Problem 72 - Calculus III

Question

Find the tangential and normal acceleration components with the position vector \(\mathbf{r}(t)=\left\langle\cos t, \sin t, e^{t}\right\rangle\).

Step-by-Step Solution

Answer

Tangential acceleration: \( \frac{e^{2t}}{\sqrt{1 + e^{2t}}} \), Normal acceleration: \( \frac{1}{\sqrt{1 + e^{2t}}} \).

Step 1: Calculate the Velocity Vector

The velocity vector is the derivative of the position vector \( \mathbf{r}(t) \). Compute \( \mathbf{v}(t) = \frac{d}{dt} \mathbf{r}(t) = \left\langle -\sin t, \cos t, e^{t} \right\rangle \).

Step 2: Calculate the Speed

The speed is the magnitude of the velocity vector. Calculate it using \( v(t) = \left| \mathbf{v}(t) \right| = \sqrt{(-\sin t)^2 + (\cos t)^2 + (e^{t})^2} = \sqrt{1 + e^{2t}} \).

Step 3: Calculate the Acceleration Vector

The acceleration vector is the derivative of the velocity vector. Compute \( \mathbf{a}(t) = \frac{d}{dt} \mathbf{v}(t) = \left\langle -\cos t, -\sin t, e^{t} \right\rangle \).

Step 4: Find the Tangential Acceleration Component

The tangential acceleration is the projection of \( \mathbf{a}(t) \) onto \( \mathbf{v}(t) \). Calculate it using \( a_T = \frac{\mathbf{a}(t) \cdot \mathbf{v}(t)}{v(t)} = \frac{(-\cos t)(-\sin t) + (-\sin t)(\cos t) + (e^{t})(e^{t})}{\sqrt{1 + e^{2t}}} = \frac{e^{2t}}{\sqrt{1 + e^{2t}}} \).

Step 5: Find the Normal Acceleration Component

Use the formula for the normal component: \( a_N = \sqrt{\left| \mathbf{a}(t) \right|^2 - a_T^2} \). First, find \( \left| \mathbf{a}(t) \right| = \sqrt{(-\cos t)^2 + (-\sin t)^2 + (e^{t})^2} = \sqrt{1 + e^{2t}} \). Then, substitute to find \( a_N = \sqrt{(\sqrt{1 + e^{2t}})^2 - \left(\frac{e^{2t}}{\sqrt{1 + e^{2t}}}\right)^2} = \frac{1}{\sqrt{1 + e^{2t}}} \).

Key Concepts

Velocity Vector

The velocity vector is a fundamental concept in physics and mathematics, representing the rate of change of position of an object over time. In this exercise, it is derived from the position vector \( \mathbf{r}(t) \) by taking the derivative with respect to time. For the given position vector \( \mathbf{r}(t) = \langle \cos t, \sin t, e^{t} \rangle \), the velocity vector \( \mathbf{v}(t) \) becomes:

• \( \frac{d}{dt} \langle \cos t, \sin t, e^{t} \rangle = \langle -\sin t, \cos t, e^{t} \rangle \)

This vector indicates how fast and in which direction the object moves at any time \(t\). Each component of the velocity vector shows the rate of change in each spatial coordinate:

• The \(-\sin t\) component for the x-direction
• The \(\cos t\) component for the y-direction
• The \(e^{t}\) component for the z-direction

Acceleration Vector

The acceleration vector describes how the velocity of an object changes over time. It is found by differentiating the velocity vector. In the given problem, the velocity vector \( \mathbf{v}(t) = \langle -\sin t, \cos t, e^{t} \rangle \) is differentiated to yield the acceleration vector:

• \( \frac{d}{dt} \langle -\sin t, \cos t, e^{t} \rangle = \langle -\cos t, -\sin t, e^{t} \rangle \)

Each component of this vector shows the rate of change of the corresponding component of the velocity vector:

• The \(-\cos t\) represents the x-direction acceleration.
• The \(-\sin t\) for the y-direction.
• The \(e^{t}\) signifies the z-direction, changing exponentially.

Notice how the acceleration vector provides insights into forces acting on the object, showing both magnitude and direction of changing velocities.

Tangential Acceleration

Tangential acceleration quantifies how the speed of an object changes due to a change in the magnitude of its velocity, specifically in the direction of the velocity. In our scenario, it is computed by projecting the acceleration vector onto the velocity vector:

• \( a_T = \frac{\mathbf{a}(t) \cdot \mathbf{v}(t)}{v(t)} \)
• The dot product \( \mathbf{a}(t) \cdot \mathbf{v}(t) = (-\cos t)(-\sin t) + (-\sin t)(\cos t) + (e^{t})(e^{t})\)
• The magnitude of the velocity vector \( v(t) = \sqrt{1 + e^{2t}} \)
• Thus, \( a_T = \frac{e^{2t}}{\sqrt{1 + e^{2t}}} \)

This component specifically highlights how an object's acceleration affects its motion along the trajectory. A high tangential acceleration means a rapidly changing speed along the path.

Normal Acceleration

Normal acceleration, on the other hand, is related to changes in the direction of the velocity. It provides insights into how an object's path curves. This type of acceleration does not affect the speed, rather, it changes how sharply the path is curving. For calculation:

• First, find the magnitude of the acceleration vector \( \left| \mathbf{a}(t) \right| = \sqrt{1 + e^{2t}} \)
• Use the formula: \( a_N = \sqrt{\left| \mathbf{a}(t) \right|^2 - a_T^2} \)
• Substitute the known values: \( a_N = \sqrt{(\sqrt{1 + e^{2t}})^2 - \left(\frac{e^{2t}}{\sqrt{1 + e^{2t}}}\right)^2} \)
• This simplifies to \( a_N = \frac{1}{\sqrt{1 + e^{2t}}} \)

Normal acceleration is always perpendicular to the velocity vector, indicating the effect on an object's path curvature rather than its speed.