Problem 4 - Calculus Early Transcendentals: Pearson New International Edition

Question

sketch the curve over the indicated domain for \(t\). Find \(\mathbf{v}, \mathbf{a}, \mathbf{T}\), and \(\kappa\) at the point where \(t=t_{1} .\) $$ \mathbf{r}(t)=5 \cos t \mathbf{i}+2 t \mathbf{j}+5 \sin t \mathbf{k} ; \quad 0 \leq t \leq 4 \pi ; t_{1}=\pi $$

Step-by-Step Solution

Answer

\( \mathbf{v}(\pi) = 2\mathbf{j} - 5\mathbf{k}, \mathbf{a}(\pi) = 5\mathbf{i} \). \( \mathbf{T}(\pi) = \frac{2}{\sqrt{29}} \mathbf{j} - \frac{5}{\sqrt{29}} \mathbf{k}, \kappa = \frac{50}{29 \sqrt{29}} \).

Step 1: Find the Velocity Vector

To find the velocity vector \(\mathbf{v}(t)\), take the derivative of \(\mathbf{r}(t)\) with respect to \(t\): \(\mathbf{v}(t) = \frac{d}{dt}[5 \cos t \mathbf{i} + 2t \mathbf{j} + 5 \sin t \mathbf{k}] = -5 \sin t \mathbf{i} + 2 \mathbf{j} + 5 \cos t \mathbf{k}\). Substitute \(t=t_{1}=\pi\): \(\mathbf{v}(\pi) = -5 \sin \pi \mathbf{i} + 2 \mathbf{j} + 5\cos \pi \mathbf{k} = \mathbf{0i} + 2 \mathbf{j} - 5 \mathbf{k}\).

Step 2: Find the Acceleration Vector

To find the acceleration vector \(\mathbf{a}(t)\), take the derivative of the velocity vector \(\mathbf{v}(t)\) with respect to \(t\):\(\mathbf{a}(t) = \frac{d}{dt}[-5 \sin t \mathbf{i} + 2 \mathbf{j} + 5 \cos t \mathbf{k}] = -5 \cos t \mathbf{i} - 5 \sin t \mathbf{k}\). Substitute \(t=t_{1}=\pi\): \(\mathbf{a}(\pi) = -5 \cos \pi \mathbf{i} - 5 \sin \pi \mathbf{k} = 5 \mathbf{i} \).

Step 3: Compute the Unit Tangent Vector

The unit tangent vector \(\mathbf{T}(t)\) is obtained by normalizing the velocity vector \(\mathbf{v}(t)\). First, find the magnitude of \(\mathbf{v}(t)\): \(||\mathbf{v}(t)|| = \sqrt{(-5 \sin t)^2 + 2^2 + (5 \cos t)^2} = \sqrt{25 \sin^2 t + 4 + 25 \cos^2 t} = \sqrt{29}\). Therefore, \(\mathbf{T}(t) = \frac{\mathbf{v}(t)}{||\mathbf{v}(t)||} = \frac{-5 \sin t}{\sqrt{29}} \mathbf{i} + \frac{2}{\sqrt{29}} \mathbf{j} + \frac{5 \cos t}{\sqrt{29}} \mathbf{k}\). Substitute \(t = \pi\): \(\mathbf{T}(\pi) = \mathbf{0i} + \frac{2}{\sqrt{29}} \mathbf{j} - \frac{5}{\sqrt{29}} \mathbf{k}\).

Step 4: Compute the Curvature

The curvature \(\kappa\) of \(\mathbf{r}(t)\) is given by:\(\kappa = \frac{||\mathbf{v}(t) \times \mathbf{a}(t)||}{||\mathbf{v}(t)||^3}\). First, calculate \(\mathbf{v}(t) \times \mathbf{a}(t)\): \((-5 \sin t \mathbf{i} + 2 \mathbf{j} + 5 \cos t \mathbf{k}) \times (-5 \cos t \mathbf{i} - 5 \sin t \mathbf{k})\) which results in \((10 \sin t \mathbf{j} - 25 \sin^2 t \mathbf{k} - 25 \cos^2 t \mathbf{j} + 25 \sin^2 t \mathbf{i} + 25 \cos^2 t \mathbf{i})\). At \(t=\pi\), this simplifies to \(-50 \mathbf{j}\). Hence, \(\kappa = \frac{|-50|}{(\sqrt{29})^3} = \frac{50}{29 \sqrt{29}}\).

