Problem 47 - Calculus Early Transcendentals: Pearson New International Edition

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 \(t=t_{1} .\) $$ \mathbf{r}(t)=(t+1) \mathbf{i}+3 t \mathbf{j}+t^{2} \mathbf{k} ; t_{1}=1 $$

Step-by-Step Solution

Answer

At \(t=1\), \(a_T = \frac{4}{\sqrt{14}}\) and \(a_N = \sqrt{\frac{20}{7}}\).

Step 1: Find the Velocity Vector

The velocity vector \(\mathbf{v}(t)\) is the derivative of the position vector \(\mathbf{r}(t)\). Compute \(\mathbf{v}(t)\):\[\mathbf{v}(t) = \frac{d}{dt}\left((t+1)\mathbf{i} + 3t\mathbf{j} + t^2\mathbf{k}\right)\]Calculating the derivatives:\[\mathbf{v}(t) = \mathbf{i} + 3\mathbf{j} + 2t\mathbf{k}\]

Step 2: Find the Acceleration Vector

The acceleration vector \(\mathbf{a}(t)\) is the derivative of the velocity vector \(\mathbf{v}(t)\). Compute \(\mathbf{a}(t)\):\[\mathbf{a}(t) = \frac{d}{dt}\left(\mathbf{i} + 3\mathbf{j} + 2t\mathbf{k}\right)\]Calculating the derivatives:\[\mathbf{a}(t) = 2\mathbf{k}\]

Step 3: Find the Magnitude of the Velocity

To compute the tangential and normal components, first find the magnitude of the velocity \(|\mathbf{v}(t)|\):\[|\mathbf{v}(t)| = \sqrt{1^2 + 3^2 + (2t)^2} = \sqrt{1 + 9 + 4t^2} = \sqrt{10 + 4t^2}\]

Step 4: Find the Tangential Component of Acceleration

The tangential component \(a_T\) is the rate of change of the speed. Calculate \(a_T\) using:\[a_T = \frac{d}{dt}|\mathbf{v}(t)|\]First, note that:\[|\mathbf{v}(t)| = (10 + 4t^2)^{1/2}\]So, \[\frac{d}{dt}|\mathbf{v}(t)| = \frac{1}{2}(10 + 4t^2)^{-1/2} \cdot 8t = \frac{4t}{\sqrt{10 + 4t^2}}\]Thus, \(a_T = \frac{4t}{\sqrt{10 + 4t^2}}\).

Step 5: Find the Normal Component of Acceleration

The normal component \(a_N\) can be found from:\[a_N = \sqrt{|\mathbf{a}(t)|^2 - a_T^2}\]Since \(|\mathbf{a}(t)| = 2\) and \(a_T = \frac{4t}{\sqrt{10 + 4t^2}}\),\[a_N = \sqrt{2^2 - \left(\frac{4t}{\sqrt{10 + 4t^2}}\right)^2} = \sqrt{4 - \frac{16t^2}{10 + 4t^2}}\]Simplify:\[a_N = \sqrt{\frac{40 + 16t^2 - 16t^2}{10 + 4t^2}} = \sqrt{\frac{40}{10 + 4t^2}}\]

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

Evaluate \(a_T\) and \(a_N\) at \(t = 1\):Calculate \(a_T(1)\):\[a_T(1) = \frac{4 \times 1}{\sqrt{10 + 4 \times 1^2}} = \frac{4}{\sqrt{14}}\]Calculate \(a_N(1)\):\[a_N(1) = \sqrt{\frac{40}{10 + 4 \times 1^2}} = \sqrt{\frac{40}{14}} = \sqrt{\frac{20}{7}}\]Thus, at \(t = 1\), \(a_T = \frac{4}{\sqrt{14}}\) and \(a_N = \sqrt{\frac{20}{7}}\).

Key Concepts

Velocity Vector

The velocity vector, often denoted as \( \mathbf{v}(t) \), is a key concept in understanding motion in a 3D space. It represents the rate of change of the position vector \( \mathbf{r}(t) \) over time. To find this vector, we compute the derivative of the position function with respect to time.

• In our example, the position function given is \( \mathbf{r}(t) = (t+1)\mathbf{i} + 3t\mathbf{j} + t^2\mathbf{k} \).
• By differentiating each component with respect to \( t \), we find the velocity vector \( \mathbf{v}(t) = \mathbf{i} + 3\mathbf{j} + 2t\mathbf{k} \).

The derivative tells us how the position changes at any given moment, giving us insight into both the direction and speed of the motion. It's crucial to understand that the velocity vector is directional, meaning it can point in different directions as the object moves through space.

Acceleration Vector

The acceleration vector, \( \mathbf{a}(t) \), plays a vital role in describing how quickly the velocity of an object is changing. It is determined by taking the derivative of the velocity vector with respect to time. Just as velocity gives us the rate of position change, acceleration tells us how velocity itself is changing.

• Starting with the velocity vector from our previous calculation, \( \mathbf{v}(t) = \mathbf{i} + 3\mathbf{j} + 2t\mathbf{k} \), we differentiate each term again.
• This produces the acceleration vector \( \mathbf{a}(t) = 2\mathbf{k} \).

Notice that in some cases, certain components of the vector might be zero, implying no change in that particular direction. Here, the lack of \( \mathbf{i} \) or \( \mathbf{j} \) components shows no acceleration in those directions, only along \( \mathbf{k} \). Understanding acceleration is crucial for predicting future motion, as it helps us determine not just where an object will be, but how fast it's speeding up or slowing down.

Magnitude of Velocity

The magnitude of the velocity vector provides the speed of an object without considering its direction, effectively answering the question, "How fast?" without regard for where it's headed. To find it, we compute the square root of the sum of the squares of its components.

• Given the velocity vector \( \mathbf{v}(t) = \mathbf{i} + 3\mathbf{j} + 2t\mathbf{k} \), its magnitude is \( |\mathbf{v}(t)| = \sqrt{1^2 + 3^2 + (2t)^2} = \sqrt{10 + 4t^2} \).

This value tells us the overall speed at which the object is moving through space at any given moment. It's analogous to the speedometer reading in a car, which gives speed irrespective of whether the car is going north or east. Understanding the magnitude is essential for calculating more complex kinetic properties like the tangential component of acceleration.

Tangential Component of Acceleration

The tangential component of acceleration, denoted \( a_T \), measures how the speed of an object is changing along its path. It represents the portion of acceleration that's directly along the direction of motion, affecting the object's speed.

• To find \( a_T \), we differentiate the magnitude of the velocity: \( a_T = \frac{d}{dt}|\mathbf{v}(t)| \).
• From our earlier magnitude of velocity calculation, we have \( |\mathbf{v}(t)| = (10 + 4t^2)^{1/2} \).
• Differentiating gives \( a_T = \frac{4t}{\sqrt{10 + 4t^2}} \).

This component is crucial when analyzing motion, as it determines how speed is increasing or decreasing. It informs us about the driver's "acceleration pedal" influence in a car, focusing solely on forward acceleration rather than changes in direction, which would relate to the normal component instead.