Question

Show that the curve determined by \(\mathbf{r}=t \mathbf{i}+t \mathbf{j}+t^{2} \mathbf{k}\) is a parabola, and find the coordinates of its focus.

Short answer

The curve is a parabola with its focus at (0, 0, 1/4).

Step-by-step solution

Rewrite the vector equation

The given vector equation is \( \mathbf{r} = t \mathbf{i} + t \mathbf{j} + t^2 \mathbf{k} \). This can be rewritten in the parametric form as \( x = t \), \( y = t \), and \( z = t^2 \). This means the curve is expressed in terms of the parameter \( t \).

Express in Cartesian form

From the parametric equations, we have \( x = t \) and \( y = t \). This implies \( x = y \). Substitute \( t = x \) into the equation for \( z \), we get \( z = x^2 \). Thus, the Cartesian equation of the curve is \( y = x \) and \( z = x^2 \).

Recognize the parabola

The equation \( z = x^2 \) with \( y = x \) indicates a parabolic curve in the \( xz \)-plane. We recognize the equation \( z = x^2 \) as a standard form of a parabola (\( z \) as a function of \( x \)).

Determine the focus of the parabola

For the parabola in the form \( z = x^2 \), the vertex is at the origin \((0, 0)\), and it opens along the \( z \)-axis. The standard formula for the focus of \( z = x^2 \) is \( (0, 0, \frac{1}{4a}) \) where \( a \) is the coefficient of \( x^2 \). In this case, \( a = 1 \), so the focus is at \( (0, 0, \frac{1}{4}) \).

Key concepts

Parametric Equations

Parametric equations provide a unique way to describe a curve using parameters. In the context of our original exercise, the vector equation \( \mathbf{r} = t \mathbf{i} + t \mathbf{j} + t^2 \mathbf{k} \) can be broken down into parametric equations: \( x = t \), \( y = t \), and \( z = t^2 \).
These equations indicate how each coordinate—\( x, y, \) and \( z \)—changes as the parameter \( t \) changes. Parametric forms are especially useful in 3D geometry because they allow for a step-by-step description of a curve's path, without directly relying on more complex Cartesian or polar coordinate forms.
In essence, parametric equations give us a comprehensive toolkit to illustrate dynamic processes, including everything from the motion of objects to the intricate paths of curves such as our parabolic trajectory.

Cartesian Form

The Cartesian form is another way to represent equations and curves, specifically by eliminating parameters and expressing relationships directly between coordinates like \( x, y, \) and \( z \).
In our exercise, starting from \( x = t \), \( y = t \), and \( z = t^2 \), we derive the relationships \( x = y \) and \( z = x^2 \).
This transformation gives us Cartesian expressions of the curve which make it easier to recognize familiar geometric shapes like lines and parabolas. The Cartesian form is essential for easily visualizing and analyzing the trajectory or path of objects in a coordinate space without handling parameters.

Parabola in 3D

When discussing parabolas in a 3-dimensional space, it's crucial to understand how they extend beyond their basic 2D characteristics. Here, our Cartesian form has derived the equation \( z = x^2 \), with \( x = y \), indicating a parabola in the \( xz \)-plane while also intersecting consistently along the line \( y = x \).
What distinguishes a parabola in 3D is its orientation and direction. In this exercise, the parabola opens upwards in the direction of the \( z \)-axis. Such a 3D representation broadens the perspective, allowing the parabola to represent potential real-world phenomena, like trajectories affected by gravitational forces.
3D parabolas are common in physics and engineering, modeling paths of particles or objects under uniform acceleration or when representing surfaces of revolutions like satellite dishes or reflective surfaces.

Focus of Parabola

The focus of a parabola is a key element defining its geometric properties. It is the point that determines the parabola's shape and direction, setting the pathway along which the structure opens.
For a parabola like \( z = x^2 \), the focus emerges from its standard form. Given that our coefficient \( a \) in \( z = ax^2 \) is \( 1 \), the focus falls at the point \( (0, 0, \frac{1}{4a}) \), translating to \( (0, 0, \frac{1}{4}) \).
The focus is instrumental in understanding not only the symmetry and orientation of a parabola but also its applications. For example, focusing properties are valuable in optics where parabolic mirrors use the focus to direct light waves toward a single point, achieving concentrated thermal or visual intensities. Understanding the geometry of focus underpins both mechanical designs and mathematical concepts.