Question

Consider the following parametric equations. a. Eliminate the parameter to obtain an equation in \(x\) and \(y\) b. Describe the curve and indicate the positive orientation. $$x=\sqrt{t}+4, y=3 \sqrt{t} ; 0 \leq t \leq 16$$

Short answer

Based on the step by step solution provided above, create a short answer. Given the parametric equations \(x=\sqrt{t}+4\) and \(y=3\sqrt{t}\) for \(0 \leq t \leq 16\), we first eliminate the parameter "t" by solving for t in one equation and substituting it into the other equation. We find the equation of the curve to be \(y^2=9(x-4)^2\), which represents an ellipse with the center at (4,0), major axis along the y-axis, and minor axis along the x-axis. The positive orientation of the ellipse is counterclockwise, starting from the point (4,0) and ending at the point (8,12) as t increases.

Step-by-step solution

Solve for t in one equation

We can solve for t in either equation, but it will be easier to solve for t in the equation \(x=\sqrt{t}+4\). $$x=\sqrt{t}+4 \Rightarrow \sqrt{t}=x-4 \Rightarrow t=(x-4)^2$$

Substitute the expression for t into the other equation

Now we'll substitute the expression we found for t into the equation for y: $$y=3\sqrt{t} \Rightarrow y=3\sqrt{(x-4)^2}$$

Eliminate the parameter

We need to rewrite the equation in terms of \(x\) and \(y\) only. Since we have a square root, we'll square both sides of the equation: $$y^2=9(x-4)^2$$ This is the equation in x and y that we were looking for.

Describe the curve

The equation \(y^2=9(x-4)^2\) is a quadratic equation with no cross terms. Therefore, the curve is a conic section. Looking more closely, we see that the equation can be rewritten as: $$\frac{y^2}{9}=\frac{(x-4)^2}{1}$$ This is the equation of an ellipse, with the center at \((4,0)\), the major axis along the y-axis, and the minor axis along the x-axis.

Indicate the positive orientation

Finally, we'll find the positive orientation of the ellipse. Since the equation of y is \(y=3 \sqrt{t}\) and the given range of t is \(0 \leq t \leq 16\), the curve starts at the point \((x,y)=(4,0)\) where \(y=3 \sqrt{0}\) when \(t=0\), and it ends at the point \((x,y)=(8,12)\) where \(y=3 \sqrt{16}\) when \(t=16\). As t increases, the y-coordinate also increases. This indicates that the positive orientation is counterclockwise along the ellipse.

Key concepts

Eliminating Parameters

When working with parametric equations, eliminating the parameter means converting the equations into a form that only involves two variables, typically \(x\) and \(y\). This allows us to analyze the relationship between these two variables without referencing the third variable, the parameter (often denoted as \(t\)). The process can simplify understanding the shape and nature of the curve described by the equations.

In the provided exercise, we have two parametric equations: \(x = \sqrt{t} + 4\) and \(y = 3\sqrt{t}\). The task is to eliminate \(t\). Start by solving one of the equations for \(t\). The equation for \(x\) is rearranged easily: \(\sqrt{t} = x - 4\), thus \(t = (x - 4)^2\).

Next, substitute the expression for \(t\) derived from the first equation into the second equation. This gives \(y = 3\sqrt{(x - 4)^2}\). To fully eliminate \(t\) and find an equation solely in \(x\) and \(y\), square both sides thus resulting in \(y^2 = 9(x - 4)^2\). This equation describes the relationship between \(x\) and \(y\) independent of the parameter \(t\).

Conic Sections

Conic sections are curves obtained by intersecting a plane with a double-napped cone. These curves include circles, ellipses, parabolas, and hyperbolas. Each type has a distinctive equation form, and understanding these forms is essential in identifying the specific curve represented by a given equation.

In the problem at hand, we derived the equation \(y^2 = 9(x - 4)^2\). To determine the kind of conic section, we can check if it matches any standard forms. Rewriting it, we have \(\frac{y^2}{9} = \frac{(x - 4)^2}{1}\). Recognizing this form indicates an ellipse, where its center is at \((4, 0)\).

For an ellipse, the general equation is \(\frac{(x - h)^2}{a^2} + \frac{(y - k)^2}{b^2} = 1\). However, our equation resembles more the pattern of a hyperbola since \(y\) and \(x\) terms are squared and equaled, indicating its shape. The axes of this conic section are oriented such that the major axis has a length of 3 units along the \(y\)-axis, and the minor axis has a length of 1 unit along the \(x\)-axis.

Positive Orientation

In the context of curves, positive orientation usually refers to the direction in which the curve is traced as the parameter increases. For two-dimensional curves, this can be clockwise or counterclockwise. Understanding the orientation helps visualize the dynamic behavior of the curve over its domain.

From the parametric equations of the exercise, \(x = \sqrt{t} + 4\) and \(y = 3\sqrt{t}\), positive orientation is determined by the change in parameter \(t\). Here, it is specified that \(t\) ranges from 0 to 16, and as \(t\) increases, both \(x\) and \(y\) increase.

Specifically, the curve described here starts at the point \((4,0)\) when \(t=0\), and ends at \((8,12)\) when \(t=16\). Observing the changes in coordinates, the movement of the curve is upward and to the right, tracing the ellipse in a counterclockwise direction. This is described as the positive orientation of the curve.