Problem 16

Question

In Problems $7-16, \mathbf{r}(t)$ is the position vector of a moving particle. Find the tangential and normal components of the acceleration at any $t$. $$ \mathbf{r}(t)=t \mathbf{i}+(2 t-1) \mathbf{j}+(4 t+2) \mathbf{k} $$

Step-by-Step Solution

Verified

Answer

Both the tangential and normal components of the acceleration are 0.

1Step 1: Find the velocity vector

To find the velocity vector, differentiate the position vector $ \mathbf{r}(t) = t \mathbf{i} + (2t-1) \mathbf{j} + (4t+2) \mathbf{k} $ with respect to $ t $. The velocity vector $ \mathbf{v}(t) $ is given by:\[ \mathbf{v}(t) = \frac{d}{dt}(t) \mathbf{i} + \frac{d}{dt}(2t-1) \mathbf{j} + \frac{d}{dt}(4t+2) \mathbf{k} = \mathbf{i} + 2\mathbf{j} + 4\mathbf{k} \].

2Step 2: Find the acceleration vector

Differentiate the velocity vector $ \mathbf{v}(t) = \mathbf{i} + 2\mathbf{j} + 4\mathbf{k} $ with respect to $ t $ to get the acceleration vector $ \mathbf{a}(t) $:\[ \mathbf{a}(t) = \frac{d}{dt}(1) \mathbf{i} + \frac{d}{dt}(2) \mathbf{j} + \frac{d}{dt}(4) \mathbf{k} = \mathbf{0} \].

3Step 3: Find the speed of the particle

The speed of the particle is the magnitude of the velocity vector. Compute $ \| \mathbf{v}(t) \| $:\[ \| \mathbf{v}(t) \| = \sqrt{1^2 + 2^2 + 4^2} = \sqrt{21} \].

4Step 4: Compute the tangential component of acceleration

The tangential component of the acceleration $ a_T $ is given by the derivative of the speed:Since $ \mathbf{a}(t) = \mathbf{0} $, the tangential component is\[ a_T = 0 \].

5Step 5: Compute the normal component of acceleration

The normal component of the acceleration $ a_N $ can be found using the formula:\[ a_N = \sqrt{\| \mathbf{a}(t) \|^2 - a_T^2} \]. Since $ \| \mathbf{a}(t) \| = 0 $ and $ a_T = 0 $, we have:\[ a_N = 0 \].

Key Concepts

Tangential AccelerationNormal AccelerationVelocity VectorAcceleration Vector

Tangential Acceleration

In vector calculus, tangential acceleration refers to the component of acceleration that acts in the direction of the particle's velocity. It's essentially how quickly the speed of the particle is changing along its path. For our given problem, the tangential component can be discovered by taking the derivative of the speed with respect to time.

The speed is the magnitude of the velocity vector.
If the acceleration vector is constant, the tangential acceleration is zero because speed is not changing.

In this exercise, since the acceleration vector $ \mathbf{a}(t) = \mathbf{0} $, the tangential acceleration, $ a_T $, is calculated to be 0. This indicates that the speed of the particle does not change, as confirmed by the zero acceleration vector.

Normal Acceleration

Normal acceleration, also known as centripetal acceleration, describes the component of acceleration that is perpendicular to the direction of the velocity vector. It helps in understanding how the direction of a moving particle changes. This is essential for particles moving along curved paths.

It is calculated using the formula $ a_N = \sqrt{\| \mathbf{a}(t) \|^2 - a_T^2} $.
This component is always directed towards the center of the particle's path.

For this problem, given that both the acceleration and the tangential acceleration are zero, the normal acceleration $ a_N $ is also zero. Hence, the direction of the particle's movement does not change.

Velocity Vector

The velocity vector provides both the speed and direction of a particle. It is the first derivative of the position vector with respect to time. In mathematical terms, it can be represented as $ \mathbf{v}(t) = \frac{d\mathbf{r}(t)}{dt} $. This shows how quickly the position vector changes over time.

For our exercise, $ \mathbf{r}(t) = t \mathbf{i} + (2t-1) \mathbf{j} + (4t+2) \mathbf{k} $.
The resulting velocity vector is $ \mathbf{v}(t) = \mathbf{i} + 2\mathbf{j} + 4\mathbf{k} $.

This constant velocity vector suggests that the particle is moving in a straight line at a constant rate, with no change in speed or direction over time.

Acceleration Vector

The acceleration vector signifies the rate of change of the velocity vector over time. It's calculated by differentiating the velocity vector. It can indicate changes in both magnitude and direction of the velocity. The mathematical representation is $ \mathbf{a}(t) = \frac{d\mathbf{v}(t)}{dt} $.

In our case, the velocity vector $ \mathbf{v}(t) = \mathbf{i} + 2\mathbf{j} + 4\mathbf{k} $ seems constant, leading to a zero acceleration vector $ \mathbf{a}(t) = \mathbf{0} $.
A zero acceleration vector means no change in the velocity over time.

Thus, as the acceleration vector is zero, the particle is not accelerating, confirming a uniform motion along a straight line.

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