Problem 32 - Essential Calculus Early Transcendentals

Question

Find the tangential and normal components of the acceleration vector. $$\mathbf{r}(t)=t \mathbf{i}+\cos ^{2} t \mathbf{j}+\sin ^{2} t \mathbf{k}$$

Step-by-Step Solution

Answer

The tangential component is \( \frac{2\sin 4t}{\sqrt{1 + 2\sin^2 2t}} \), and the normal component is \( \sqrt{4 - \frac{4\sin^2 4t}{1 + 2\sin^2 2t}} \)."

Step 1: Find the Velocity Vector

The velocity vector \( \mathbf{v}(t) \) is the first derivative of the position vector \( \mathbf{r}(t) \). Differentiate each component individually:- \( \frac{d}{dt}(t) = 1 \)- \( \frac{d}{dt}(\cos^2 t) = -2\cos t \sin t \)- \( \frac{d}{dt}(\sin^2 t) = 2\sin t \cos t \)Thus, the velocity vector is \( \mathbf{v}(t) = \mathbf{i} - 2\cos t \sin t \mathbf{j} + 2\sin t \cos t \mathbf{k} \).

Step 2: Find the Acceleration Vector

The acceleration vector \( \mathbf{a}(t) \) is the derivative of the velocity vector \( \mathbf{v}(t) \). Differentiate each component:- The derivative of \(1\) with respect to \(t\) is \(0\).- The derivative of \(-2\cos t \sin t\) is \(-2(\cos^2 t - \sin^2 t)\).- The derivative of \(2\sin t \cos t\) is \(2(\cos^2 t - \sin^2 t)\).Thus, \( \mathbf{a}(t) = 0 \mathbf{i} + 2(\sin^2 t - \cos^2 t) \mathbf{j} + 2(\cos^2 t - \sin^2 t) \mathbf{k} \).

Step 3: Find Speed

The speed \( v(t) \) is the magnitude of the velocity vector \( \mathbf{v}(t) \). Compute it by finding the square root of the sum of squares:\[ v(t) = \sqrt{1^2 + (-2\cos t \sin t)^2 + (2\sin t \cos t)^2} = \sqrt{1 + 4\cos^2 t \sin^2 t + 4\sin^2 t \cos^2 t} = \sqrt{1 + 8\cos^2 t \sin^2 t} \]Simplify using the identity \( \sin 2t = 2\sin t \cos t \):\[ v(t) = \sqrt{1 + 2\sin^2 2t} \].

Step 4: Find Tangential Component of Acceleration

The tangential component of the acceleration \( a_T \) is given by the formula \( a_T(t) = \frac{d}{dt}(v(t)) \). First, differentiate \( v(t) = \sqrt{1 + 2\sin^2 2t} \) with respect to \(t\):Use the chain rule:\[ \frac{d}{dt}\left(\sqrt{1 + 2\sin^2 2t}\right) = \frac{1}{2}\frac{4\sin 2t \cos 2t}{\sqrt{1 + 2\sin^2 2t}} \]This simplifies to:\[ a_T(t) = \frac{2\sin 4t}{\sqrt{1 + 2\sin^2 2t}} \].

Step 5: Find Normal Component of Acceleration

The normal component \( a_N \) can be found using \( a_N = \sqrt{\|\mathbf{a}(t)\|^2 - a_T(t)^2} \). First, find \( \|\mathbf{a}(t)\| \):\[ \|\mathbf{a}(t)\| = \sqrt{(2(\sin^2 t - \cos^2 t))^2 + (2(\cos^2 t - \sin^2 t))^2} = 2\]Then,\[ a_N = \sqrt{2^2 - \left(\frac{2\sin 4t}{\sqrt{1 + 2\sin^2 2t}}\right)^2} \]Ultimately:\[ a_N = \sqrt{4 - \frac{4\sin^2 4t}{1 + 2\sin^2 2t}} \]

Key Concepts

Tangential Component of Acceleration

When we talk about the tangential component of acceleration, we're focusing on how fast the speed of an object changes as it moves along a path. In simpler terms, this is the part of acceleration that aligns with the direction of velocity. In our exercise, after finding the speed function, the derivative of this function with respect to time is taken to find the tangential component of acceleration, denoted by \( a_T(t) \). This means we are seeing how quickly the object's speed is changing at any time \( t \). The formula we use is:

• \( a_T(t) = \frac{d}{dt}(v(t)) \)

By substituting \( v(t) = \sqrt{1 + 2\sin^2 2t} \) into this formula and using the chain rule during differentiation, we find:

• \( a_T(t) = \frac{2\sin 4t}{\sqrt{1 + 2\sin^2 2t}} \)

This captures how different portions of the motion are accelerating along the path, but remember that it doesn’t account for any changes in direction.

Normal Component of Acceleration

The normal component of acceleration is all about the change in direction. It considers how the object's path curves. Even if an object moves at constant speed, if it's changing direction, it has a normal acceleration. This is perpendicular to the tangential component. For this exercise, the normal component of acceleration, labeled as \( a_N \), can be calculated using the formula:

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

First, we find the magnitude of the acceleration vector \( \mathbf{a}(t) \), which gives us \( 2 \). Then, we plug in our previously found \( a_T(t) \) into the formula to solve for \( a_N \):

• \( a_N = \sqrt{4 - \frac{4\sin^2 4t}{1 + 2\sin^2 2t}} \)

This component indicates how much the object’s path is curving at any point \( t \), independent of speed changes. So, even with constant speed, the normal acceleration can cause the object to move in a curved path.

Velocity and Speed Calculations

Understanding the difference between velocity and speed is crucial. Velocity is a vector quantity, meaning it has both magnitude and direction, while speed is a scalar, which means it only has magnitude. In our exercise, the velocity vector is obtained by differentiating the position vector \( \mathbf{r}(t) \) with respect to \( t \), leading to:

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

Here, each term reflects a component of how the object moves along each axis.
To find the speed, we calculate the magnitude of this velocity vector. We use the formula for the magnitude of a vector, which is the square root of the sum of the squares of its components:

• \( v(t) = \sqrt{1 + 2\sin^2 2t} \)

The speed calculation gives us a sense of how fast the object is moving at any given point in its journey, regardless of the direction it’s taking. It's like checking how fast your car goes, irrespective of whether you're heading north or east.