Question

In what plane does the curve \(\mathbf{r}(t)=t \mathbf{i}+t^{2} \mathbf{k}\) lie?

Short answer

Answer: The curve lies in the xz-plane, represented by the equation $y = 0$.

Step-by-step solution

Find the tangent vector

Differentiate the given curve vector with respect to the parameter 't' to find the tangent vector. $$ \frac{d\mathbf{r}(t)}{dt} = \frac{d}{dt} (t \mathbf{i}+t^{2} \mathbf{k}) $$ Now, compute the derivative: $$ \frac{d\mathbf{r}(t)}{dt} = \mathbf{i} + 2t \mathbf{k} $$ So, the tangent vector to the curve is \(\mathbf{T}(t) = \mathbf{i} + 2t \mathbf{k}\).

Identify the normal vector

Since there's no \(\mathbf{j}\) component in the curve vector and tangent vector, the normal vector will have only \(\mathbf{j}\) component and no \(\mathbf{i}\) and \(\mathbf{k}\) components. Let's denote the normal vector as \(\mathbf{N} = A \mathbf{j}\), where A is a non-zero constant.

Find the equation of the plane

The general equation of a plane is given by: $$ A(x-x_{0})+B(y-y_{0})+C(z-z_{0})=0 $$ Where \((A, B, C)\) is the normal vector to the plane, and \((x_{0}, y_{0}, z_{0})\) is a point on the plane. Since \(\mathbf{N} = A \mathbf{j}\), we have \((A, B, C) = (0, A, 0)\). We can take a point on the curve, say \((0,0,0)\) when \(t = 0\). Substitute these values into the equation to get: $$ A(0 - 0) + A(y - 0) + 0(0 - 0) = 0 $$ Which simplifies to: $$ Ay = 0 \quad \text{or} \quad y = 0 $$ So, the equation of the plane containing the curve is \(y = 0\). This indicates that the curve lies in the xz-plane.

Key concepts

Tangent Vector

When studying calculus and geometry, the idea of a tangent vector plays a crucial role in understanding how curves behave. A tangent vector represents the direction and speed of a curve's movement at any given point. It's akin to placing a tiny arrow along the path a particle would follow, pointing in the direction it's traveling at that moment.

As seen in the problem where we differentiate the curve \(\mathbf{r}(t)=t \mathbf{i}+t^{2} \mathbf{k}\) with respect to \(t\), the resulting tangent vector \(\mathbf{T}(t) = \mathbf{i} + 2t \mathbf{k}\) captures the instantaneous rate of change of the position vector. Think of this as a snapshot of the curve's motion: at any instant \(t\), the vector \(\mathbf{i} + 2t \mathbf{k}\) will point to the direction in which the curve is heading.

In physical terms, if you were traveling along the curve, your velocity vector at time \(t\) would be exactly that tangent vector. It's essential not only for visualizing the path but also for finding other properties of the curve, like curvature or the equation of the plane it lies in, as explored further in this problem.

Normal Vector

Now, let's focus on the notion of the normal vector. A normal vector, sometimes called the perpendicular vector, is a vector that is perpendicular to the tangent vector at a given point on a curve or surface. In the context of planes and curves, it's used to define the orientation of a plane in three-dimensional space.

In our exercise, the absence of a \(\mathbf{j}\) component in both the curve and its tangent vector indicated that the normal vector should be in the \(y\)-direction. By identifying \(\mathbf{N} = A \mathbf{j}\) where \(A\) is a constant, we establish that the normal vector is perpendicular to both the \(x\)- and \(z\)-axes in this case.

The importance of the normal vector springs from its role in descriptions of planes. Not only does it help in determining the orientation of a plane, but it's also a key player in calculating reflections and refractions, determining lighting in computer graphics, and more. Being able to identify and utilize normal vectors is foundational when working with three-dimensional shapes and their properties.

Equation of a Plane

The equation of a plane is an algebraic way to express all points that lie in that plane. In its general form, the equation \(A(x-x_{0})+B(y-y_{0})+C(z-z_{0})=0\) utilizes a point on the plane \((x_{0}, y_{0}, z_{0})\) and the components \(A, B, C\) of a normal vector to the plane.

The process unfolded in our solution by applying this general formula using the normal vector \(\mathbf{N} = A \mathbf{j}\), leading to \(y = 0\), which is the equation of the xz-plane. There's an elegance to this equation; every point \((x, 0, z)\) fits this description perfectly, confirming that the curve, described by \(\mathbf{r}(t)\), indeed lies in that plane.

The plane's equation is a powerful tool—it's not confined to this context. It's used to define work planes in design software, to analyze geometric problems in physics, and to demarcate territories in geographical mapping. Recognizing how to define and use the equation of a plane is invaluable for students tackling multidimensional calculus and spatial analysis.

Vector Differentiation

Vector differentiation is the process of finding the derivative of a vector function with respect to a variable, typically time or space. It's a higher-dimensional analog to taking the derivative of a scalar function, but with a focus on how each component of the vector changes. In our original exercise, we performed vector differentiation on the curve vector \(\mathbf{r}(t)\) with respect to \(t\), yielding the tangent vector \(\mathbf{T}(t)\).

Understanding vector differentiation is crucial for anyone delving into physics, engineering, or advanced mathematics. It enables one to describe the behavior of moving objects, the flow of fluids, changes in fields, and much more. Mastery of vector differentiation allows for the exploration of dynamic systems, the optimization of functions in multiple dimensions, and even the groundwork for understanding higher-level concepts such as vector fields and differential equations.

Students embarking on this journey need to be comfortable with applying differentiation rules to each component, understanding the geometric and physical interpretation of derivatives, and using these skills to solve concrete problems, such as finding the plane in which a curve lies.