Problem 46 - Calculus III

Question

Find the arc length of the curve on the given interval.Find the equations of the normal plane and the osculating plane of the curve \(\mathbf{r}(t)=\langle 2 \sin (3 t), t, 2 \cos (3 t)\rangle\) at point \((0, \pi,-2)\)

Step-by-Step Solution

Answer

Normal plane: \(3x + y = \pi\); Osculating plane: \(x = 0\).

Step 1: Find Corresponding 't' Value

We need to find the value of \( t \) when \( \mathbf{r}(t) = (0, \pi, -2) \). Given \( 2 \sin(3t) = 0 \), \( t = \pi \), and \( 2 \cos(3t) = -2 \), apply these conditions. From \( 2 \sin(3t) = 0 \), \( \sin(3t) = 0 \), so \( 3t = n\pi \) for integer \( n \). From \( 2 \cos(3t) = -2 \), \( \cos(3t) = -1 \). The solution is valid when \( 3t = (2m+1)\pi \) for integer \( m \). Therefore, the matching values for both conditions give \( t = \frac{\pi}{3} \).

Step 2: Calculate Derivatives

Compute the first and second derivatives of the vector function \( \mathbf{r}(t) \).The first derivative is \( \mathbf{r}'(t) = \langle 6\cos(3t), 1, -6\sin(3t) \rangle \).The second derivative is \( \mathbf{r}''(t) = \langle -18\sin(3t), 0, -18\cos(3t) \rangle \).

Step 3: Determine Tangent Vector

The tangent vector \( \mathbf{T}(t) \) is the normalized \( \mathbf{r}'(t) \).At \( t = \frac{\pi}{3} \), \( \mathbf{r}'(\frac{\pi}{3}) = \langle 6(\frac{1}{2}), 1, 0 \rangle = \langle 3, 1, 0 \rangle \).Normalize by dividing by its magnitude: \( \mathbf{T}(\frac{\pi}{3}) = \frac{\langle 3, 1, 0 \rangle}{\sqrt{10}} \).

Step 4: Find Principal Normal Vector

Compute the principal normal vector \( \mathbf{N}(t) \) as the derivative of \( \mathbf{T}(t) \) divided by its magnitude.The derivative \( \mathbf{T}'(t) = \langle -9\sin(3t)/\sqrt{10}, 0, -9\cos(3t)/\sqrt{10} \rangle \).At \( t = \frac{\pi}{3} \), \( \mathbf{T}'(\frac{\pi}{3}) = \langle -\frac{9\sqrt{3}}{20}, 0, -\frac{9}{10} \rangle \), which can be normalized.

Step 5: Determine Binormal Vector

The binormal vector \( \mathbf{B}(t) \) is found using \( \mathbf{B}(t) = \mathbf{T}(t) \times \mathbf{N}(t) \).At \( t = \frac{\pi}{3} \), perform the cross product between \( \mathbf{T}(\frac{\pi}{3}) \) and \( \mathbf{N}(\frac{\pi}{3}) \) to find \( \mathbf{B}(\frac{\pi}{3}) \).

Step 6: Normal Plane Equation

The equation of the normal plane at the point is given by the plane vector form using \( \mathbf{T}(t) \).Its vector equation will be \( 3x + y = \pi \), substituting the given point and direction.

Step 7: Osculating Plane Equation

The osculating plane is spanned by \( \mathbf{T}(t) \) and \( \mathbf{N}(t) \).Find the equation using the normal vector to this plane, \( \mathbf{B}(t) \), resulting in \( x = 0 \) when evaluated at the given point. Therefore, the osculating plane equation is: \( 3x + y + 0 \cdot z = \pi \).

Key Concepts

Normal Plane

In the study of vector-valued functions and curves in space, the concept of the normal plane is crucial in understanding the orientation of the curve at a specific point. Every point on a space curve is associated with a unique plane termed the "normal plane." This plane is positioned perpendicular to the tangent vector at that particular point on the curve. It includes both the principal normal vector and the binormal vector.
This plane provides a snapshot of the curve's tendency to change direction in an instant and helps in identifying the plane's intersection with the direction the curve is heading.

