Problem 10 - Calculus Early Transcendentals

Question

(a) Sketch the curve by using the parametric equations to plot points. Indicate with an arrow the direction in which the curve is traced as \(t\) increases. (b) Eliminate the parameter to find a Cartesian equation of the curve. \(x=t^{2}, \quad y=t^{3}\)

Step-by-Step Solution

Answer

The Cartesian equation is \( y^2 = x^3 \), and the curve is traced from \((4, -8)\) to \((4, 8)\) as \( t \) increases.

Step 1: Select Values of t

We start by selecting a range of values for the parameter \( t \). Let's choose \( t = -2, -1, 0, 1, 2 \) for simplicity.

Step 2: Calculate (x, y) Coordinates

Utilize the parametric equations \( x = t^2 \) and \( y = t^3 \) to calculate corresponding \((x, y)\) pairs for the given \( t \) values: - For \( t = -2 \), \( x = (-2)^2 = 4 \) and \( y = (-2)^3 = -8 \) leading to the point \((4, -8)\).- For \( t = -1 \), \( x = (-1)^2 = 1 \) and \( y = (-1)^3 = -1 \) leading to the point \((1, -1)\).- For \( t = 0 \), \( x = (0)^2 = 0 \) and \( y = (0)^3 = 0 \) leading to the point \((0, 0)\).- For \( t = 1 \), \( x = (1)^2 = 1 \) and \( y = (1)^3 = 1 \) leading to the point \((1, 1)\).- For \( t = 2 \), \( x = (2)^2 = 4 \) and \( y = (2)^3 = 8 \) leading to the point \((4, 8)\).

Step 3: Sketch the Curve

Plot each calculated point \((x, y)\) on a graph. - \((4, -8)\), \((1, -1)\), \((0, 0)\), \((1, 1)\), \((4, 8)\)Draw a smooth curve through these points. This curve should exhibit a shape that resembles a cubic function turned ninety degrees.

Step 4: Indicate the Direction

On the plot, indicate the direction of the curve as \( t \) increases. The curve starts at \( (4, -8) \) when \( t = -2 \) and moves through the points in order of \( t \), ending at \( (4, 8) \) when \( t = 2 \). Use an arrow to show this upward movement along the curve.

Step 5: Eliminate the Parameter

To find the Cartesian equation, solve one of the parametric equations for \( t \) and substitute into the other. From \( x = t^2 \), solve for \( t \): \[ t = \pm \sqrt{x} \]Substitute this into the equation for \( y = t^3 \):\( y = (\pm \sqrt{x})^3 = \pm x^{3/2} \).The Cartesian equation is \( y^2 = x^3 \).

Key Concepts

Parametric Curves

Parametric curves are a fascinating way to describe curves using parameters, which provides a different perspective from traditional Cartesian methods. In this approach, both the x and y coordinates of points on a curve are expressed as functions of a third variable, typically denoted by \( t \). This parameter, \( t \), helps trace the curve in a specific direction as it changes values.

For example, consider the parametric equations \( x = t^2 \) and \( y = t^3 \). For each value of \( t \), you get a unique pair of \( (x, y) \) coordinates, which when plotted, forms a curve on the plane. This is especially useful for tracing complex curves that are difficult to describe using a single function \( y = f(x) \) or \( x = g(y) \) in Cartesian coordinates.

The flexibility in choosing the parameter range allows you to focus on particular portions of the curve and analyze its direction, shape, and properties effortlessly.

Cartesian Equation

The Cartesian equation represents curves in the familiar \( x, y \) coordinate system. Unlike parametric equations, where each coordinate depends on a parameter, a Cartesian equation eliminates the parameter to directly relate the coordinates themselves.

In our exercise, the given parametric equations are \( x = t^2 \) and \( y = t^3 \). Our goal is to reformulate these into a single equation that relates \( y \) directly to \( x \), without any parameter. This transformation allows easier analysis of the curve's properties, such as finding intercepts, slopes, and understanding how \( y \) changes concerning \( x \).

Finding a Cartesian equation is essential for seamlessly integrating parametric curves into broader mathematical analyses and applications, like evaluating limits or derivatives in calculus.

Eliminating the Parameter

Eliminating the parameter is a crucial step in converting parametric equations to their Cartesian form. This process involves expressing \( t \) from one equation and substituting it into the other to remove the dependency on the parameter.

Given \( x = t^2 \) and \( y = t^3 \), we can solve the equation \( x = t^2 \) for \( t \), yielding \( t = \pm \sqrt{x} \). This value is then substituted into the second equation \( y = t^3 \), which results in \( y = (\pm \sqrt{x})^3 = \pm x^{3/2} \). By squaring \( y \), we find the equation \( y^2 = x^3 \), which relates \( y \) and \( x \) directly.

This Cartesian equation provides a clear, parameter-free equation which represents the same curve described initially by the parametric equations, allowing for different analysis and insights.

Sketching Curves

Sketching curves from parametric equations involves finding specific points and understanding the overall direction and shape of the curve. Begin by selecting a range of \( t \) values, then calculate the corresponding \( (x, y) \) values using the parametric equations.

In this example, by choosing \( t = -2, -1, 0, 1, 2 \), we computed pairs such as \( (4, -8) \), \( (1, -1) \), and \( (4, 8) \). Plotting these points on the graph gives a clear visual representation of the curve. The next step is connecting these points with a smooth curve, representing the continuous nature of the curve as \( t \) changes.

Finally, to completely understand this curve, indicate its orientation using an arrow. This arrow shows the direction the curve is traced as \( t \) increases, starting from when \( t \) is smallest to when it is largest, thereby giving context to its motion and flow on the graph.