Question

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

Short answer

Answer: The unit binormal vector for the given curve is \(\mathbf{B}(t) = <-1, 0, 0>\) and the torsion is 0.

Step-by-step solution

Find the position vector

Given curve: $$\mathbf{r}(t) = \langle 12t, 5\cos t, 5\sin t\rangle$$

Differentiate the position vector

The derivative of the position vector with respect to time \(t\) is given by: $$\mathbf{v}(t) = \mathbf{r'}(t) = \left\langle \frac{d(12t)}{dt}, \frac{d(5\cos t)}{dt}, \frac{d(5\sin t)}{dt} \right\rangle$$ Which is: $$\mathbf{v}(t) = \langle 12, -5\sin t, 5\cos t\rangle$$

Find the unit tangent vector

The unit tangent vector \(\mathbf{T}(t)\) is defined as the normalized velocity vector: $$\mathbf{T}(t) = \frac{\mathbf{v}(t)}{||\mathbf{v}(t)||}$$ First, let's compute the magnitude of the velocity vector: $$||\mathbf{v}(t)|| = \sqrt{(12)^2 + (-5\sin t)^2 + (5\cos t)^2} = \sqrt{144 + 25\sin^2t + 25\cos^2t} = \sqrt{144 + 25(\sin^2t + \cos^2t)} = \sqrt{144 + 25} = 13$$ Now we can compute the unit tangent vector: $$\mathbf{T}(t) = \frac{1}{13}\langle 12, -5\sin t, 5\cos t\rangle$$

Differentiate the unit tangent vector

Now differentiate the unit tangent vector with respect to time \(t\): $$\mathbf{T}'(t) = \left\langle \frac{d(12/13)}{dt}, \frac{d(-5\sin t/13)}{dt}, \frac{d(5\cos t/13)}{dt} \right\rangle$$ Which is: $$\mathbf{T}'(t) = \langle 0, -\frac{5}{13}\cos t, -\frac{5}{13}\sin t \rangle$$

Find the unit binormal vector

The unit binormal vector \(\mathbf{B}(t)\) is defined as the cross product of the unit tangent vector \(\mathbf{T}(t)\) and the derivative of the unit tangent vector \(\mathbf{T}'(t)\), normalized: $$\mathbf{B}(t) = \frac{\mathbf{T}(t) \times \mathbf{T}'(t)}{||\mathbf{T}(t) \times \mathbf{T}'(t)||}$$ Compute the cross product of \(\mathbf{T}(t)\) and \(\mathbf{T}'(t)\): $$\mathbf{T}(t) \times \mathbf{T}'(t) = \left\langle \frac{12}{13}, -\frac{5\sin t}{13}, \frac{5\cos t}{13} \right\rangle \times \left\langle 0, -\frac{5\cos t}{13}, -\frac{5\sin t}{13} \right\rangle = \left\langle\frac{-25(\sin^2t+\cos^2t)}{13^2}, 0, 0\right\rangle = \left\langle\frac{-25}{169}, 0, 0\right\rangle$$ Now, let's find its magnitude: $$||\mathbf{T}(t) \times \mathbf{T'}(t)|| = \sqrt{\left(\frac{-25}{169}\right)^2+0^2+0^2} = \frac{25}{169}$$ So, the unit binormal vector \(\mathbf{B}(t)\) is: $$\mathbf{B}(t) = \frac{1}{\frac{25}{169}}\left\langle\frac{-25}{169}, 0, 0\right\rangle = <-1, 0, 0>$$

Find the torsion

Since \(\mathbf{B}(t)\) is not a function of \(t\), the torsion is 0. So, the unit binormal vector is \(\mathbf{B}(t) = <-1, 0, 0>\), and the torsion is 0.

Key concepts

Tangent Vector

In calculus and vector analysis, the tangent vector is a fundamental concept used to describe the direction in which a curve is heading at a particular point. For a given curve defined by a position vector \( \mathbf{r}(t) \), the tangent vector \( \mathbf{v}(t) \) is found by differentiating the position vector with respect to the parameter \( t \). This derivative gives us the velocity vector, \( \mathbf{v}(t) = \mathbf{r}'(t) \), that tells us how the position changes over time.

• The tangent vector shows the direction of the curve, pointing along the path of the curve.
• To find the unit tangent vector, which has a magnitude of 1, we normalize the tangent vector by dividing it by its magnitude.
• In the example, the tangent vector is \( \langle 12, -5\sin t, 5\cos t \rangle \).

By understanding the tangent vector, you get insight into the curvature and orientation of a path at any point along the curve.

Cross Product

The cross product is a mathematical operation on two vectors in three-dimensional space, resulting in another vector perpendicular to both of the original vectors. It is particularly useful in defining rotational aspects like angular momentum and torque. The cross product occurs in the context of binormal vectors, which are useful for understanding the geometry of curves.

• For two vectors \( \mathbf{A} \) and \( \mathbf{B} \), the cross product \( \mathbf{A} \times \mathbf{B} \) generates a vector orthogonal to both.
• In our solution, we find the unit binormal vector \( \mathbf{B}(t) \) by taking the cross product of the tangent vector \( \mathbf{T}(t) \) and its derivative \( \mathbf{T}'(t) \).
• The result of this cross product is \( \left\langle \frac{-25}{169}, 0, 0 \right\rangle \), which then needs normalization to form the unit binormal vector.

The cross product is central to determining the orientation of the normal plane to a given point on the curve.

Torsion

Torsion is a concept that describes how a curve twists out of its osculating plane as it moves along. While curvature measures how much a curve deviates from being a straight line, torsion measures the deviation from being planar.

• Torsion is calculated based on the derivatives of the tangent and binormal vectors.
• In our example, since the resulting binormal vector is constant over time, torsion turns out to be zero. This indicates that the curve itself lies in a plane and doesn’t twist.
• When torsion is zero, it implies the curve doesn't spiral out of its initial plane.

Understanding torsion allows you to fully grasp the spatial orientation and properties of complex curves beyond a simple world of planar geometries.

Position Vector

The position vector is a vector that represents the location of a point in space as a function of a parameter, typically time \( t \). It's foundational for describing the path of an object or a curve, giving us a concrete way to express motion or the geometry of a shape.

• The position vector is often written as \( \mathbf{r}(t) = \langle x(t), y(t), z(t) \rangle \), designating three-dimensional space coordinates.
• In the given solution, the position vector is \( \mathbf{r}(t) = \langle 12t, 5\cos t, 5\sin t \rangle \), providing a parametric description of the curve.
• Understanding how to derive the position vector and interpret its components is key to analyzing the nature of curves.

Once you grasp the position vector, it's easier to delve into more complex concepts, such as tangent vectors and curvature, using the framework it provides.