Problem 27 - Thomas Calculus

Question

Rounding the answers to four decimal places, use a CAS to find v, a, speed, \(\mathbf{T}, \mathbf{N}, \mathbf{B}, \kappa, \tau,\) and the tangential and normal components of acceleration for the curves in Exercises \(27-30\) 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 \(t=\sqrt{3}\), calculate each vector and scalar quantity by evaluating the expressions derived from the derivative operations and vector magnitudes.

Step 1: Find the velocity vector \(\mathbf{v}(t)\)

Differentiate \(\mathbf{r}(t)=(t \cos t) \mathbf{i}+(t \sin t) \mathbf{j}+t \mathbf{k}\) with respect to \(t\). This gives \(\mathbf{v}(t)=(\cos t - t\sin t)\mathbf{i} + (\sin t + t\cos t)\mathbf{j} + \mathbf{k}\).

Step 2: Evaluate \(\mathbf{v}(t)\) at \(t=\sqrt{3}\)

Substitute \(t=\sqrt{3}\) into \(\mathbf{v}(t)=(\cos t - t\sin t)\mathbf{i} + (\sin t + t\cos t)\mathbf{j} + \mathbf{k}\) to find the velocity vector at that time. Compute \(\mathbf{v}(\sqrt{3})\).

Step 3: Calculate the acceleration vector \(\mathbf{a}(t)\)

Differentiate the velocity vector \(\mathbf{v}(t)=(\cos t - t\sin t)\mathbf{i} + (\sin t + t\cos t)\mathbf{j} + \mathbf{k}\) with respect to \(t\) to get \(\mathbf{a}(t)=-(2\sin t + t\cos t)\mathbf{i} + (2\cos t - t\sin t)\mathbf{j}\).

Step 4: Evaluate \(\mathbf{a}(t)\) at \(t=\sqrt{3}\)

Substitute \(t=\sqrt{3}\) into \(\mathbf{a}(t)=-(2\sin t + t\cos t)\mathbf{i} + (2\cos t - t\sin t)\mathbf{j}\) to find the acceleration vector at that time. Compute \(\mathbf{a}(\sqrt{3})\).

Step 5: Calculate the speed \(v(t)\)

Find the magnitude of the velocity vector: \(v(t)=\|\mathbf{v}(t)\|=\sqrt{(\cos t - t\sin t)^2 + (\sin t + t\cos t)^2 + 1}\).

Step 6: Evaluate speed at \(t=\sqrt{3}\)

Substitute \(t=\sqrt{3}\) into the speed formula and evaluate. Compute \(v(\sqrt{3})\).

Step 7: Find the unit tangent vector \(\mathbf{T}(t)\)

Divide the velocity vector by the speed: \(\mathbf{T}(t)=\frac{\mathbf{v}(t)}{v(t)}\).

Step 8: Evaluate \(\mathbf{T}(t)\) at \(t=\sqrt{3}\)

Calculate \(\mathbf{T}(t)\) using the velocity vector \(\mathbf{v}(\sqrt{3})\) and the speed \(v(\sqrt{3})\).

Step 9: Find the principal normal vector \(\mathbf{N}(t)\)

Differentiate \(\mathbf{T}(t)\) with respect to \(t\), and then normalize the resultant vector to find \(\mathbf{N}(t)\).

Step 10: Evaluate \(\mathbf{N}(t)\) at \(t=\sqrt{3}\)

Use the expression for \(\mathbf{T}(t)\) to find \(\mathbf{N}(t)\) at \(t=\sqrt{3}\).

Step 11: Find the binormal vector \(\mathbf{B}(t)\)

Use the cross product of \(\mathbf{T}(t)\) and \(\mathbf{N}(t)\): \(\mathbf{B}(t)=\mathbf{T}(t) \times \mathbf{N}(t)\).

Step 12: Evaluate \(\mathbf{B}(t)\) at \(t=\sqrt{3}\)

Calculate \(\mathbf{B}(t)\) using the previously found \(\mathbf{T}(\sqrt{3})\) and \(\mathbf{N}(\sqrt{3})\).

Step 13: Calculate curvature \(\kappa\)

Use the formula \(\kappa=\frac{\|\mathbf{a}(t) \times \mathbf{v}(t)\|}{v(t)^3}\).

