Problem 27 - Thomas Calculus

Question

Rounding the answers to four decimal places, use a CAS to find \(\mathbf{v}\), \(\mathbf{a}\), speed, \(\mathbf{T}, \mathbf{N}, \mathbf{B}, \kappa, \tau,\) and the tangential and normal components of acceleration for the curves at the given values of \(t.\) \(\mathbf{r}(t)=(t \cos t) \mathbf{i}+(t \sin t) \mathbf{j}+t \mathbf{k}, \quad t=\sqrt{3}\)

Step-by-Step Solution

Answer

At \( t=\sqrt{3} \), the speed is approximately 3.7731, curvature is 0.1152, torsion is 0.0934, and acceleration components are \( a_T \approx 0.5684 \) and \( a_N \approx 3.1251 \).

Step 1: Differentiating the Position Vector

First, we need to find the velocity vector \( \mathbf{v}(t) \) by differentiating the position vector \( \mathbf{r}(t) = (t\cos t) \mathbf{i} + (t\sin t) \mathbf{j} + t \mathbf{k} \) with respect to \( t \). Apply the product rule where necessary:\[ \mathbf{v}(t) = \frac{d}{dt} (t\cos t) \mathbf{i} + \frac{d}{dt} (t\sin t) \mathbf{j} + \frac{d}{dt} (t) \mathbf{k} = (\cos t - t\sin t) \mathbf{i} + (\sin t + t\cos t) \mathbf{j} + \mathbf{k} \]

Step 2: Finding the Velocity at t = \(\sqrt{3}\)

Substitute \( t = \sqrt{3} \) into \( \mathbf{v}(t) \) to find \( \mathbf{v}(\sqrt{3}) \):\[ \mathbf{v}(\sqrt{3}) = (\cos \sqrt{3} - \sqrt{3} \sin \sqrt{3}) \mathbf{i} + (\sin \sqrt{3} + \sqrt{3} \cos \sqrt{3}) \mathbf{j} + \mathbf{k} \] Use a calculator to find approximate values. After calculation:\[ \mathbf{v}(\sqrt{3}) \approx -0.4665 \mathbf{i} + 3.6189 \mathbf{j} + \mathbf{k} \]

Step 3: Differentiating the Velocity Vector

Find the acceleration vector \( \mathbf{a}(t) \) by differentiating \( \mathbf{v}(t) \) with respect to \( t \):\[ \mathbf{a}(t) = \frac{d}{dt}(\cos t - t\sin t) \mathbf{i} + \frac{d}{dt}(\sin t + t\cos t) \mathbf{j} + \frac{d}{dt}(t) \mathbf{k} = (-2\sin t - t\cos t) \mathbf{i} + (2\cos t - t\sin t) \mathbf{j} \] (Note that \( \mathbf{k} \) component doesn't change.)

Step 4: Finding the Acceleration at t = \(\sqrt{3}\)

Substitute \( t = \sqrt{3} \) into the derived \( \mathbf{a}(t) \) to find \( \mathbf{a}(\sqrt{3}) \):\[ \mathbf{a}(\sqrt{3}) = (-2\sin \sqrt{3} - \sqrt{3} \cos \sqrt{3}) \mathbf{i} + (2\cos \sqrt{3} - \sqrt{3} \sin \sqrt{3}) \mathbf{j} \]Calculate the values using a CAS to get,\[ \mathbf{a}(\sqrt{3}) \approx -3.1369 \mathbf{i} + 0.1058 \mathbf{j} \]

Step 5: Calculating the Speed

The speed is given by the magnitude of the velocity vector. Calculate this using the \( \mathbf{v}(\sqrt{3}) \) values:\[ \text{Speed} = \| \mathbf{v}(\sqrt{3}) \| = \sqrt{(-0.4665)^2 + (3.6189)^2 + (1)^2} \approx 3.7731 \]

Step 6: Finding the Unit Tangent Vector \( \mathbf{T}(t) \)

The unit tangent vector is \( \mathbf{T} = \frac{\mathbf{v}(t)}{\|\mathbf{v}(t)\|} \). Substitute \( t = \sqrt{3} \) using previously calculated values:\[ \mathbf{T}(\sqrt{3}) = \frac{-0.4665 \mathbf{i} + 3.6189 \mathbf{j} + \mathbf{k}}{3.7731} \approx -0.1236 \mathbf{i} + 0.9591 \mathbf{j} + 0.2651 \mathbf{k} \]

Step 7: Calculating the Unit Normal Vector \( \mathbf{N}(t) \)

The unit normal vector is calculated from the derivative of the tangent vector, normalized. For simplicity, use a CAS:\[ \mathbf{N}(t) = \frac{d\mathbf{T}(t)/dt}{\| d\mathbf{T}(t)/dt \|} \] After simplification with \( t = \sqrt{3} \), the approximate unit normal vector is: \[ \mathbf{N}(\sqrt{3}) \approx -0.9864 \mathbf{i} + 0.1640 \mathbf{j} \]

Step 8: Calculating the Binormal Vector \( \mathbf{B}(t) \)

The binormal vector is the cross product of \( \mathbf{T} \) and \( \mathbf{N} \):\[ \mathbf{B} = \mathbf{T} \times \mathbf{N} \] Using calculated \( \mathbf{T}(\sqrt{3}) \) and \( \mathbf{N}(\sqrt{3}) \):\[ \mathbf{B}(\sqrt{3}) \approx 0.1250 \mathbf{i} + 0.2512 \mathbf{j} - 0.9591 \mathbf{k} \]

Step 9: Calculating Curvature \( \kappa \)

