Problem 6 - 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 \(t=t_{1}\). \(\mathbf{r}(t)=\frac{t^{2}}{4} \mathbf{i}+2 \cos t \mathbf{j}+2 \sin t \mathbf{k} ; \quad 0 \leq t \leq 4 \pi ; t_{1}=\pi\)

Step-by-Step Solution

Answer

Calculate \( \mathbf{v}(\pi) \), \( \mathbf{a}(\pi) \), \( \mathbf{T}(\pi) \), and \( \kappa(\pi) \). Sketch shows a spiral opening along \( x \).

Step 1: Sketch the Curve

To sketch the curve, analyze the vector function \( \mathbf{r}(t) = \frac{t^2}{4} \mathbf{i} + 2 \cos t \mathbf{j} + 2 \sin t \mathbf{k} \). This function describes a parametric curve in 3D. The component \( \frac{t^2}{4} \) indicates a parabolic shape along the x-axis and keeps increasing as \( t \) increases. The components \( 2 \cos t \) and \( 2 \sin t \) describe a circle in the plane perpendicular to the x-axis, with radius 2.

Step 2: Calculate the Velocity Vector \( \mathbf{v}(t) \)

The velocity vector \( \mathbf{v}(t) \) is the derivative of the position vector \( \mathbf{r}(t) \). Compute: \[ \mathbf{v}(t) = \frac{d}{dt}\left( \frac{t^2}{4} \mathbf{i} + 2 \cos t \mathbf{j} + 2 \sin t \mathbf{k} \right) = \frac{t}{2} \mathbf{i} - 2 \sin t \mathbf{j} + 2 \cos t \mathbf{k} \].

Step 3: Calculate the Acceleration Vector \( \mathbf{a}(t) \)

The acceleration vector \( \mathbf{a}(t) \) is obtained by differentiating the velocity vector \( \mathbf{v}(t) \): \[ \mathbf{a}(t) = \frac{d}{dt}\left( \frac{t}{2} \mathbf{i} - 2 \sin t \mathbf{j} + 2 \cos t \mathbf{k} \right) = \frac{1}{2} \mathbf{i} - 2 \cos t \mathbf{j} - 2 \sin t \mathbf{k} \].

Step 4: Evaluate the Tangent Vector \( \mathbf{T}(t) \) at \( t_1 = \pi \)

The unit tangent vector \( \mathbf{T}(t) \) is taken by normalizing \( \mathbf{v}(t) \). At \( t_1 = \pi \), \[ \mathbf{v}(\pi) = \frac{\pi}{2} \mathbf{i} + 0 \mathbf{j} - 2 \mathbf{k} \]Calculate the magnitude: \[ ||\mathbf{v}(\pi)|| = \sqrt{\left(\frac{\pi}{2}\right)^2 + 0 + (-2)^2} \]Normalize \( \mathbf{v}(\pi) \):\[ \mathbf{T}(\pi) = \frac{1}{||\mathbf{v}(\pi)||} \mathbf{v}(\pi) \].

Step 5: Find the Curvature \( \kappa(t) \) and Evaluate at \( t_1 = \pi \)

Curvature \( \kappa(t) \) is calculated as the magnitude of the derivative of the unit tangent vector with respect to arc length \( s \): \[ \kappa(t) = \frac{|| \mathbf{T}'(t) ||}{|| \mathbf{v}(t) ||} \]Compute \( \mathbf{T}'(t) \) and evaluate \( \kappa \) at \( t_1 = \pi \).

Key Concepts

Parametric Curves

Parametric curves are created when multiple variables define the coordinates of points on a curve, using another set of variable(s), typically denoted as parameter(s). In many cases, the curve is defined with respect to time, like our vector function \( \mathbf{r}(t) \). The idea here is to express each coordinate (x, y, z) as a function of one or several parameters.
In our scenario, the parameter is \( t \) and the expression for the parametric curve is \( \mathbf{r}(t) = \frac{t^2}{4} \mathbf{i} + 2 \cos t \mathbf{j} + 2 \sin t \mathbf{k} \). This provides a curve in three dimensions:

• \( \frac{t^2}{4} \): The x-coordinate forms a parabolic shape.
• \( 2 \cos t \text{ and } 2 \sin t \): These describe circular motion in the yz-plane, with radius 2.

By varying \( t \), we can trace the complete trajectory of the curve in 3D space, illustrating the motion pattern of the given vector function.

Tangent Vector

The tangent vector at any point on a parametric curve like \( \mathbf{r}(t) \) provides a direction in which the curve proceeds at that point. It is essentially the derivative of the position vector with respect to the parameter \( t \). To find the tangent vector \( \mathbf{v}(t) \), we differentiate each component of \( \mathbf{r}(t) \):

• The derivative of \( \frac{t^2}{4} \) is \( \frac{t}{2} \).
• The derivative of \( 2 \cos t \) is \( -2 \sin t \).
• The derivative of \( 2 \sin t \) is \( 2 \cos t \).

Thus, \( \mathbf{v}(t) = \frac{t}{2} \mathbf{i} - 2 \sin t \mathbf{j} + 2 \cos t \mathbf{k} \), which gives the direction of the curve at any value of \( t \). This tangent vector is crucial for understanding the motion along the curve, as it tells us both the speed and direction of travel.

Velocity and Acceleration

To understand how an object moves along a parametric curve, we look at velocity and acceleration vectors. The velocity vector \( \mathbf{v}(t) \) tells us the instantaneous rate of change of position, which involves both speed and direction. Once you've computed \( \mathbf{v}(t) \), find acceleration \( \mathbf{a}(t) \) by differentiating \( \mathbf{v}(t) \). So,

• \( \mathbf{a}(t) = \frac{1}{2} \mathbf{i} - 2 \cos t \mathbf{j} - 2 \sin t \mathbf{k} \).

It's like imagining how a car's speedometer is moving up or down. The acceleration vector indicates how the speed is changing as you follow the curve. While velocity gives the first derivative of position, acceleration is the next step—it’s the derivative of velocity.
At \( t_1 = \pi \), plug \( t_1 \) into \( \mathbf{v}(t) \) and \( \mathbf{a}(t) \) to get the velocity and acceleration at that specific point:

• Velocity: \( \mathbf{v}(\pi) = \frac{\pi}{2} \mathbf{i} + 0 \mathbf{j} - 2 \mathbf{k} \)
• Acceleration: simple substitution will give you \( \mathbf{a}(\pi) \) as well.

Curvature

Curvature is a measure of how sharply a curve bends at a particular point. It's a crucial aspect when studying the geometric properties of parametric curves. The higher the curvature, the faster the direction of the curve is changing at that point.
To calculate curvature \( \kappa(t) \), you need the magnitude of the derivative of the unit tangent vector \( \mathbf{T}(t) \), with respect to the arc length. The unit tangent vector is derived by normalizing the velocity vector, \( \mathbf{v}(t) \), at each point:

• First, find the magnitude of the velocity vector \( ||\mathbf{v}(t)|| \).
• Then, normalize \( \mathbf{v}(t) \) to get \( \mathbf{T}(t) \).
• Calculate the derivative \( \mathbf{T}'(t) \) to find its change with respect to \( s \).

At \( t = \pi \), compute these values:

• Evaluate these steps to obtain \( \kappa(\pi) \).

This process gives a sense of how much the path's direction is changing at \( t = \pi \), further enhancing the understanding of the path's geometric behavior.