Problem 22 - 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. $$ x=\frac{1}{\sqrt{t+1}}, \quad y=\frac{t}{1+t}, t>-1 $$

Step-by-Step Solution

Answer

The rectangular form is \( y = \frac{1-x^2}{x^2} \). The domain is \( x \in \mathbb{R} \backslash \{0\} \).

Step 1: Solve for t in terms of x

The given parametric equation for x is \( x = \frac{1}{\sqrt{t+1}} \). Let's solve for \( t \) in terms of \( x \). Start by multiplying both sides by \( \sqrt{t+1} \), which gives \( \sqrt{t+1} \cdot x = 1 \). Then, square both sides to eliminate the square root: \( t + 1 = \frac{1}{x^2} \). Finally, solve for \( t \) by subtracting 1: \( t = \frac{1}{x^2} - 1 \).

Step 2: Substitute t in terms of x into the equation for y

Using the expression for \( t \) we found in Step 1, substitute it into the equation for \( y \). The parametric equation for \( y \) is \( y = \frac{t}{1+t} \). By substitution, \( y = \frac{ \frac{1}{x^2} - 1}{1 + \left(\frac{1}{x^2} - 1\right) } \). Simplify the denominator: \( 1 + \frac{1}{x^2} - 1 = \frac{1}{x^2} \). Thus, \( y = \frac{\frac{1}{x^2} - 1}{\frac{1}{x^2}} \).

Step 3: Simplify the expression for y

Simplify the result from Step 2. The expression \( y = \frac{\frac{1}{x^2} - 1}{\frac{1}{x^2}} \) simplifies to \( y = \frac{1 - x^2}{x^2} \). Thus, the rectangular form of the curve is \( y = \frac{1 - x^2}{x^2} \).

Step 4: Determine the domain of the rectangular form

The expression for \( y \) is \( y = \frac{1 - x^2}{x^2} \), which is undefined for \( x = 0 \). Additionally, since \( t > -1 \) and \( t = \frac{1}{x^2} - 1 \), \( \frac{1}{x^2} > 0 \) implies that \( x eq 0 \). Thus, the domain of the rectangular form is all real numbers except \( x = 0 \), i.e., \( x \in \mathbb{R} \backslash \{0\} \).

Key Concepts

Rectangular Form

Parametric equations, like the given \( x = \frac{1}{\sqrt{t+1}} \) and \( y = \frac{t}{1+t} \), describe a curve using a third variable, \( t \), which you can think of as time or another parameter. To understand the shape of the curve in a more familiar way, we convert these equations into a rectangular form, which uses only \( x \) and \( y \). This generally makes it easier to work with the equations and to visualize the shape of the curve directly.
The main method to convert from parametric to rectangular form is to eliminate the parameter \( t \). This involves solving one of the equations for \( t \) and substituting it into the other equation. In our exercise, by solving \( x = \frac{1}{\sqrt{t+1}} \) for \( t \), we get \( t = \frac{1}{x^2} - 1 \).
Substituting this expression into the equation for \( y \), yields: \( y = \frac{ \left( \frac{1}{x^2} - 1 \right) }{1 + \left( \frac{1}{x^2} - 1 \right) } \).After simplifying, we arrive at the rectangular form of the equation: \( y = \frac{1 - x^2}{x^2} \), which describes the relationship between \( x \) and \( y \) without the need for \( t \).

Domain of a Function

The domain of a function is all the values that \( x \) can take without making the function undefined. With our rectangular form equation \( y = \frac{1 - x^2}{x^2} \), we must consider points where the function fails to exist.

• First, notice the denominator \( x^2 \). Since division by zero is undefined, \( x = 0 \) is not included in the domain. Hence, \( x \) cannot be zero.
• Next, check restrictions from the original parametric form. For \( t > -1 \), the derived condition \( \frac{1}{x^2} - 1 > -1 \) implies that \( \frac{1}{x^2} > 0 \). This confirms \( x eq 0 \).

Combining these restrictions, the domain of the function is all real numbers except \( x = 0 \). Thus, we can represent the domain as: \( x \in \mathbb{R} \setminus \{0\} \). This means \( x \) can be any real number except zero.

Algebraic Manipulation

Algebraic manipulation is all about rearranging equations and expressions so you can simplify or solve them. It is a fundamental skill in math that allows us to convert parametric equations into rectangular form.
In the original exercise, we see multiple examples of algebraic manipulation:

• First, solving \( x = \frac{1}{\sqrt{t+1}} \) involves rearranging the equation to isolate \( t \). Multiply both sides by \( \sqrt{t+1} \) then square both sides to get \( t + 1 = \frac{1}{x^2} \). Finally, subtract 1 to solve for \( t \).
• Then, substitute the expression for \( t \) into \( y = \frac{t}{1 + t} \), and simplify by combining terms to get \( y = \frac{1 - x^2}{x^2} \).

Through such steps, algebraic manipulation helps us convert complex parametric forms into more manageable rectangular forms, while ensuring that all manipulations are valid to retain the equation's equality. Mastering these skills is essential for tackling various mathematical problems efficiently.