91Ó°ÊÓ

In calculus, the position vector is a key concept used to determine a particle's location in a plane or space. The position vector \( \mathbf{r}(t) \) describes how the position of a particle changes over time. It is represented as a vector function, and in our exercise, it is given by:\[ \mathbf{r}(t) = (\cos t + t \sin t) \mathbf{i} + (\sin t - t \cos t) \mathbf{j}. \]

• The component \( (\cos t + t \sin t) \mathbf{i} \) represents the horizontal movement.
• The component \( (\sin t - t \cos t) \mathbf{j} \) represents the vertical movement.

Understanding the position vector is essential because it forms the foundation from which we derive other important vectors like velocity and acceleration.

The velocity vector \( \mathbf{v}(t) \) provides information about the particle's speed and direction at any given time. It is the derivative of the position vector \( \mathbf{r}(t) \) with respect to time \( t \). This derivative gives us an insight on how the position of the particle changes per unit time:\[ \mathbf{v}(t) = \frac{d}{dt}(\cos t + t \sin t) \mathbf{i} + \frac{d}{dt}(\sin t - t \cos t) \mathbf{j}. \]When we calculate this, we find:\[ \mathbf{v}(t) = t \cos t \mathbf{i} + t \sin t \mathbf{j}. \]

• The term \( t \cos t \mathbf{i} \) refers to the horizontal component of velocity.
• The term \( t \sin t \mathbf{j} \) refers to the vertical component of velocity.

The magnitude of this vector gives us the speed of the particle, confirming that\( \frac{ds}{dt} = t \), verifying that the speed increases directly with time as indicated in the solution.

The acceleration vector \( \mathbf{a}(t) \) describes how the velocity of a particle changes over time. This vector is the derivative of the velocity vector \( \mathbf{v}(t) \):\[ \mathbf{a}(t) = \frac{d}{dt}(t \cos t) \mathbf{i} + \frac{d}{dt}(t \sin t) \mathbf{j}. \]After differentiation, we derive:\[ \mathbf{a}(t) = (\cos t - t \sin t) \mathbf{i} + (\sin t + t \cos t) \mathbf{j}. \]

• \( \cos t - t \sin t \) represents the horizontal component of acceleration.
• \( \sin t + t \cos t \) represents the vertical component of acceleration.

This vector provides crucial information about the forces influencing the particle and how its path is altered over time.

The tangential and normal components of acceleration further dissect the overall acceleration into two insightful parts:

• Tangential Component \( (a_T) \): This measures the rate of change of speed, essentially the derivative of the magnitude of the velocity vector. For our exercise, since \( \|\mathbf{v}(t)\| = t \), this rate is constant at 1, confirming \( a_T = 1 \).
• Normal Component \( (a_N) \): This captures the rate at which the direction of the velocity vector changes, often referred to as the radial or centripetal acceleration. It's computed using:\[ a_N = \sqrt{\|\mathbf{a}(t)\|^2 - a_T^2} \]which simplifies to \( a_N = t \) in our scenario.

The tangent component influences the speed, while the normal affects the trajectory, together defining the motion dynamics of the particle.