Problem 16 - Thomas Calculus

Question

\({r}(t)\) is the position of a particle in space at time \(t .\) Find the angle between the velocity and acceleration vectors at time \(t=0 .\) \begin{equation} \mathbf{r}(t)=\left(\frac{\sqrt{2}}{2} t\right) \mathbf{i}+\left(\frac{\sqrt{2}}{2} t-16 t^{2}\right) \mathbf{j} \end{equation}

Step-by-Step Solution

Answer

Angle at \(t=0\) is \(135^\circ\) or \(\frac{3\pi}{4}\) radians.

Step 1: Find the Velocity Vector

The velocity vector \( \mathbf{v}(t) \) is the derivative of the position vector \( \mathbf{r}(t) \). Differentiate \( \mathbf{r}(t) \) with respect to \( t \): \[ \mathbf{v}(t) = \frac{d}{dt}\left( \frac{\sqrt{2}}{2} t \right) \mathbf{i} + \frac{d}{dt}\left( \frac{\sqrt{2}}{2} t - 16 t^2 \right) \mathbf{j} \] Calculating the derivatives, we get \[ \mathbf{v}(t) = \frac{\sqrt{2}}{2} \mathbf{i} + \left( \frac{\sqrt{2}}{2} - 32t \right) \mathbf{j} \]

Step 2: Find the Acceleration Vector

The acceleration vector \( \mathbf{a}(t) \) is the derivative of the velocity vector \( \mathbf{v}(t) \). Differentiate \( \mathbf{v}(t) \) with respect to \( t \): \[ \mathbf{a}(t) = \frac{d}{dt}\left( \frac{\sqrt{2}}{2} \right) \mathbf{i} + \frac{d}{dt}\left( \frac{\sqrt{2}}{2} - 32t \right) \mathbf{j} \] Calculating the derivatives, we find \[ \mathbf{a}(t) = 0 \mathbf{i} - 32 \mathbf{j} \] So, \( \mathbf{a}(t) = -32 \mathbf{j} \).

Step 3: Evaluate Vectors at t=0

Substitute \( t = 0 \) into \( \mathbf{v}(t) \) and \( \mathbf{a}(t) \): \[ \mathbf{v}(0) = \frac{\sqrt{2}}{2} \mathbf{i} + \frac{\sqrt{2}}{2} \mathbf{j} \] \[ \mathbf{a}(0) = -32 \mathbf{j} \]

Step 4: Calculate the Dot Product

The dot product of the vectors \( \mathbf{v}(0) \) and \( \mathbf{a}(0) \) is: \[ \mathbf{v}(0) \cdot \mathbf{a}(0) = \left( \frac{\sqrt{2}}{2} \right)(0) + \left( \frac{\sqrt{2}}{2} \right)(-32) = -16\sqrt{2} \]

Step 5: Calculate Magnitudes

Calculate the magnitudes of \( \mathbf{v}(0) \) and \( \mathbf{a}(0) \): \[ \| \mathbf{v}(0) \| = \sqrt{\left( \frac{\sqrt{2}}{2} \right)^2 + \left( \frac{\sqrt{2}}{2} \right)^2} = 1 \] \[ \| \mathbf{a}(0) \| = \sqrt{(-32)^2} = 32 \]

Step 6: Find the Angle

Use the dot product formula \( \mathbf{v} \cdot \mathbf{a} = \| \mathbf{v} \| \| \mathbf{a} \| \cos(\theta) \) to find the angle: \[ -16\sqrt{2} = 1 \times 32 \times \cos(\theta) \] \[ \cos(\theta) = \frac{-16\sqrt{2}}{32} = -\frac{\sqrt{2}}{2} \] \( \theta = 135^\circ \) or \( \theta = \frac{3\pi}{4} \) radians.

Key Concepts

Velocity and Acceleration Vectors

In vector calculus, a particle's movement in space can be described using vectors. One common task is finding the velocity and acceleration vectors from a particle's position vector. Here, the position vector \( \mathbf{r}(t) \) is dependent on time \( t \). The velocity vector \( \mathbf{v}(t) \) is obtained by differentiating the position vector with respect to \( t \). Simply put, we find the rate of change of position over time. For the given function \( \mathbf{r}(t) = \left( \frac{\sqrt{2}}{2} t \right) \mathbf{i} + \left( \frac{\sqrt{2}}{2} t - 16t^2 \right) \mathbf{j} \), differentiate each component separately. This results in the velocity vector \( \mathbf{v}(t) = \frac{\sqrt{2}}{2} \mathbf{i} + \left( \frac{\sqrt{2}}{2} - 32t \right) \mathbf{j} \). Next, by differentiating the velocity vector, we obtain the acceleration vector \( \mathbf{a}(t) \). This represents how the velocity changes over time. In this exercise, \( \mathbf{a}(t) = -32 \mathbf{j} \), indicating a constant acceleration in the \( \mathbf{j} \) direction. Both of these vectors are essential for understanding the motion of the particle.

Dot Product

The dot product is a fundamental operation in vector calculus, which measures the degree of alignment between two vectors. It is calculated by multiplying corresponding components of the vectors and summing these products. This exercise requires calculating the dot product between the velocity vector \( \mathbf{v}(0) \) and the acceleration vector \( \mathbf{a}(0) \) at time \( t = 0 \). The vectors at this moment are \( \mathbf{v}(0) = \frac{\sqrt{2}}{2} \mathbf{i} + \frac{\sqrt{2}}{2} \mathbf{j} \) and \( \mathbf{a}(0) = -32 \mathbf{j} \). To find the dot product:

• Multiply the \( \mathbf{i} \) components: \( \frac{\sqrt{2}}{2} \times 0 = 0 \)
• Multiply the \( \mathbf{j} \) components: \( \frac{\sqrt{2}}{2} \times (-32) = -16\sqrt{2} \)

Sum these results to get the dot product: \(-16\sqrt{2}\). The significance of a negative dot product suggests that the vectors point in somewhat opposite directions.

Angle Between Vectors

Finding the angle between two vectors involves using the dot product formula, which relates the dot product to the magnitudes of the vectors and the cosine of the angle between them. The formula is: \( \mathbf{v} \cdot \mathbf{a} = \| \mathbf{v} \| \| \mathbf{a} \| \cos(\theta) \). Given the calculated dot product \( -16\sqrt{2} \), and the magnitudes, \( \| \mathbf{v}(0) \| = 1 \) and \( \| \mathbf{a}(0) \| = 32 \), substitute these values into the formula to solve for \( \cos(\theta) \).The calculations yield \( \cos(\theta) = -\frac{\sqrt{2}}{2} \). Since the cosine of an angle in a right triangle also represents an angle measure, this result corresponds to probabilities for \( \theta \) being \( 135^\circ \) or \( \frac{3\pi}{4} \) radians. Geometrically, this means the velocity and acceleration vectors form an obtuse angle, highlighting their tendency to act in opposing directions over time.