Problem 5 - Advanced Engineering Mathematics

Question

In Problems, graph the curve traced by the given vector function. \(\mathbf{r}(t)=\left\langle e^{t}, e^{2 t}\right\rangle\)

Step-by-Step Solution

Answer

Graph the right half of the parabola \( y = x^2 \) for \( x > 0 \).

Step 1: Understand the Vector Function

The given vector function \( \mathbf{r}(t) = \langle e^t, e^{2t} \rangle \) describes a parametric curve in the plane. \( e^t \) is the x-component and \( e^{2t} \) is the y-component of the vector function.

Step 2: Eliminate the Parameter

To graph the curve, we often eliminate the parameter to find a relationship between \( x \) and \( y \). Let \( x = e^t \). Thus, \( y = e^{2t} = (e^t)^2 = x^2 \). The equation of the curve is \( y = x^2 \).

Step 3: Identify the Range and Plot Points

Since \( x = e^t \), \( x > 0 \). Choose values for \( t \) such as 0, 1, and -1 to get corresponding points. When \( t=0 \), \( x=1 \), \( y=1 \); when \( t=1 \), \( x=e \), \( y=e^2 \); and when \( t=-1 \), \( x=\frac{1}{e} \), \( y=\frac{1}{e^2} \).

Step 4: Sketch the Graph

Plot these points and note that the curve follows the parabola \( y = x^2 \) but only for \( x > 0 \). The curve starts from rightward at the coordinate \( (1, 1) \) and as \( t \) increases, \( x \) and \( y \) increase exponentially. Conversely, for negative \( t \), \( x \) and \( y \) rapidly approach zero but remain positive.

Key Concepts

Vector Functions

Vector functions extend the idea of regular functions by using vectors to describe relations. Instead of just associating a single output with each input, vector functions deal with a vector of values. These are often used to depict curves, surfaces, or paths in two or three dimensions. The point on the graph at any parameter value can be determined using vector functions.

Let's take the function \( \mathbf{r}(t) = \langle e^t, e^{2t} \rangle \). Here, every value of the parameter \( t \) results in a vector, comprising the x-value of \( e^t \) and the y-value of \( e^{2t} \). This allows us to visualize the path made by each component—creating a parametric curve in the plane.

In real-world applications, vector functions are crucial, often representing real phenomena like motion paths or the rotation of objects. Understanding them opens a gateway to deeper insights into multidimensional concepts.

Parametric Equations

Parametric equations provide another perspective on understanding curves. Instead of relying solely on standard x-y functions, they employ parameters to describe each coordinate. The parameter, often symbolized by \( t \), helps to trace the path taken by a point moving along a curve.

The function \( \mathbf{r}(t) = \langle e^t, e^{2t} \rangle \) illustrates this vividly. Here, \( e^t \) and \( e^{2t} \) are parametric equations showing how the x and y coordinates evolve with \( t \). As \( t \) varies, the resulting values provide a complete map of the curve's trajectory.

Parametric equations are powerful tools especially in physics and engineering, offering a dynamic way of interpreting mathematical models. They allow us to describe continuous motion and make complex animations or simulations attainable.

Eliminating Parameters

Eliminating parameters involves removing the parameter \( t \) to directly link the x and y coordinates. This process allows us to express the curve in a more familiar form of a standard equation. Generally, this helps in simplifying analyses or sketching graphs.

In the exercise, to eliminate the parameter from \( \mathbf{r}(t) = \langle e^t, e^{2t} \rangle \), we set \( x = e^t \). Then, naturally, \( y = e^{2t} = (e^t)^2 = x^2 \). This resulted in the equation \( y = x^2 \), which is a parabola.

This conversion means we can use known methods and intuition about parabolas to analyze and graph the curve. Now featuring a more straightforward format, our interpretation becomes effortless. Whether in theoretical contexts or practical applications, eliminating parameters frequently simplifies our work and aids in clearer comprehension.