• The normal plane is defined using the tangent vector \(\mathbf{T}(t)\) at that point.
• If \(\mathbf{r}(t)\) is the position vector of the curve, then the normal plane can be mathematically represented as holding the equation determined by \(\mathbf{T}(t) \cdot \mathbf{(r} - \mathbf{r}_0) = 0\), where \(\mathbf{r}_0\) is a position vector of a specific point on the curve.
• This plane effectively slices through the curve perpendicularly, offering a cross-sectional view of changes in direction.

Understanding this concept enhances one's ability to grasp the geometrical nature of space curves and addresses how they interact with their surrounding environment.

Osculating Plane

Closely related to the normal plane, the osculating plane offers a deeper insight into the curvature of a curve in space. It is described as a plane that "kisses" or "osculates" the curve at a specific point. This plane contains both the tangent vector and the principal normal vector. It captures the curve's main directional turn at that point.
The osculating plane serves as a depiction of the immediate path the curve is following and describes how sharply the curve is turning.

• Mathematically, the osculating plane is defined by the cross product of the tangent vector \(\mathbf{T}(t)\) and the principal normal vector \(\mathbf{N}(t)\).
• This plane is the one that best approximates the curve near that point, showing the primary travel direction of the curve.
• The normal vector to this plane is often the binormal vector, serving as a perpendicular reference to the osculating surface.

With the osculating plane, one can visualize the precise way in which the curve bends locally around a specific point, offering a more refined look at its geometry.

Tangent Vector

The tangent vector plays a fundamental role in understanding the direction and speed at which a particle moves along a space curve. At any given point along this curve, the tangent vector is derived by differentiating the position vector \(\mathbf{r}(t)\) of a point on the curve.
It points in the direction of the steepest ascent from the point on the curve and is normalized to give a unit vector that solely indicates direction without concern for magnitude.

• The tangent vector \(\mathbf{T}(t)\) is computed as the derivative \(\mathbf{r}'(t)\) of the curve's position vector, divided by its magnitude to produce a unit vector.
• It is defined as \(\mathbf{T}(t) = \frac{\mathbf{r}'(t)}{||\mathbf{r}'(t)||}\) for normalization.
• The tangent vector is essential across physics and engineering disciplines as it can represent velocity direction and help track the trajectory of moving objects.

Interpreting the tangent vector correctly enables students and professionals to trace paths and predict future movements indirectly indicated by these mathematical directions.

Principal Normal Vector

The principal normal vector, often just called the normal vector, is central to comprehending how a space curve is changing its direction at any given point. This vector is perpendicular to the tangent vector and lies in the osculating plane.
Its significance lies in its ability to provide insights into the instantaneous change of direction that the tangent vector undergoes.
The steps for finding a principal normal vector are relatively straightforward:

• The principal normal vector \(\mathbf{N}(t)\) is obtained by taking the derivative of the tangent vector \(\mathbf{T}'(t)\).
• It's then normalized to a unit vector, similar to the tangent vector: \(\mathbf{N}(t) = \frac{\mathbf{T}'(t)}{||\mathbf{T}'(t)||}\).
• This vector essentially captures the amount of curving or deviation from a straight path as experienced by the curve.

Examining the principal normal vector provides one with the capacity to intuitively understand how curves deviate and how sharply their trajectory alters.

Binormal Vector

The binormal vector is an essential part of the Frenet-Serret frame, complementing the tangent and principal normal vectors to form a complete set of orthogonal vectors known as the TNB frame. This vector is perpendicular to both the tangent and principal normal vectors and is extremely useful in concluding a three-dimensional reference frame for curves.
Understanding the binormal vector involves the following components:

• The binormal vector \(\mathbf{B}(t)\) is computed by taking the cross product of the tangent vector \(\mathbf{T}(t)\) and the principal normal vector \(\mathbf{N}(t)\).
• It is expressed as \(\mathbf{B}(t) = \mathbf{T}(t) \times \mathbf{N}(t)\) and lies perpendicular to the plane formed by them.
• The binormal vector helps in defining the orientation of the curve in space and is an integral part of justifying the curve's spatial dynamics.

With all three vectors—the tangent, principal normal, and binormal—together, they fully describe any spatial curve's behavior and orientation and enable the geometric transformation of curves to more tangible interpretations.