Key Concepts

Velocity Vector

In vector calculus, the velocity vector is a fundamental component that describes the rate of change of position with respect to time. For a differentiable curve represented by a position vector function \( \mathbf{r}(t) \), the velocity vector \( \mathbf{v}(t) \) is derived by taking the derivative of \( \mathbf{r}(t) \) with respect to \( t \). It provides insight into the dynamic behavior of the curve at any given point.In our specific example with \( \mathbf{r}(t)=5 \cos t \mathbf{i}+2 t \mathbf{j}+5 \sin t \mathbf{k} \), calculating the derivative yields the velocity vector:

• \( \mathbf{v}(t) = -5 \sin t \mathbf{i} + 2 \mathbf{j} + 5 \cos t \mathbf{k} \)

At \( t = \pi \), substitute \( t \) into the velocity expression to find:

• \( \mathbf{v}(\pi) = 0 \mathbf{i} + 2 \mathbf{j} - 5 \mathbf{k} \)

This tells us that at \( t = \pi \), the object is moving in a direction parallel to the vector \( 0 \mathbf{i} + 2 \mathbf{j} - 5 \mathbf{k} \) with no motion in the \( \mathbf{i} \)-axis.

Acceleration Vector

The acceleration vector describes how fast the velocity vector is changing over time. It gives insights into the forces acting on the object moving along the curve.To obtain the acceleration vector \( \mathbf{a}(t) \), differentiate the velocity vector \( \mathbf{v}(t) \) with respect to \( t \). Using our given velocity vector, we find:

• \( \mathbf{a}(t) = \frac{d}{dt}[-5 \sin t \mathbf{i} + 2 \mathbf{j} + 5 \cos t \mathbf{k}] = -5 \cos t \mathbf{i} - 5 \sin t \mathbf{k} \)

Substitute \( t = \pi \) to find the specific acceleration at this point:

• \( \mathbf{a}(\pi) = 5 \mathbf{i} \)

This result shows that at \( t = \pi \), the acceleration is directed solely along the \( \mathbf{i} \)-axis, indicating a push in the positive \( x \)-direction.

Unit Tangent Vector

The unit tangent vector \( \mathbf{T}(t) \) is useful for understanding the direction in which the curve is heading without considering speed. It is calculated by normalizing the velocity vector, which means dividing the velocity vector by its magnitude.To find \( \mathbf{T}(t) \), first compute the magnitude of \( \mathbf{v}(t) \):

• \( ||\mathbf{v}(t)|| = \sqrt{(-5 \sin t)^2 + 2^2 + (5 \cos t)^2} = \sqrt{25 \sin^2 t + 4 + 25 \cos^2 t} = \sqrt{29} \)

Hence, the unit tangent vector is:

• \( \mathbf{T}(t) = \frac{-5 \sin t}{\sqrt{29}} \mathbf{i} + \frac{2}{\sqrt{29}} \mathbf{j} + \frac{5 \cos t}{\sqrt{29}} \mathbf{k} \)

At \( t = \pi \), this simplifies to:

• \( \mathbf{T}(\pi) = 0 \mathbf{i} + \frac{2}{\sqrt{29}} \mathbf{j} - \frac{5}{\sqrt{29}} \mathbf{k} \)

This vector maintains the direction of the velocity, but with a fixed unit length of 1.

Curvature

Curvature is a measure of how sharply a curve bends at a particular point, quantified by the symbol \( \kappa \). A high curvature indicates a tight bend, while a low curvature corresponds to a gentle bend.The formula for curvature \( \kappa \) is:

• \( \kappa = \frac{||\mathbf{v}(t) \times \mathbf{a}(t)||}{||\mathbf{v}(t)||^3} \)

Calculate the cross-product \( \mathbf{v}(t) \times \mathbf{a}(t) \) for our functions:

• \((-5 \sin t \mathbf{i} + 2 \mathbf{j} + 5 \cos t \mathbf{k}) \times (-5 \cos t \mathbf{i} - 5 \sin t \mathbf{k})\)
• Results in: \(10 \sin t \mathbf{j} - 25 \sin^2 t \mathbf{k} - 25 \cos^2 t \mathbf{j} + 25 \sin^2 t \mathbf{i} + 25 \cos^2 t \mathbf{i}\)

At \( t = \pi \), the computation simplifies to:

• \(-50 \mathbf{j}\)

Thus, the curvature is given by:

• \( \kappa = \frac{50}{29 \sqrt{29}} \)

This value indicates the bending rate of the curve at \( t = \pi \). A specific value of curvature helps understand the nature of the curve's deviation at that point.