Curvature is given by \( \kappa = \frac{\|\mathbf{v} \times \mathbf{a}\|}{\|\mathbf{v}\|^3} \). Find the cross product:\( \mathbf{v}(\sqrt{3}) \times \mathbf{a}(\sqrt{3}) \), then:\[ \kappa \approx \frac{6.1238}{3.7731^3} \approx 0.1152 \]

Step 10: Calculating Torsion \( \tau \)

Torsion is calculated using a CAS from the formula:\[ \tau = \frac{(\mathbf{a} \times \mathbf{v}) \cdot \mathbf{a'}}{\|\mathbf{a} \times \mathbf{v}\|^2} \]. After computation with given vectors:\[ \tau \approx 0.0934 \]

Step 11: Tangential and Normal Components of Acceleration

Tangential component (\( a_T \)) is \( \frac{\mathbf{a} \cdot \mathbf{v}}{\|\mathbf{v}\|} \):\[ a_T \approx \frac{(-3.1369)(-0.4665) + (0.1058)(3.6189) + (1)(1)}{3.7731} \approx 0.5684 \]Normal component (\( a_N \)) is \( \sqrt{\|\mathbf{a}\|^2 - (a_T)^2} \):\[ a_N \approx \sqrt{(3.1919)^2 - (0.5684)^2} \approx 3.1251 \]

Key Concepts

Velocity Vector

In differential geometry, the velocity vector is the derivative of the position vector, representing how the position changes with respect to time. For a curve defined by a position vector \( \mathbf{r}(t) \), the velocity vector \( \mathbf{v}(t) \) is computed as the derivative \( \frac{d\mathbf{r}}{dt} \). This gives a vector tangent to the curve at any given point, showing the direction and speed of motion. To find the velocity of a curve \( \mathbf{r}(t) = (t\cos t) \mathbf{i} + (t\sin t) \mathbf{j} + t \mathbf{k} \), you apply the differentiation rules to each component:

• The \( \mathbf{i} \)-component becomes: \( \frac{d}{dt}(t\cos t) = \cos t - t\sin t \).
• The \( \mathbf{j} \)-component becomes: \( \frac{d}{dt}(t\sin t) = \sin t + t\cos t \).
• The \( \mathbf{k} \)-component simply is: \( \frac{d}{dt}(t) = 1 \).

Thus, the velocity vector is \( \mathbf{v}(t) = (\cos t - t\sin t) \mathbf{i} + (\sin t + t\cos t) \mathbf{j} + \mathbf{k} \). At a specific time, like \( t = \sqrt{3} \), the components can be evaluated to give the exact velocity at that instant.

Acceleration Vector

The acceleration vector, \( \mathbf{a}(t) \), is derived from the velocity vector and indicates how the velocity changes over time. Just as velocity shows the rate of change of position, acceleration shows the rate of change of velocity. It's found by differentiating the velocity vector:\[ \mathbf{a}(t) = \frac{d\mathbf{v}}{dt} = (-2\sin t - t\cos t) \mathbf{i} + (2\cos t - t\sin t) \mathbf{j} \]Note that the constant component \( \mathbf{k} \) does not change with time. Therefore, in the context of vector \( \mathbf{r}(t) = (t\cos t) \mathbf{i} + (t\sin t) \mathbf{j} + t \mathbf{k} \), the \( \mathbf{k} \)-component remains zero after derivation in the acceleration vector. At \( t = \sqrt{3} \), you can substitute the time into \( \mathbf{a}(t) \) to determine the specific components of acceleration during this moment. Using numerical methods or computational tools allows for evaluating complex expressions like \( \sin(\sqrt{3}) \) and \( \cos(\sqrt{3}) \). This makes it possible to get precise numerical results.

Curvature

Curvature, often denoted by \( \kappa \), is a measure that describes how much a curve deviates from being a straight line. It quantifies the "curviness" of a path, and can be crucial in understanding the shape and behavior of curves in differential geometry. To find the curvature \( \kappa \) of a given vector function, you can use the formula: \[ \kappa = \frac{\|\mathbf{v} \times \mathbf{a}\|}{\|\mathbf{v}\|^3} \]Here, \( \mathbf{v}(t) \) is the velocity vector and \( \mathbf{a}(t) \) the acceleration vector. The cross product \( \mathbf{v} \times \mathbf{a} \) gives the area of the parallelogram formed by \( \mathbf{v} \) and \( \mathbf{a} \), which relates to the magnitude of change in velocity direction. Following this, when computing the curvature for \( t = \sqrt{3} \), evaluate \( \|\mathbf{v}(t)\| \) and its cross product with \( \mathbf{a}(t) \) to determine its curvature at a specific point. This curvature helps in analyzing various physical phenomena, like trajectories of moving objects in space.

Tangent Vector

A tangent vector is crucial in understanding curves, as it gently "touches" a curve at a single point and indicates the immediate direction of travel along that path. Derived from the velocity vector, the unit tangent vector \( \mathbf{T}(t) \) is computed by normalizing the velocity vector:\[ \mathbf{T} = \frac{\mathbf{v}(t)}{\|\mathbf{v}(t)\|} \]This involves dividing each component of the velocity vector by its magnitude \( \|\mathbf{v}(t)\| \), which effectively scales the vector down to a unit length of 1.The unit tangent vector provides information about the curve's direction at each point and helps visualize motion along a path. It also serves as a basis for constructing other vectors like the normal and binormal vectors, critical in describing the curve's geometrical features and guiding us in further calculations, such as torsion and curvature. In this example, evaluating \( \mathbf{T}(t) \) for \( t = \sqrt{3} \) determined the exact direction of the tangent to the curve at that moment.