Step 14: Evaluate \(\kappa\) at \(t=\sqrt{3}\)

Substitute the evaluated vectors into the curvature formula and compute \(\kappa(\sqrt{3})\).

Step 15: Calculate torsion \(\tau\)

Use the formula \(\tau=-\frac{(\mathbf{a}(t) \cdot \mathbf{B}(t))}{\|\mathbf{a}(t) \times \mathbf{v}(t)\|}\).

Step 16: Evaluate \(\tau\) at \(t=\sqrt{3}\)

Plug in the values and evaluate the expression for torsion at \(t=\sqrt{3}\).

Step 17: Calculate tangential and normal components of acceleration

The tangential component is given by \(a_T=\mathbf{a}(t) \cdot \mathbf{T}(t)\), and the normal component by \(a_N=\|\mathbf{a}(t) \times \mathbf{T}(t)\|\).

Step 18: Evaluate components at \(t=\sqrt{3}\)

Compute both the tangential and normal components of the acceleration using the expressions from Step 17 and substitute \(t=\sqrt{3}\).

Key Concepts

Velocity Vector

In vector calculus, the velocity vector provides information about the direction and speed of a particle along a path. When working with vector-valued functions, such as \( \mathbf{r}(t) = (t \cos t) \mathbf{i} + (t \sin t) \mathbf{j} + t \mathbf{k} \), the velocity vector is obtained by differentiating this position vector with respect to \( t \). For our exercise, by differentiating the given function, we find that the velocity vector becomes \( \mathbf{v}(t) = (\cos t - t\sin t)\mathbf{i} + (\sin t + t\cos t)\mathbf{j} + \mathbf{k} \). The velocity vector is a key concept as it tells us how fast the particle's position is changing with time. Plugging in specific values of \( t \), such as \( t=\sqrt{3} \), gives us the velocity at that particular instant, allowing us to understand the motion's dynamics.

Acceleration Vector

The acceleration vector is crucial in understanding how the velocity of the particle changes over time. This is found by differentiating the velocity vector with respect to time \( t \). In our example, the acceleration vector can be obtained by differentiating the previously found velocity vector, yielding \( \mathbf{a}(t) = -(2\sin t + t\cos t)\mathbf{i} + (2\cos t - t\sin t)\mathbf{j} \). The acceleration vector gives us insights into changes in speed and direction of the particle. Evaluating this at a specific point, such as \( t=\sqrt{3} \), enables us to determine how acceleration is contributing to the trajectory of the particle at that time.

Curvature and Torsion

Curvature and torsion are measures that describe the bending and twisting of a path in space. Curvature \( \kappa \) indicates how quickly a curve changes direction and is calculated using the formula: \[ \kappa = \frac{\|\mathbf{a}(t) \times \mathbf{v}(t)\|}{v(t)^3} \] This captures how sharply the curve bends.Torsion \( \tau \) represents the rate of twist of the curve around its principal normal vector, calculated by: \[ \tau = -\frac{(\mathbf{a}(t) \cdot \mathbf{B}(t))}{\|\mathbf{a}(t) \times \mathbf{v}(t)\|} \] Torsion provides an understanding of the spatial aspect of the curve.By categorizing these attributes at \( t=\sqrt{3} \), we gain a full understanding of the path's spatial dynamics. Curvature and torsion together help in visualizing the three-dimensional behavior of the trajectory.

Tangential and Normal Components of Acceleration

Acceleration in vector calculus can be broken down into tangential and normal components. These components give insight into how the total acceleration vector affects the speed and the direction of the velocity vector. The tangential component of acceleration \( a_T \) is the projection of the acceleration vector along the velocity vector, indicating how it affects the speed. It can be found as: \[ a_T = \mathbf{a}(t) \cdot \mathbf{T}(t) \] where \( \mathbf{T}(t) \) is the unit tangent vector.The normal component \( a_N \) is perpendicular to the velocity and indicates how acceleration changes the direction of motion. It is calculated as: \[ a_N = \|\mathbf{a}(t) \times \mathbf{T}(t)\| \]By evaluating these components at specific values of \( t \), such as \( t=\sqrt{3} \), we can see the distinct influences on the speed and directional change of the particle at that moment.