91Ó°ÊÓ

The position vector \(\mathbf{r}\) describes the position of a particle in space as a function of time \(t\). In curvilinear motion, this vector can change direction and magnitude over time. For our case, the position vector is given by:
\[ \mathbf{r} = (t^3 - t^2) \mathbf{i} + t^4 \mathbf{j} \] Here, the components \(t^3 - t^2\) and \(t^4\) provide the particle's position in the x (\mathbf{i}) and y (\mathbf{j}) directions respectively. The term 'curvilinear' implies that the particle doesn't move in a straight line but follows a curve, hence the position vector isn't linear in terms of \(t\). Each component must be understood separately, and changes over time, giving different positions at each instant. Thus, it’s important to take derivatives to find velocity and acceleration.

The velocity vector \(\mathbf{v}\) is the first derivative of the position vector with respect to time, representing how fast and in what direction the position is changing. Mathematically,
\[ \mathbf{v} = \frac{d\mathbf{r}}{dt} \] For the given \(\mathbf{r}\), computing the derivative:
\[ \mathbf{r} = (t^3 - t^2) \mathbf{i} + t^4 \mathbf{j} \implies \mathbf{v} = \frac{d}{dt} [(t^3 - t^2) \mathbf{i} + t^4 \mathbf{j}] = (3t^2 - 2t)i + 4t^3 \mathbf{j} \] This vector gives us the rate of change of both the x and y components of the position. At any instant \(t\), \(\mathbf{v}\) tells how fast and in which direction the particle is moving.

The acceleration vector \(\mathbf{a}\) is the derivative of the velocity vector with respect to time, representing the rate of change of velocity itself. Mathematically,
\[ \mathbf{a} = \frac{d\mathbf{v}}{dt} \] For the velocity vector,
\[ \mathbf{v} = (3t^2 - 2t) \mathbf{i} + 4t^3 \mathbf{j} \] Taking the derivative:
\[ \mathbf{a} = \frac{d}{dt} [(3t^2 - 2t) \mathbf{i} + 4t^3 \mathbf{j}] = (6t - 2) \mathbf{i} + 12t^2 \mathbf{j} \] This formula gives a measure of how the velocity is changing over time, including changes in magnitude and direction.

The dot product (or scalar product) between two vectors measures how much one vector extends in the direction of the other. It’s given by:
\[ \mathbf{u} \cdot \mathbf{v} = u_x v_x + u_y v_y \] For our vectors \(\mathbf{v} = \mathbf{i} + 4\mathbf{j}\) and \(\mathbf{a} = 4\mathbf{i} + 12\mathbf{j}\), the dot product is calculated as,
\[ \mathbf{v} \cdot \mathbf{a} = (1)(4) + (4)(12) = 4 + 48 = 52 \] This result is key in finding the angle between two vectors, showing how their directions align.

To find the angle \(\theta\) between two vectors, we use the dot product formula:
\[ \mathbf{v} \cdot \mathbf{a} = \|\mathbf{v}\| \|\mathbf{a}\| \cos(\theta) \] We already found the dot product \(\mathbf{v} \cdot \mathbf{a} = 52\). Next, calculate the magnitudes:
\[ \|\mathbf{v}\| = \sqrt{1^2 + 4^2} = \sqrt{17} \]
\[ \|\mathbf{a}\| = \sqrt{4^2 + 12^2} = 4\sqrt{10} \]
Hence,
\[ \cos(\theta) = \frac{52}{\sqrt{17} \cdot 4\sqrt{10}} = \frac{13}{\sqrt{170}} = \frac{13 \sqrt{170}}{170} \]
Finally, \ \theta = \cos^{-1}\left(\frac{13 \sqrt{170}}{170}\right) \ gives the angle in degrees.