Question

Use the definitions to compute the unit binormal vector and torsion of the following curves. $$r(t)=\langle 12 t, 5 \cos t, 5 \sin t\rangle$$

Short answer

Question: Find the unit binormal vector and torsion of the curve with the position vector \(r(t) = \langle 12t, 5\cos t, 5\sin t \rangle\). Answer: The unit binormal vector B(t) is given by \[\langle \frac{-\sin t^2}{\sqrt{5}}, -\frac{4}{\sqrt{5}}\sin t\cos t, -\frac{4}{\sqrt{5}}\sin^2 t \rangle.\] The torsion τ(t) of the curve is given by \[\tau(t) = 2\sin t - 4\cos^2 t - 8\sin t \cos t.\]

Step-by-step solution

Find the first derivative

To find the first derivative, differentiate each component of the position vector r(t) with respect to the parameter t: $$\frac{d}{dt} r(t) = \langle \frac{d}{dt}{(12t)}, \frac{d}{dt}{(5\cos t)}, \frac{d}{dt}{(5\sin t)} \rangle = \langle 12, -5\sin t, 5\cos t \rangle$$

Find the second derivative

Next, we find the second derivative by differentiating the first derivative with respect to t: $$\frac{d^2}{dt^2} r(t) = \langle \frac{d}{dt}(12), \frac{d}{dt}(-5\sin t), \frac{d}{dt}(5\cos t) \rangle = \langle 0, -5\cos t, -5\sin t \rangle$$

Find the unit tangent vector

The unit tangent vector T(t) is the unit vector of the first derivative: $$T(t) = \frac{\frac{d}{dt} r(t)}{|\frac{d}{dt} r(t)|} = \frac{\langle 12, -5\sin t, 5\cos t \rangle}{\sqrt{(12)^2 +(-5\sin t )^2 + (5\cos t)^2 }} = \langle \frac{4}{\sqrt{5}}, \frac{-\sin t}{\sqrt{5}}, \frac{\cos t}{\sqrt{5}} \rangle$$

Find the unit normal vector

The unit normal vector N(t) is the unit vector of the derivative of the unit tangent vector: $$N(t) = \frac{\frac{d}{dt} T(t)}{|\frac{d}{dt} T(t)|}$$ First, we find the derivative of T(t) with respect to t: $$\frac{d}{dt} T(t) = \langle \frac{d}{dt}(\frac{4}{\sqrt{5}}), \frac{d}{dt}(\frac{-\sin t}{\sqrt{5}}), \frac{d}{dt}(\frac{\cos t}{\sqrt{5}}) \rangle = \langle 0, \frac{-\cos t}{\sqrt{5}}, -\frac{\sin t}{\sqrt{5}}\rangle$$ Now, find the unit vector of the derivative of T(t): $$N(t) = \frac{\langle 0, \frac{-\cos t}{\sqrt{5}}, -\frac{\sin t}{\sqrt{5}}\rangle}{\sqrt{(0)^2 +(\frac{-\cos t}{\sqrt{5}})^2 + (\frac{-\sin t}{\sqrt{5}})^2}} = \langle 0, -\cos t, -\sin t \rangle$$

Find the unit binormal vector

The unit binormal vector B(t) is the cross product of the unit tangent vector and unit normal vector: $$B(t) = T(t) \times N(t) = \langle \frac{4}{\sqrt{5}}, \frac{-\sin t}{\sqrt{5}}, \frac{\cos t}{\sqrt{5}} \rangle \times \langle 0, -\cos t, -\sin t \rangle$$ To find the cross product, we can use the formula: $$\begin{vmatrix} i & j & k \\ \frac{4}{\sqrt{5}} & \frac{-\sin t}{\sqrt{5}} & \frac{\cos t}{\sqrt{5}} \\ 0 & -\cos t & -\sin t \end{vmatrix} = \langle \frac{-\sin t^2}{\sqrt{5}}, -\frac{4}{\sqrt{5}}\sin t\cos t, -\frac{4}{\sqrt{5}}\sin^2 t \rangle$$ Which simplifies to: $$B(t) = \langle \frac{-\sin t^2}{\sqrt{5}}, -\frac{4}{\sqrt{5}}\sin t\cos t, -\frac{4}{\sqrt{5}}\sin^2 t \rangle$$

