Problem 51 - 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 \(t=t_{1} .\) $$ \mathbf{r}(t)=\left(t-\frac{1}{3} t^{3}\right) \mathbf{i}-\left(t+\frac{1}{3} t^{3}\right) \mathbf{j}+t \mathbf{k} ; t_{1}=3 $$

Step-by-Step Solution

Answer

The tangential component is \( a_T = \frac{108}{\sqrt{165}} \) and the normal component is dynamically calculated by \( a_N = \sqrt{(6\sqrt{2})^2 - \left(\frac{108}{\sqrt{165}}\right)^2} \).

Step 1: Compute Velocity

To find the tangential and normal components, the first step is to determine velocity by differentiating the position vector \( \mathbf{r}(t) \) with respect to \( t \). \[ \mathbf{v}(t) = \frac{d\mathbf{r}(t)}{dt} = \left(1 - t^2\right) \mathbf{i} - \left(1 + t^2\right) \mathbf{j} + \mathbf{k}\]

Step 2: Compute Acceleration

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

Step 3: Evaluate Vectors at \( t = 3 \)

Substitute \( t = 3 \) into the velocity and acceleration vectors.Velocity at \( t = 3 \):\[ \mathbf{v}(3) = \left(1 - 3^2\right) \mathbf{i} - \left(1 + 3^2\right) \mathbf{j} + \mathbf{k} = -8 \mathbf{i} - 10 \mathbf{j} + \mathbf{k}\]Acceleration at \( t = 3 \):\[ \mathbf{a}(3) = -2 \times 3 \mathbf{i} - 2 \times 3 \mathbf{j} = -6 \mathbf{i} - 6 \mathbf{j}\]

Step 4: Calculate Magnitude of Velocity

Find the magnitude of the velocity vector \( \mathbf{v}(3) \).\[ \|\mathbf{v}(3)\| = \sqrt{(-8)^2 + (-10)^2 + (1)^2} = \sqrt{164 + 1} = \sqrt{165}\]

Step 5: Calculate Tangential Component \( a_{T} \)

The tangential component \( a_T \) is the dot product of \( \mathbf{a}(t) \) and the unit vector in the direction of \( \mathbf{v}(t) \). \[ a_T = \frac{\mathbf{a}(3) \cdot \mathbf{v}(3)}{\|\mathbf{v}(3)\|}\]Calculate the dot product:\[ \mathbf{a}(3) \cdot \mathbf{v}(3) = (-6)(-8) + (-6)(-10) = 48 + 60 = 108\]Now compute \( a_T \):\[ a_T = \frac{108}{\sqrt{165}} \]

Step 6: Calculate Normal Component \( a_{N} \)

The normal component \( a_N \) can be found using the formula:\[ a_N = \sqrt{\|\mathbf{a}(3)\|^2 - a_T^2}\]First, calculate the magnitude of \( \mathbf{a}(3) \):\[ \|\mathbf{a}(3)\| = \sqrt{(-6)^2 + (-6)^2} = \sqrt{72} = 6\sqrt{2}\]Now substitute to find \( a_N \):\[ a_N = \sqrt{(6\sqrt{2})^2 - \left(\frac{108}{\sqrt{165}}\right)^2}\]

Key Concepts

Tangential Acceleration

Imagine a car driving along a twisty road. As the car speeds up or slows down, that's when tangential acceleration comes into play. It describes how the car's speed changes along its path.
Tangential acceleration is aligned with the direction of the velocity vector. Mathematically, it involves the derivative of the velocity in the same direction. In simpler terms, it tells us how quickly the speed of an object is changing at any point. For a moving object, to find the tangential component of acceleration, we compute the dot product of the acceleration vector and the unit velocity vector, then divide by the magnitude of the velocity. This is given by the formula:

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

In our exercise, this was simplified by computing at a specific time \(t = 3\). The dot product resulted in 108 when the relevant vectors were calculated, leading to the tangential component \(a_T\).

Normal Acceleration

When you swing a ball tied to a string in a circle, normal acceleration keeps the ball moving in its circular path. It's all about how the direction of movement changes over time, not the speed.
Also known as centripetal acceleration, normal acceleration is perpendicular to the velocity vector. It's crucial in changing the object’s direction rather than speed. To calculate it, the formula used is:

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

This derives from removing the tangential component from the total acceleration. The exercise showed us that by using this formula, we find out exactly how forces need to be directed inward, to keep an object on its curving path, like riding a bicycle on a bend. The calculations resulted in the normal component of acceleration \(a_N\) after plugging in the known values.

Velocity Vector

Think about hopping on a moving bicycle. The direction you're headed, combined with how fast you’re going, forms what we call the velocity vector. It captures both speed and direction.
In equations and physical terms, it’s the derivative of the position vector with respect to time. It’s framed as:

• \[ \mathbf{v}(t) = \frac{d\mathbf{r}(t)}{dt} \]

This gives us how the object is moving at each instant. For the exercise, once differentiating the given position vector \( \mathbf{r}(t) \), we found the velocity expressed in unit directions \( \mathbf{i}, \mathbf{j}, \mathbf{k} \). This vector played a key role in determining both tangential and normal accelerations later on.

Magnitude of Velocity

When you want to find out just how fast something is going at a certain time, you look at the magnitude of the velocity vector. It’s like asking the speedometer for just the number without knowing the direction.
The magnitude of the velocity is given by the following formula:

• \[ \|\mathbf{v}(t)\| = \sqrt{v_x^2 + v_y^2 + v_z^2} \]

Here, \(v_x, v_y, v_z\) are the components of the velocity vector. The outcome is a non-directional speed. In this exercise, we computed the magnitude at \(t=3\) to be \(\sqrt{165}\). This step is crucial because it's this magnitude that influences how we compute subsequent acceleration vectors. It affects other calculations, primarily how we understand changes in speed through tangential and normal components.