Question

find the tangential and normal components \(\left(a_{T}\right.\) and \(\left.a_{N}\right)\) of the acceleration vector at \(t .\) Then evaluate at \(t=t_{1} .\) $$ \mathbf{r}(t)=t \mathbf{i}+\frac{1}{3} t^{3} \mathbf{j}+t^{-1} \mathbf{k}, t>0 ; t_{1}=1 $$

Short answer

At \(t = 1\), \(a_T = 0\) and \(a_N = \sqrt{8}\).

Step-by-step solution

Find the velocity vector

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

Find the acceleration vector

Differentiate the velocity vector \( \mathbf{v}(t) = \mathbf{i} + t^2 \mathbf{j} - t^{-2} \mathbf{k} \) with respect to \( t \) to find the acceleration vector \( \mathbf{a}(t) \):\[ \mathbf{a}(t) = \frac{d}{dt} (\mathbf{i} + t^2 \mathbf{j} - t^{-2} \mathbf{k}) = 2t \mathbf{j} + 2t^{-3} \mathbf{k}. \]

Compute the magnitude of the velocity vector

Find the magnitude of \( \mathbf{v}(t) \):\[ ||\mathbf{v}(t)|| = \sqrt{1^2 + (t^2)^2 + (-t^{-2})^2} = \sqrt{1 + t^4 + t^{-4}}. \]

Compute the tangential component of acceleration

The tangential component of acceleration \( a_T \) is found by:\[ a_T = \frac{\mathbf{a}(t) \cdot \mathbf{v}(t)}{||\mathbf{v}(t)||}, \]where \( \cdot \) denotes the dot product, meaning:\[ \mathbf{a}(t) \cdot \mathbf{v}(t) = (2t \mathbf{j} + 2t^{-3} \mathbf{k}) \cdot (\mathbf{i} + t^2 \mathbf{j} - t^{-2} \mathbf{k}) = 2t^3 - 2t^{-5}, \]thus\[ a_T = \frac{2t^3 - 2t^{-5}}{\sqrt{1 + t^4 + t^{-4}}}. \]

Compute the normal component of acceleration

The normal component of acceleration \( a_N \) is calculated using:\[ a_N = \sqrt{||\mathbf{a}(t)||^2 - a_T^2}, \]where \[ ||\mathbf{a}(t)|| = \sqrt{0^2 + (2t)^2 + (2t^{-3})^2} = \sqrt{4t^2 + 4t^{-6}}. \]Then, substitute \( a_T \) to find \( a_N \).

Evaluate at \(t = t_1 = 1\)

Substitute \( t = 1 \) into the expressions for \( a_T \) and \( a_N \):\[ a_T = \frac{2(1)^3 - 2(1)^{-5}}{\sqrt{1 + (1)^4 + (1)^{-4}}} = \frac{2 - 2}{\sqrt{3}} = 0, \]and the magnitude of acceleration at \( t = 1 \) is:\[ ||\mathbf{a}(1)|| = \sqrt{4(1)^2 + 4(1)^{-6}} = \sqrt{4 + 4} = \sqrt{8}. \]Thus,\[ a_N = \sqrt{8 - 0} = \sqrt{8}. \]

Key concepts

Acceleration vector

In calculus, understanding how changes in position occur over time involves working with vectors. The acceleration vector is a vital concept that describes how the velocity of an object changes over time. It provides the rate at which an object's speed and direction change, directly affecting its trajectory.
The acceleration vector, typically denoted as \( \mathbf{a}(t) \), is obtained by differentiating the velocity vector \( \mathbf{v}(t) \). In our specific problem, after differentiating the velocity vector \( \mathbf{v}(t) = \mathbf{i} + t^2 \mathbf{j} - t^{-2} \mathbf{k} \), we find the acceleration vector:

• \( \mathbf{a}(t) = 2t \mathbf{j} + 2t^{-3} \mathbf{k} \)

This vector shows how the object's motion changes with respect to time in the \( \mathbf{j} \) and \( \mathbf{k} \) directions.

Tangential component

The tangential component of acceleration, represented as \( a_T \), describes how the speed of an object increases or decreases along its path. It effectively measures the acceleration component parallel to the velocity vector.
In mathematical terms, the tangential component can be computed using the dot product:

• \( a_T = \frac{\mathbf{a}(t) \cdot \mathbf{v}(t)}{||\mathbf{v}(t)||} \)

The dot product \( \mathbf{a}(t) \cdot \mathbf{v}(t) \) highlights the portion of acceleration aligned with velocity, providing insight into changes in speed. In our exercise, using values at \( t = 1 \), the tangential component was calculated to be 0.
This indicates that at this specific point, the object is not accelerating or decelerating along its path.

Normal component

The normal component of acceleration, denoted \( a_N \), addresses how rapidly an object is changing direction. This component acts perpendicular to the velocity vector, influencing the object's curvature of motion.
To find \( a_N \), the relationship between the total acceleration magnitude and the tangential component is used:

• \( a_N = \sqrt{||\mathbf{a}(t)||^2 - a_T^2} \)

For the given problem, after calculating the total magnitude and subtracting the tangential component squared, we determine the normal component \( a_N \) at \( t = 1 \). It was found to be \( \sqrt{8} \), indicating a significant change in direction of the object's motion at that time.

Velocity vector

The velocity vector \( \mathbf{v}(t) \) is a crucial concept as it defines the speed and direction of an object at any instant. It is derived by taking the derivative of the position vector with respect to time.
In our example, the position vector is given by \( \mathbf{r}(t) = t \mathbf{i} + \frac{1}{3} t^3 \mathbf{j} + t^{-1} \mathbf{k} \). Differentiating this with respect to \( t \), we obtain:

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

This vector gives us information about how fast and in what direction the object is moving at any time \( t \). Calculating its magnitude helps us determine the object's speed, which is necessary for finding acceleration components.

Dot product

The dot product is a mathematical operation that plays a key role in vector calculus, especially when analyzing vector relationships. It is defined as the sum of the products of the corresponding components of two vectors, \( \mathbf{u} \) and \( \mathbf{v} \).
The formula for a dot product is given by:

• \( \mathbf{u} \cdot \mathbf{v} = u_1v_1 + u_2v_2 + u_3v_3 \)

This product is pivotal in determining the angle between vectors or in finding projections. In the context of our problem, the dot product was used to calculate the tangential component \( a_T \) of the acceleration vector, highlighting its parallel relationship with the velocity vector. This simple yet powerful tool facilitates understanding how two vectors interact.