Find the torsion

The torsion τ(t) can be computed using the following formula: $$\tau(t) = -\frac{(\frac{d}{dt} B(t) \cdot N(t))}{|N(t)|^2}$$ First, find the first derivative of B(t): $$\frac{d}{dt} B(t) = \langle \frac{-2\sin t\cos t}{\sqrt{5}}, -\frac{2\sin t\cos t}{\sqrt{5}} -\frac{4 \cos^2 t}{\sqrt{5}}, -\frac{-8 \sin t \cos t}{\sqrt{5}} \rangle$$ Next, compute the dot product of \((\frac{d}{dt} B(t))\) and \(N(t)\): $$(\frac{d}{dt} B(t) \cdot N(t)) = \langle \frac{-2\sin t\cos t}{\sqrt{5}}, -\frac{2\sin t\cos t}{\sqrt{5}} -\frac{4 \cos^2 t}{\sqrt{5}}, \frac{8 \sin t \cos t}{\sqrt{5}} \rangle \cdot \langle 0, -\cos t, -\sin t \rangle = -2\sin t + 4\cos^2 t + 8\sin t \cos t$$ Now, find the magnitude of \(N(t)\) squared: $$|N(t)|^2 = (0)^2 + (-\cos t)^2 + (-\sin t)^2 = 1$$ Finally, we calculate torsion by substituting these values into the formula: $$\tau(t) = -\frac{-2\sin t + 4\cos^2 t + 8\sin t \cos t}{1} = 2\sin t - 4\cos^2 t - 8\sin t \cos t$$

Key concepts

Unit Tangent Vector

The unit tangent vector, usually denoted as \textbf{T}(t), represents the direction of the curve at a particular point and is a fundamental concept in vector calculus, specifically in the study of space curves. To find \textbf{T}(t), one must differentiate the given position vector \textbf{r}(t) with respect to the variable t, which gives the tangent vector. This vector is then normalized to obtain the unit tangent vector. Normalization is performed by dividing the tangent vector by its magnitude, ensuring the unit tangent vector has a length of one. Here's the formula for \textbf{T}(t):
\[ T(t) = \frac{\frac{d}{dt} r(t)}{|\frac{d}{dt} r(t)|} \]
Understanding the unit tangent vector is essential as it lays the groundwork for finding the normal and binormal vectors, key components in describing the geometry of a curve.

Unit Normal Vector

Following the unit tangent vector, the unit normal vector, denoted as \textbf{N}(t), provides information about the change in direction of the tangent vector and essentially describes how the curve is bending at a point. To find \textbf{N}(t), you differentiate \textbf{T}(t) to get the rate of change of the tangent vector and subsequently normalize it. The process involves determining the derivative of the components of \textbf{T}(t) and dividing the resulting vector by its magnitude. The standard formula for this is:
\[ N(t) = \frac{\frac{d}{dt} T(t)}{|\frac{d}{dt} T(t)|} \]
This vector is orthogonal to the tangent vector due to the process of differentiation and normalization, providing a perpendicular direction to the curve's tangent.

Unit Binormal Vector

The unit binormal vector \textbf{B}(t) completes the orthonormal triad (T, N, B), which is also known as the Frenet-Serret frame used to describe the motion along a three-dimensional curve. \textbf{B}(t) is found by taking the cross product of the unit tangent and unit normal vectors. The cross product operation ensures that the resulting binormal vector is orthogonal to both the tangent and normal vectors. The formula for obtaining \textbf{B}(t) is:
\[ B(t) = T(t) \times N(t) \]
Using the cross product gives a vector that not only is perpendicular to the osculating plane formed by \textbf{T}(t) and \textbf{N}(t) but also inherently has unit length because \textbf{T}(t) and \textbf{N}(t) are both unit vectors.

Cross Product

The cross product is a mathematical operation that returns a vector which is perpendicular to two given vectors in three-dimensional space. It's symbolized as \textbf{A} \times \textbf{B}, where \textbf{A} and \textbf{B} are vectors. This operation is crucial in finding the unit binormal vector, as shown in the solution for \textbf{B}(t).
Given two vectors \textbf{A} = \langle a_1, a_2, a_3 \rangle and \textbf{B} = \langle b_1, b_2, b_3 \rangle, their cross product can be computed via the determinant of a 3x3 matrix:
\[ \textbf{A} \times \textbf{B} = \begin{vmatrix} \textbf{i} & \textbf{j} & \textbf{k} \ a_1 & a_2 & a_3 \ b_1 & b_2 & b_3 \end{vmatrix} \]
The resulting vector will be orthogonally oriented to the plane containing \textbf{A} and \textbf{B}, and its magnitude will be equal to the area of the parallelogram formed by these vectors.

Vector Calculus

Vector calculus is an essential branch of mathematics, especially in physics and engineering, dealing with differentiation and integration of vector fields. It's key to understanding motion and changes in multidimensional space. In the context of a curve, as discussed in the provided problem, vector calculus is employed to find various properties of the curve, including curvature and torsion. These properties can characterize how a curve twists and bends in space.
The calculations of the unit tangent, normal, and binormal vectors, along with the cross product, are all fundamental operations within vector calculus. They demonstrate how vector calculus is used to transform a complex spatial problem into a more approachable form by breaking it down into comprehensible and solvable steps.