Problem 52 - 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 \(t=t_{1}\). \(\mathbf{r}(t)=t \mathbf{i}+\frac{1}{3} t^{3} \mathbf{j}+t^{-1} \mathbf{k}, t>0 ; t_{1}=1\)

Step-by-Step Solution

Answer

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

Step 1: Find the Velocity Vector

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

Step 2: Find the Acceleration Vector

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

Step 3: Compute Speed

Find the speed \(v\) by taking the magnitude of the velocity vector \(\mathbf{v}(t)\):\[v = \| \mathbf{v}(t) \| = \sqrt{(1)^2 + (t^2)^2 + (-t^{-2})^2}\]Solving gives:\[v = \sqrt{1 + t^4 + t^{-4}}\]

Step 4: Tangential Component of Acceleration

The tangential component of the acceleration \(a_T\) is given by the derivative of speed:\[a_T = \frac{d}{dt}(v)\]Differentiate \(v\) with respect to \(t\) using the expression \(v = \sqrt{1 + t^4 + t^{-4}}\).

Step 5: Normal Component of Acceleration

The normal component of the acceleration \(a_N\) is found using the formula:\[a_N = \sqrt{\| \mathbf{a}(t) \|^2 - a_T^2}\]Where \(\| \mathbf{a}(t) \| = \sqrt{(2t)^2 + (2t^{-3})^2}\).

Step 6: Evaluate at \(t = t_1 = 1\)

Substitute \(t = 1\) into the expressions for \(a_T\) and \(\mathbf{a}(t)\):- Find \(\mathbf{a}(1)\) which results in vector \(2\mathbf{j} + 2\mathbf{k}\).- Use in formulas of \(a_T\) and \(a_N\) to evaluate the components at \(t = 1\).

Key Concepts

Velocity Vector

The velocity vector is an essential concept when studying motion along a path. It represents the direction and rate of change of position of an object at any given moment in time. To calculate this vector, we differentiate the position vector \( \mathbf{r}(t) \) with respect to the time variable \( t \). This process involves taking the derivative of each component of the position vector, typically denoted in \( \mathbf{i}, \mathbf{j}, \) and \( \mathbf{k} \) directions.

For example, given \( \mathbf{r}(t) = t \mathbf{i} + \frac{1}{3} t^3 \mathbf{j} + t^{-1} \mathbf{k} \):

• Differentiating \( t \) with respect to \( t \) gives \( \mathbf{i} \).
• Differentiating \( \frac{1}{3} t^3 \) yields \( t^2 \mathbf{j} \).
• Differentiating \( t^{-1} \) results in \( -t^{-2} \mathbf{k} \).

Thus, our velocity vector \( \mathbf{v}(t) \) becomes \( \mathbf{i} + t^2 \mathbf{j} - t^{-2} \mathbf{k} \).
This provides not only the speed but also the direction in which the object is moving at any time \( t \).

Acceleration Vector

Acceleration is a crucial aspect of motion that tells us how an object's velocity changes over time. The acceleration vector is obtained by differentiating the velocity vector with respect to time \( t \). It incorporates the rate of change of each component of velocity.

For the given velocity vector \( \mathbf{v}(t) = \mathbf{i} + t^2 \mathbf{j} - t^{-2} \mathbf{k} \), we differentiate each term separately:

• The derivative of the constant \( \mathbf{i} \) is zero, implying no change in the \( \mathbf{i} \)-direction.
• The derivative of \( t^2 \) gives \( 2t \mathbf{j} \).
• The derivative of \( -t^{-2} \) results in \( 2t^{-3} \mathbf{k} \).

As a result, the acceleration vector \( \mathbf{a}(t) \) becomes \( 2t \mathbf{j} + 2t^{-3} \mathbf{k} \). This vector highlights how speed and direction in the \( \mathbf{j} \) and \( \mathbf{k} \) components are changing over time.

Tangential Component

The tangential component of acceleration \( a_T \) measures how much an object's speed along its path increases or decreases. It's found by taking the derivative of the speed \( v \), which is the magnitude of the velocity vector.

First, calculate the speed \( v \) by finding the magnitude of \( \mathbf{v}(t) = \mathbf{i} + t^2 \mathbf{j} - t^{-2} \mathbf{k} \):\[ v = \| \mathbf{v}(t) \| = \sqrt{1 + t^4 + t^{-4}} \]
Differentiating this expression with respect to \( t \), we obtain the tangential component. This analysis tells us how fast the object is speeding up or slowing down along its path, helping us understand more about its motion's overall dynamics.

Normal Component

The normal component of acceleration \( a_N \) deals with how the path direction changes, rather than the speed along the path. It indicates how much centrifugal force the object experiences as it travels along its curve.

The normal component is determined using: \[ a_N = \sqrt{\| \mathbf{a}(t) \|^2 - a_T^2} \] where \( \| \mathbf{a}(t) \| \) is the magnitude of the acceleration vector.

For our function, compute the magnitude of the acceleration vector \( \mathbf{a}(t) = 2t \mathbf{j} + 2t^{-3} \mathbf{k} \):

• Calculate \( \| \mathbf{a}(t) \| = \sqrt{(2t)^2 + (2t^{-3})^2} \).

This helps us grasp how the object curves along its path and the forces required to maintain such curvilinear motion.