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 \(t=t_{1} .\) $$ \mathbf{r}(t)=a \cos t \mathbf{i}+a \sin t \mathbf{j} ; t_{1}=\pi / 6 $$

Short answer

At \( t = \frac{\pi}{6} \), \( a_T = 0 \) and \( a_N = a \).

Step-by-step solution

Find the Velocity Vector

To find the velocity vector \( \mathbf{v}(t) \), differentiate the position vector \( \mathbf{r}(t) = a \cos t \, \mathbf{i} + a \sin t \, \mathbf{j} \) with respect to \( t \). \[ \mathbf{v}(t) = \frac{d}{dt}(a \cos t \, \mathbf{i} + a \sin t \, \mathbf{j}) = -a \sin t \, \mathbf{i} + a \cos t \, \mathbf{j} \]

Find the Speed

The speed is the magnitude of the velocity vector \( \mathbf{v}(t) \). Calculate it as follows:\[ v(t) = \| \mathbf{v}(t) \| = \sqrt{(-a \sin t)^2 + (a \cos t)^2} = \sqrt{a^2 \sin^2 t + a^2 \cos^2 t} = a \]

Find the Acceleration Vector

To find the acceleration vector \( \mathbf{a}(t) \), differentiate the velocity vector \( \mathbf{v}(t) \).\[ \mathbf{a}(t) = \frac{d}{dt}(-a \sin t \, \mathbf{i} + a \cos t \, \mathbf{j}) = -a \cos t \, \mathbf{i} - a \sin t \, \mathbf{j} \]

Tangential Component of Acceleration

The tangential component of acceleration \( a_T \) is given by the derivative of the speed with respect to time.\[ a_T = \frac{d}{dt}(a) = 0 \]Since \( a \) is a constant, \( a_T = 0 \).

Normal Component of Acceleration

The normal component of acceleration \( a_N \) can be found using the formula:\[ a_N = \sqrt{ \| \mathbf{a}(t) \|^2 - a_T^2 } \]Compute the magnitude of acceleration \( \| \mathbf{a}(t) \| \):\[ \| \mathbf{a}(t) \| = \sqrt{(-a \cos t)^2 + (-a \sin t)^2} = a \]Thus,\[ a_N = \sqrt{a^2 - 0^2} = a \]

Evaluate at \( t = \frac{\pi}{6} \)

Both \( a_T \) and \( a_N \) have been calculated as constants, independent of \( t \). Therefore, at \( t = \frac{\pi}{6} \), the results are:- \( a_T = 0 \)- \( a_N = a \)

Key concepts

Understanding the Velocity Vector

The velocity vector is fundamental in understanding how the position of an object changes over time. In calculus, the velocity vector is derived by taking the derivative of the position vector with respect to time. Given the position vector \( \mathbf{r}(t) = a \cos t \, \mathbf{i} + a \sin t \, \mathbf{j} \), the velocity vector is calculated by differentiating each component:

• The derivative of \( a \cos t \) is \( -a \sin t \).
• The derivative of \( a \sin t \) is \( a \cos t \).

These derivatives form the velocity vector: \( \mathbf{v}(t) = -a \sin t \, \mathbf{i} + a \cos t \, \mathbf{j} \). Understanding this vector helps in predicting the direction and speed of the object at any given time.

Simplifying Speed Calculation

Once you have the velocity vector, the next step is to find the speed, which is simply the magnitude of the velocity vector. The speed tells us how fast an object is moving, irrespective of its direction. You can calculate speed \( v(t) \) from the velocity vector \( \mathbf{v}(t) = -a \sin t \, \mathbf{i} + a \cos t \, \mathbf{j} \) using the formula:

• \( v(t) = \sqrt{(-a \sin t)^2 + (a \cos t)^2} \)
• This simplifies to \( v(t) = \sqrt{a^2 \sin^2 t + a^2 \cos^2 t} \)

Using the Pythagorean identity \( \sin^2 t + \cos^2 t = 1 \), you can further simplify it to \( v(t) = a \). Hence, the speed remains constant over time.

Analyzing the Acceleration Vector

The acceleration vector provides insight into how the velocity of an object changes with time. To find it, differentiate the velocity vector with respect to time. For our velocity vector \( \mathbf{v}(t) = -a \sin t \, \mathbf{i} + a \cos t \, \mathbf{j} \), the differentiation will be:

• Derivative of \( -a \sin t \) is \( -a \cos t \).
• Derivative of \( a \cos t \) is \( -a \sin t \).

Thus, the acceleration vector is: \( \mathbf{a}(t) = -a \cos t \, \mathbf{i} - a \sin t \, \mathbf{j} \). This vector is crucial for understanding how both the speed and direction of the object are changing.

Differentiation in Calculus and its Role

Differentiation is a core concept in calculus that allows us to understand changes. It's the process of finding the derivative, which represents the rate at which a quantity changes. In this context, differentiation helps:

• Derive the velocity vector from the position vector, indicating how fast and in what direction an object moves.
• Calculate the acceleration vector from the velocity vector, revealing how the velocity is changing over time.
• Calculate the tangential component of acceleration, which involves differentiating speed.

Through these derivatives, differentiation provides a comprehensive picture of an object's motion, as shown in the example of calculating the tangential \( a_T \) and normal \( a_N \) components of acceleration.