Question

Alternative derivation of the curvature Derive the computational formula for curvature using the following steps. a. Use the tangential and normal components of the acceleration to show that \(\left.\mathbf{v} \times \mathbf{a}=\kappa|\mathbf{v}|^{3} \mathbf{B} . \text { (Note that } \mathbf{T} \times \mathbf{T}=\mathbf{0} .\right)\) b. Solve the equation in part (a) for \(\kappa\) and conclude that \(\kappa=\frac{|\mathbf{v} \times \mathbf{a}|}{\left|\mathbf{v}^{3}\right|},\) as shown in the text.

Short answer

Question: Derive an alternative computational formula for curvature using tangential and normal components of acceleration. Answer: The alternative computational formula for the curvature is given by \(\kappa = \frac{|\mathbf{v} \times \mathbf{a}|}{|\mathbf{v}|^3}\), where \(\kappa\) is the curvature, \(\mathbf{v}\) is the velocity vector, and \(\mathbf{a}\) is the acceleration vector.

Step-by-step solution

Express Acceleration in Terms of Tangential and Normal Components

To begin the derivation, we need to express the acceleration vector in terms of its tangential and normal components. Recall that the tangential and normal components of acceleration are given by: \(\mathbf{a} = a_T \mathbf{T} + a_N \mathbf{N}\) where \(a_T\) is the tangential component of acceleration, \(a_N\) is the normal component of acceleration, \(\mathbf{T}\) is the unit tangent vector, and \(\mathbf{N}\) is the unit normal vector.

Calculate the Cross Product Between Velocity and Acceleration Vectors

Now, we calculate the cross product \(\mathbf{v} \times \mathbf{a}\): \(\mathbf{v} \times \mathbf{a} = \mathbf{v} \times (a_T \mathbf{T} + a_N \mathbf{N})\) Since the cross product is distributive, we can write: \(\mathbf{v} \times \mathbf{a} = a_T (\mathbf{v} \times \mathbf{T}) + a_N (\mathbf{v} \times \mathbf{N})\) Notice that \(\mathbf{v} \times \mathbf{T} = 0\), because \(\mathbf{v}\) and \(\mathbf{T}\) are parallel. Thus, we are left with: \(\mathbf{v} \times \mathbf{a} = a_N (\mathbf{v} \times \mathbf{N})\) We know that \(\mathbf{B} = \mathbf{T} \times \mathbf{N}\), and since \(\kappa = \frac{a_N}{|\mathbf{v}|}\), we can write \(a_N = \kappa |\mathbf{v}|\). Substitute \(a_N\) and \(\mathbf{B}\) into the equation: \(\mathbf{v} \times \mathbf{a} = \kappa |\mathbf{v}| (\mathbf{v} \times \mathbf{B})\)

Solve for Curvature (\(\kappa\))

In order to find the expression for \(\kappa\), we take the magnitude of both sides of the equation: \(|\mathbf{v} \times \mathbf{a}| = \kappa |\mathbf{v}| |(\mathbf{v} \times \mathbf{B})|\) Since \(|\mathbf{B}|\) is always 1 (as a binormal vector), the equation becomes: \(|\mathbf{v} \times \mathbf{a}| = \kappa |\mathbf{v}| |\mathbf{v} \times \mathbf{B}|\) Divide both sides by \(|\mathbf{v}|^3\): \(\kappa = \frac{|\mathbf{v} \times \mathbf{a}|}{|\mathbf{v}|^3}\) This is the alternative expression for the curvature that we were tasked to derive.

Key concepts

Acceleration Components

When discussing acceleration in calculus, it's important to understand its breakdown into different components. Acceleration can be expressed in terms of tangential and normal components. This is crucial when analyzing motion in a curved path, like in our exercise about curvature.

Let's break it down:

• Tangential Component (\(a_T\)): This refers to the change in the speed along the path. If an object is speeding up or slowing down, this component is non-zero.
• Normal Component (\(a_N\)): This measures the change in direction. It acts perpendicular to the path and relates to how tightly the path is curved and how quickly the direction changes.

Combining these two, acceleration can be written as \(\mathbf{a} = a_T \mathbf{T} + a_N \mathbf{N} \), where \(\mathbf{T} \) is the unit tangent vector, and \(\mathbf{N} \) is the unit normal vector. Understanding these components is essential in analyzing the forces acting on an object moving along a curved path.

Cross Product

The cross product is a vector operation in three-dimensional space. It’s used to find a vector perpendicular to two given vectors. In the context of our solution, the cross product helps in deriving a formula for curvature.

Here are some crucial points:

• Distributive Property: The cross product is distributive, meaning \((\mathbf{v} \times (a_T \mathbf{T} + a_N \mathbf{N}) = a_T(\mathbf{v} \times \mathbf{T}) + a_N(\mathbf{v} \times \mathbf{N}))\).
• Perpendicular Vector: The result of a cross product is a vector that is perpendicular to both vectors involved. For instance, \(\mathbf{B} = \mathbf{T} \times \mathbf{N}\) gives us the binormal vector, which is crucial in analyzing three-dimensional motion.
• Magnitude: The magnitude of the cross product is affected by the sine of the angle between the two vectors, which is important when solving for \(\kappa\).

The cross product allows us to link the vectors of velocity and acceleration in order to derive properties like curvature, providing insights into the nature of the motion.

Unit Tangent Vector

The unit tangent vector, denoted as \(\mathbf{T}\), represents the direction of the path at a particular point and is essential when defining motion along a curve. The direction of this vector aligns with the velocity of the object.

Key features include:

• Normalized Vector: The vector is of unit length; \(|\mathbf{T}| = 1\). It defines direction, not magnitude.
• Relation to Path: It is parallel to the path, indicating the direction of movement.
• Dependency on Derivatives: It is the derivative of the position function, normalized: \(\mathbf{T} = \frac{\mathbf{v}}{|\mathbf{v}|}\), where \(\mathbf{v}\) is the velocity vector.

The unit tangent vector plays a pivotal role in describing the geometry of curves and aids in the breakdown of acceleration components, making it indispensable in calculations involving curvature.

Binormal Vector

The binormal vector, \(\mathbf{B}\), is another vital component when analyzing the geometry of curves in three-dimensional space. It is essential in understanding the complete three-dimensional orientation of a curve.

Here are the details:

• Perpendicularity: \(\mathbf{B}\) is perpendicular to both the tangent (\(\mathbf{T}\)) and normal (\(\mathbf{N}\)) vectors.
• Cross Product Definition: It is defined as the cross product of the tangent and normal vectors: \(\mathbf{B} = \mathbf{T} \times \mathbf{N}\).
• Unit Norm: Like \(\mathbf{T}\) and \(\mathbf{N}\), the binormal vector is also a unit vector (\(|\mathbf{B}| = 1\)), maintaining the geometric integrity of the curve.

In the context of curvature, understanding \(\mathbf{B}\) is crucial as it provides the complete spatial frame to understand how a curve twists in space. It is through vectors like \(\mathbf{B}\) that we can fully describe the properties of a curve and solve complex spatial problems in calculus.