Problem 12 - Thomas Calculus

Question

Find \(\mathbf{T}, \mathbf{N},\) and \(\kappa\) for the space curves $$ \mathbf{r}(t)=(6 \sin 2 t) \mathbf{i}+(6 \cos 2 t) \mathbf{j}+5 t \mathbf{k} $$

Step-by-Step Solution

Answer

\( \mathbf{T}(t) = \frac{12 \cos 2t}{13}\mathbf{i} - \frac{12 \sin 2t}{13}\mathbf{j} + \frac{5}{13}\mathbf{k} \), \( \mathbf{N}(t) = -\sin 2t \mathbf{i} - \cos 2t \mathbf{j} \), and \( \kappa = \frac{24}{169} \).

Step 1: Compute r'(t)

First, find the derivative of the given curve with respect to t. The curve is \(\mathbf{r}(t) = (6 \sin 2t) \mathbf{i} + (6 \cos 2t) \mathbf{j} + 5t \mathbf{k}\).Calculate the derivative: \[ \mathbf{r}'(t) = \frac{d}{dt}((6 \sin 2t) \mathbf{i} + (6 \cos 2t) \mathbf{j} + 5t \mathbf{k}) = (12 \cos 2t) \mathbf{i} - (12 \sin 2t) \mathbf{j} + 5 \mathbf{k} \]

Step 2: Calculate |r'(t)|

Next, find the magnitude of \( \mathbf{r}'(t) \). This is needed to normalize the vector.Calculate the magnitude:\[ |\mathbf{r}'(t)| = \sqrt{(12 \cos 2t)^2 + (-12 \sin 2t)^2 + 5^2} = \sqrt{144 \cos^2 2t + 144 \sin^2 2t + 25} = \sqrt{169} = 13 \]

Step 3: Calculate T(t)

The unit tangent vector \( \mathbf{T}(t) \) is \[ \mathbf{T}(t) = \frac{\mathbf{r}'(t)}{|\mathbf{r}'(t)|} \]Substitute \( \mathbf{r}'(t) \) and \( |\mathbf{r}'(t)| \) to get:\[ \mathbf{T}(t) = \frac{(12 \cos 2t) \mathbf{i} - (12 \sin 2t) \mathbf{j} + 5 \mathbf{k}}{13} = \frac{12 \cos 2t}{13}\mathbf{i} - \frac{12 \sin 2t}{13}\mathbf{j} + \frac{5}{13}\mathbf{k} \]

Step 4: Compute T'(t)

Differentiate \( \mathbf{T}(t) \) to find \( \mathbf{T}'(t) \).\[ \mathbf{T}'(t) = \frac{d}{dt}\left(\frac{12 \cos 2t}{13}\mathbf{i} - \frac{12 \sin 2t}{13}\mathbf{j} + \frac{5}{13}\mathbf{k}\right) = \left(-\frac{24 \sin 2t}{13}\right)\mathbf{i} - \left(\frac{24 \cos 2t}{13}\right)\mathbf{j} \]

Step 5: Calculate |T'(t)|

Find the magnitude of \( \mathbf{T}'(t) \).\[ |\mathbf{T}'(t)| = \sqrt{\left(-\frac{24 \sin 2t}{13}\right)^2 + \left(-\frac{24 \cos 2t}{13}\right)^2} = \frac{24}{13} \]

Step 6: Calculate N(t)

The unit normal vector \( \mathbf{N}(t) \) is\[ \mathbf{N}(t) = \frac{\mathbf{T}'(t)}{|\mathbf{T}'(t)|} \]Substitute \( \mathbf{T}'(t) \) and \( |\mathbf{T}'(t)| \) to get:\[ \mathbf{N}(t) = \frac{-24 \sin 2t}{24}\mathbf{i} - \frac{24 \cos 2t}{24}\mathbf{j} \]Simplifying:\[ \mathbf{N}(t) = -\sin 2t \mathbf{i} - \cos 2t \mathbf{j} \]

Step 7: Calculate Curvature \(\kappa \)

The curvature \( \kappa (t) \) is given by \[ \kappa (t) = \frac{|\mathbf{T}'(t)|}{|\mathbf{r}'(t)|} \]Substitute the magnitudes:\[ \kappa (t) = \frac{\frac{24}{13}}{13} = \frac{24}{169} \]

Key Concepts

Unit Tangent Vector

The unit tangent vector, denoted as \( \mathbf{T}(t) \), is a key component in understanding the geometry of a curve in space. It represents the direction in which the curve is heading at any given point and has a length of 1, making it a 'unit' vector. To find \( \mathbf{T}(t) \), you first compute the derivative of the position vector \( \mathbf{r}(t) \), which gives the velocity vector \( \mathbf{r}'(t) \). The velocity vector indicates not just the direction but also the speed at which you're moving along the curve.

However, we're solely interested in the direction, hence the need to "unitize" the vector. This is done by dividing \( \mathbf{r}'(t) \) by its magnitude \( |\mathbf{r}'(t)| \). This process normalizes the vector to obtain \( \mathbf{T}(t) = \frac{\mathbf{r}'(t)}{|\mathbf{r}'(t)|} \).

Through this, \( \mathbf{T}(t) \) provides an insightful visualization of the movement along the curve, acting as a compass pointing out the curve's path without concern for speed.

Unit Normal Vector

Once we have the unit tangent vector, the next step is determining the unit normal vector \( \mathbf{N}(t) \). This vector is perpendicular to the unit tangent vector and points towards the curve's center of curvature, essentially pointing in the direction that the curve is "turning."

To find \( \mathbf{N}(t) \), differentiate the unit tangent vector \( \mathbf{T}(t) \) to get \( \mathbf{T}'(t) \), which shows how \( \mathbf{T}(t) \) is changing. A nonzero \( \mathbf{T}'(t) \) indicates a change in direction along the curve, hence the curvature.

The magnitude of this change, \( | \mathbf{T}'(t) | \), is used to normalize \( \mathbf{T}'(t) \), resulting in \( \mathbf{N}(t) = \frac{\mathbf{T}'(t)}{| \mathbf{T}'(t) |} \). This indicates the sharpness of the curve and provides another layer of understanding of the curve's geometric properties.

Curvature

Curvature \( \kappa(t) \) is a fundamental geometric property that quantifies how sharply a space curve bends at a given point. A high curvature value indicates a sharp turn, while a low curvature suggests a gentle bend.

To calculate curvature, we rely on the previously determined vectors. Using the formula \( \kappa(t) = \frac{|\mathbf{T}'(t)|}{|\mathbf{r}'(t)|} \), we relate the rate at which the direction of the tangent is changing \( |\mathbf{T}'(t)| \) to the speed along the curve \( |\mathbf{r}'(t)| \). This helps adjust the curvature value for how fast you are moving along the curve, depicting a more accurate measure of how much the curve is bending.

Understanding curvature is crucial as it provides insights into the behavior of the path in various fields such as physics, engineering, and computer graphics, enabling us to predict and modify how objects travel along curves.