) \). Break it into two separate log terms: \( y = \ln(e^{2x}) - \ln(25) \). This further simplifies to \( y = 2x - \, \ln(25) \).

Step 4: Determine the domain of the rectangular form

Given that \( 1 \leq t \leq e \), substitute \( t = \frac{e^x}{5} \) into the inequality. This results in \( 1 \leq \frac{e^x}{5} \leq e \), which simplifies to \( 5 \leq e^x \leq 5e \). Taking the natural logarithm of each part: \( \ln(5) \leq x \leq \ln(5e) \). Simplifying gives \( \ln(5) \leq x \leq \ln(5) + 1 \).

Step 5: Conclusion

The rectangular form of the curve is \( y = 2x - \ln(25) \), with a domain of \( \ln(5) \leq x \leq \ln(5) + 1 \).

Key Concepts

Rectangular Form

When dealing with parametric equations, a common task is to convert them into rectangular form. This means expressing one variable in terms of the other without the parameter. In our problem, the parameters are expressed as:

• \( x = \ln(5t) \)
• \( y = \ln(t^2) \)

To convert these into rectangular form, we first express the parameter \( t \) in terms of \( x \). From \( x = \ln(5t) \), we rearrange to find \( t \) by reversing the logarithmic operation using an exponential function:\[ 5t = e^x \]\[ t = \frac{e^x}{5} \]Next, substitute \( t \) in \( y = \ln(t^2) \), resulting in:\[ y = \ln\left(\left(\frac{e^x}{5}\right)^2\right) \]Finally, simplify to get the rectangular form \( y = 2x - \ln(25) \).
Rectangular forms are important because they provide a direct relationship between \( x \) and \( y \), making it easier to understand the behavior of the curve in a familiar coordinate system.

Domains

The domain of a function refers to the set of all possible values that the independent variable can take on in a given equation. When converting from parametric equations to rectangular form, special attention is required to identify the domain for the new variable relationships. Here, the original parameter \( t \) satisfies the range \( 1 \leq t \leq e \).After converting to rectangular form, the bound values of \( t \) need to be expressed in terms of \( x \). We use the substitution \( t = \frac{e^x}{5} \) and incorporate this into inequalities:

• From \( 1 \leq \frac{e^x}{5} \), we deduce \( 5 \leq e^x \); thus \( x \geq \ln(5) \).
• From \( \frac{e^x}{5} \leq e \), we deduce \( e^x \leq 5e \); thus \( x \leq \ln(5e) \).

This gives us the domain of our rectangular form as \( \ln(5) \leq x \leq \ln(5) + 1 \). By establishing appropriate domains, we ensure the function behaves properly within intended constraints.

Natural Logarithms

Natural logarithms, denoted \( \ln \), are the logarithms to the base \( e \), where \( e \) is an irrational constant approximately equal to 2.718. They are particularly useful in solving equations involving growth and decay, such as those found in natural processes.

• The equation \( x = \ln(5t) \) uses the natural logarithm to relate \( x \) with \( 5t \). By reversing this operation through exponentiation, we uncover the relationship \( t = \frac{e^x}{5} \).
• Similarly, \( y = \ln(t^2) \) uses natural logarithms to express \( y \) as a function of \( t^2 \). This allows easy manipulation later on when simplifying to rectangular form:

\[ y = \ln(e^{2x}) - \ln(25) \]Natural logarithms are crucial in linking algebraic expressions with their exponential counterparts, simplifying transformations between forms.

Exponential Functions

Exponential functions are the inverse of logarithmic functions and appear frequently in equations involving growth or decay processes. The basic form is \( e^x \), where \( e \) is a constant approximately equal to 2.718.In our exercise, exponential functions help transition from parametric to rectangular form:

• The expression \( 5t = e^x \) comes from reversing \( \ln(5t) = x \) using exponentiation, allowing us to isolate \( t \) as \( t = \frac{e^x}{5} \).
• Exponentiation also aids in simplifying the natural logarithm of powers in \( y = \ln((t)^2) \). By converting, we attain:\[ y = \ln\left(\frac{e^{2x}}{25}\right) \]

This transformation is critical as it allows us to handle complex algebraic manipulations, facilitating conversion from one mathematical form to another while retaining intrinsic properties.