Problem 6 - Essential Calculus Early Transcendentals

Question

\(5-12=\) Sketch the curve with the given vector equation. Indicate with an arrow the direction in which \(t\) increases. $$ \mathbf{r}(t)=\left\langle t^{3}, t^{2}\right\rangle $$

Step-by-Step Solution

Answer

The curve extends in the first quadrant starting at the origin, with an increasing t direction indicated by an arrow away from the origin.

Step 1: Understand the Components of the Vector Equation

The vector equation given is \( \mathbf{r}(t)=\left\langle t^{3}, t^{2} \right\rangle \). The first component, \( t^3 \), is the expression for the x-coordinate, while the second component, \( t^2 \), is the expression for the y-coordinate for each value of \( t \).

Step 2: Determine the Parametric Equations

Extract the parametric equations from the vector equation: \( x(t) = t^3 \) and \( y(t) = t^2 \). These equations tell us how the x and y coordinates change with respect to \( t \).

Step 3: Analyze the Relationship Between x and y

To sketch the curve, find the relationship between \( x \) and \( y \). From \( y = t^2 \), solve for \( t \) in terms of \( y \): \( t = \sqrt{y} \) (for \( t \geq 0 \)) and \( t = -\sqrt{y} \) (for \( t \leq 0 \)). Substitute into \( x = t^3 \) to get \( x = y^{3/2} \). This is the equation of the curve.

Step 4: Sketch the Curve

The equation \( x = y^{3/2} \) represents a curve. Because \( y\geq 0 \) (since \( y = t^2 \)), this curve will only exist in the first quadrant of the Cartesian plane. Sketch this curve by noting that for each \( y \), only positive \( x \) values will result.

Step 5: Indicate the Direction of Increasing t

Since \( x(t) = t^3 \) and \( y(t) = t^2 \), as \( t \) increases (from negative to positive), \( x(t) \) and \( y(t) \) increase following their respective plots. The curve starts from \( t = 0 \) when \( x = 0 \) and \( y = 0 \), and extends in the positive x and y direction. Indicate this direction on the curve with an arrow pointing away from the origin.

Key Concepts

Parametric Equations and Their Components

In vector calculus, parametric equations are essential tools used to describe a curve by parametrizing the coordinates of each point on the curve. For our example, the vector equation is given as \( \mathbf{r}(t) = \langle t^3, t^2 \rangle \). This means:

• \(x(t) = t^3\) maps out the x-coordinate based on the parameter \(t\).
• \(y(t) = t^2\) maps out the y-coordinate based on \(t\).

These parametric equations allow us to understand how the x and y coordinates interact and change as \(t\) varies. By breaking down a vector equation into its component parametric equations, we can assess how both dimensions function and relate to the parameter \(t\). This is particularly useful for analyzing and sketching curves in space. Notice that as \(t\) changes, it generates a set of \((x, y)\) points that form the curve on the plane.

Understanding the Vector Equation

A vector equation, like \( \mathbf{r}(t) = \langle t^3, t^2 \rangle \), is a mathematical representation capturing both direction and magnitude in two or three-dimensional space. Here:

• The vector's components, \(t^3\) and \(t^2\), determine the movement along the x-axis and y-axis respectively.
• The curve drawn by this vector equation is characterized by the parametric path \(t\) takes from negative infinity to infinity.

Analyzing the structure of such vector equations reveals how each component depends on the parameter \(t\). This dependency unfolds the behavior of the vector path across the coordinate plane. As \(t\) progresses, it shows how each part of the vector contributes to tracing the curve's trajectory. Understanding vector equations is pivotal for visualizing paths and directions, aiding in the construction of graphical representations for mathematical and engineering applications.

Curve Sketching - Visualizing the Parametric Path

Once the parametric equations \(x(t) = t^3\) and \(y(t) = t^2\) are established, the next step is to sketch the curve they create. By analyzing their relationship, \(y = t^2\) gives us \(t = \pm \sqrt{y}\). Substituting \(t\) in \(x = t^3\) yields \(x = y^{3/2}\). This reveals the equation of the curve:

• It lies within the first quadrant since \(y \geq 0\).
• Only positive \(x\) values emerge from the non-negative \(y\) values.

To sketch, start from the origin \((0,0)\) and extend towards positive values of \(x\) and \(y\) as \(t\) increases. It's crucial to indicate the direction of the curve, which begins at the origin and moves away as \(t\) becomes more positive. Visualization helps comprehend complex equations, turning abstract mathematical descriptions into concrete, interpretable graphics.