Problem 47 - Calculus Early Transcendentals: Pearson New International Edition

Question

Find the tangential and normal components \(\left(a_{T}\right.\) and \(a_{N}\) ) of the acceleration vector at \(t\). Then evaluate at \(t=t_{1}\). \(\mathbf{r}(t)=(t+1) \mathbf{i}+3 t \mathbf{j}+t^{2} \mathbf{k} ; t_{1}=1\)

Step-by-Step Solution

Answer

At \( t = 1\), \( a_T = \frac{4}{\sqrt{14}} \) and \( a_N = \frac{2\sqrt{5}}{\sqrt{7}} \).

Step 1: Find the Velocity Vector

To find the velocity vector, differentiate the position vector \( \mathbf{r}(t) = (t+1) \mathbf{i} + 3t \mathbf{j} + t^2 \mathbf{k} \) with respect to \( t \). \[ \mathbf{v}(t) = \frac{d}{dt} \mathbf{r}(t) = \mathbf{i} + 3 \mathbf{j} + 2t \mathbf{k} \]

Step 2: Calculate the Speed

Speed is the magnitude of the velocity vector \( \mathbf{v}(t) \). \[ \text{Speed } = \| \mathbf{v}(t) \| = \sqrt{(1)^2 + (3)^2 + (2t)^2} = \sqrt{1 + 9 + 4t^2} = \sqrt{10 + 4t^2} \]

Step 3: Find the Acceleration Vector

The acceleration vector is the derivative of the velocity vector with respect to \( t \). \[ \mathbf{a}(t) = \frac{d}{dt} \mathbf{v}(t) = 0 \mathbf{i} + 0 \mathbf{j} + 2 \mathbf{k} = 2 \mathbf{k} \]

Step 4: Determine Tangential Component \(a_T\)

The tangential component of acceleration \( a_T \) is found using the formula \( a_T = \frac{\mathbf{v} \cdot \mathbf{a} }{ \|\mathbf{v}\|} \). Compute the dot product \( \mathbf{v}(t) \cdot \mathbf{a}(t) = (1) \cdot 0 + (3) \cdot 0 + (2t) \cdot 2 = 4t \) Thus, \[ a_T = \frac{4t}{\sqrt{10 + 4t^2}} \] Substitute \( t = 1 \):\[ a_T(1) = \frac{4 \cdot 1}{\sqrt{10 + 4 \cdot 1^2}} = \frac{4}{\sqrt{14}} \]

Step 5: Determine Normal Component \(a_N\)

The normal component of acceleration is given by \( a_N = \sqrt{ \| \mathbf{a} \|^2 - a_T^2 } \). First, find the magnitude of acceleration: \[ \|\mathbf{a}(t)\| = \sqrt{0^2 + 0^2 + 2^2} = 2 \] Then calculate \( a_T^2 \):\[ a_T^2 = \left( \frac{4}{\sqrt{14}} \right)^2 = \frac{16}{14} = \frac{8}{7} \]Now compute \( a_N \): \[ a_N = \sqrt{2^2 - \frac{8}{7}} = \sqrt{4 - \frac{8}{7}} = \sqrt{\frac{28}{7} - \frac{8}{7}} = \sqrt{\frac{20}{7}} = \frac{\sqrt{20}}{\sqrt{7}} = \frac{2\sqrt{5}}{\sqrt{7}} \]

Step 6: Conclusion

At \( t = 1 \), the tangential component of acceleration \( a_T \) is \( \frac{4}{\sqrt{14}} \) and the normal component \( a_N \) is \( \frac{2\sqrt{5}}{\sqrt{7}} \). These components describe how the acceleration breaks down into directions tangential and normal to the path of the particle.

Key Concepts

Acceleration Vector

The acceleration vector is a key concept in understanding how an object’s velocity changes over time. It's defined as the derivative of the velocity vector with respect to time. Essentially, this vector gives us insight into how the speed and direction of an object's motion are changing. For the given exercise, the acceleration vector is derived from the velocity vector:

• Differentiate the velocity vector to find the acceleration vector.
• The given velocity vector is \(\mathbf{v}(t) = \mathbf{i} + 3 \mathbf{j} + 2t \mathbf{k}.\)
• Therefore, the acceleration vector is \( \mathbf{a}(t) = 2 \mathbf{k},\)indicating a consistent change in velocity along the k-axis only.

In real-world applications, acceleration vectors can show forces acting upon the particle, moving it out of its original path.

Velocity Vector

Understanding the velocity vector of an object provides insights into its speed and direction at any given moment. This vector is simply the derivative of the position vector with respect to time, reflecting the object's rate of change of position:

• The formula to find the velocity vector is \( \mathbf{v}(t) = \frac{d}{dt} \mathbf{r}(t).\)
• In the exercise, the position vector is \( \mathbf{r}(t) = (t+1) \mathbf{i} + 3t \mathbf{j} + t^2 \mathbf{k}.\)
• Deriving this gives the velocity vector as \( \mathbf{v}(t) = \mathbf{i} + 3 \mathbf{j} + 2t \mathbf{k}.\)

By examining the velocity vector, one can determine both the speed and direction in which an object is moving at any given point in time.

Magnitude of Velocity

The magnitude of velocity, often referred to as speed, is a scalar that represents the rate at which an object covers distance. It is crucial to distinguish this from velocity, which includes direction. To find the magnitude:

• Compute the square root of the sum of the squares of the components of the velocity vector.
• In this exercise, the velocity vector is \( \mathbf{v}(t) = \mathbf{i} + 3 \mathbf{j} + 2t \mathbf{k}.\)
• The magnitude is calculated as \( \|\mathbf{v}(t)\| = \sqrt{1^2 + 3^2 + (2t)^2} = \sqrt{10 + 4t^2}.\)

Thus, the magnitude of velocity tells us how fast the particle is moving along its path without considering its direction.

Tangential Acceleration

Tangential acceleration relates to how quickly an object's speed is changing along its path. It's a component of the acceleration vector that aligns with the direction of the velocity vector. The formula for tangential acceleration is:

• \( a_T = \frac{\mathbf{v} \cdot \mathbf{a} }{ \|\mathbf{v}\|} \)
• This is calculated through the dot product of the velocity and acceleration vectors divided by the magnitude of the velocity.
• In the exercise, \\( a_T(t) = \frac{4t}{\sqrt{10 + 4t^2}}.\)
• At \( t = 1,\) the tangential acceleration \( a_T(1) = \frac{4}{\sqrt{14}}.\)

This component crucially affects how an object speeds up or slows down along its trajectory.

Normal Acceleration

Normal acceleration, or centripetal acceleration, describes how an object’s direction is changing as it moves along a curved path. It acts perpendicular to the velocity vector, steering the object toward the center of the curve:

• To find normal acceleration, use \( a_N = \sqrt{ \| \mathbf{a} \|^2 - a_T^2 }.\)
• This is derived from the Pythagorean theorem applied within the velocity and acceleration vector framework.
• In the exercise, the normal component at \( t = 1 \) is \( a_N = \frac{2\sqrt{5}}{\sqrt{7}}.\)

Understanding normal acceleration helps in analyzing forces that change the direction of motion without altering speed, such as a car turning a corner at a constant speed.