Problem 200 - Calculus Volume 3

Question

Suppose that the position function for an object in three dimensions is given by the equation \(\mathbf{r}(t)=t \cos (t) \mathbf{i}+t \sin (t) \mathbf{j}+3 t \mathbf{k}\) Find the angle between the velocity and acceleration vectors when \(t=1.5\).

Step-by-Step Solution

Answer

The angle between the velocity and acceleration vectors at \( t = 1.5 \) is computed using the inverse cosine of the dot product over the magnitudes.

Step 1: Find the Velocity Vector

To find the velocity vector, differentiate the position function \( \mathbf{r}(t) = t \cos(t) \mathbf{i} + t \sin(t) \mathbf{j} + 3t \mathbf{k} \) with respect to \(t\). Using the product rule and chain rule, we get:\[\mathbf{v}(t) = \frac{d\mathbf{r}}{dt} = \left( \cos(t) - t \sin(t) \right) \mathbf{i} + \left( \sin(t) + t \cos(t) \right) \mathbf{j} + 3 \mathbf{k}\]

Step 2: Compute the Velocity at t=1.5

Substitute \(t = 1.5\) into the velocity vector:\[\mathbf{v}(1.5) = \left( \cos(1.5) - 1.5 \sin(1.5) \right) \mathbf{i} + \left( \sin(1.5) + 1.5 \cos(1.5) \right) \mathbf{j} + 3 \mathbf{k}\]

Step 3: Find the Acceleration Vector

Differentiate the velocity vector to get the acceleration vector. Applying the product rule and chain rule again, we get:\[\mathbf{a}(t) = \frac{d\mathbf{v}}{dt} = \left( -2\sin(t) - t \cos(t) \right) \mathbf{i} + \left( 2\cos(t) - t \sin(t) \right) \mathbf{j} \]

Step 4: Compute the Acceleration at t=1.5

Substitute \(t = 1.5\) into the acceleration vector:\[\mathbf{a}(1.5) = \left( -2\sin(1.5) - 1.5\cos(1.5) \right) \mathbf{i} + \left( 2\cos(1.5) - 1.5\sin(1.5) \right) \mathbf{j}\]

Step 5: Calculate the Dot Product of Velocity and Acceleration

Compute the dot product \( \mathbf{v}(1.5) \cdot \mathbf{a}(1.5) \). Use the components found by substituting into \( \mathbf{v}(t) \) and \( \mathbf{a}(t) \) at \( t = 1.5 \). Calculate the product of corresponding components and sum them.

Step 6: Find Magnitudes of Vectors

Calculate the magnitudes of \(\mathbf{v}(1.5)\) and \(\mathbf{a}(1.5)\) using the formula \( \| \mathbf{x} \| = \sqrt{x_1^2 + x_2^2 + x_3^2} \).

Step 7: Apply Cosine Formula for Angle

The cosine of the angle \( \theta \) between two vectors \( \mathbf{v} \) and \( \mathbf{a} \) is given by the formula:\[ \cos(\theta) = \frac{ \mathbf{v} \cdot \mathbf{a} }{ \| \mathbf{v} \| \| \mathbf{a} \| } \]Solve for \( \theta = \cos^{-1}\left( \frac{ \mathbf{v}(1.5) \cdot \mathbf{a}(1.5) }{ \| \mathbf{v}(1.5) \| \| \mathbf{a}(1.5) \| } \right) \).

Step 8: Conclusion and Calculation

After performing the necessary arithmetic for the dot product and magnitudes, calculate the value from the cosine formula to find \( \theta \). This value represents the angle between the velocity and acceleration vectors at \( t = 1.5 \).

Key Concepts

Position Function

The position function is a fundamental concept in vector calculus. It describes the location of an object in space as a function of time. In our exercise, the position function is given by \( \mathbf{r}(t) = t \cos(t) \mathbf{i} + t \sin(t) \mathbf{j} + 3t \mathbf{k} \). Each term in this function contributes to the object's position along the \( x \), \( y \), and \( z \) axes, respectively.

• \( t \cos(t) \mathbf{i} \): Describes motion along the \( x \)-axis.
• \( t \sin(t) \mathbf{j} \): Describes motion along the \( y \)-axis.
• \( 3t \mathbf{k} \): Represents constant motion along the \( z \)-axis, scaling with time.

Understanding the position function allows you to determine the path of an object as it moves through three-dimensional space.

Velocity Vector

The velocity vector represents the rate of change of the position function with respect to time. Essentially, it indicates how fast and in which direction the position of an object is changing.To find the velocity vector from the position function, we differentiate \( \mathbf{r}(t) \). For our exercise, the velocity vector is derived as:\[\mathbf{v}(t) = \left( \cos(t) - t \sin(t) \right) \mathbf{i} + \left( \sin(t) + t \cos(t) \right) \mathbf{j} + 3 \mathbf{k}\]The components of this vector consist of:

• \( \cos(t) - t \sin(t) \) for the \( x \)-axis.
• \( \sin(t) + t \cos(t) \) for the \( y \)-axis.
• Constant value \( 3 \) for the \( z \)-axis, showing steady upward motion.

The velocity informs us of how the object is moving in its path.

Acceleration Vector

The acceleration vector is found by differentiating the velocity vector. It describes how the velocity of an object changes with time. This vector provides critical insights into any changes in the speed and direction of motion.In our problem, the acceleration vector is:\[\mathbf{a}(t) = \left( -2\sin(t) - t \cos(t) \right) \mathbf{i} + \left( 2\cos(t) - t \sin(t) \right) \mathbf{j} \]Key components:

• \( -2\sin(t) - t \cos(t) \) along the \( x \)-axis.
• \( 2\cos(t) - t \sin(t) \) along the \( y \)-axis.

Interestingly, our acceleration vector indicates varying changes in motion along \( x \) and \( y \), without a constant acceleration for \( z \).

Dot Product

The dot product of two vectors is a scalar representing the magnitude of their interaction. For vectors \( \mathbf{v} \) and \( \mathbf{a} \), used in our calculation:\[\mathbf{v} \cdot \mathbf{a} = (v_i \cdot a_i) + (v_j \cdot a_j) + (v_k \cdot a_k)\]It involves summing the products of corresponding components of the vectors. This product is crucial in finding angles between vectors:

• Helps in calculating cosine of the angle.
• Highlights parallelism (if zero) or orthogonality between vectors.

Here, computing \( \mathbf{v}(1.5) \cdot \mathbf{a}(1.5) \) is vital in assessing the relationship between velocity and acceleration.

Angle Between Vectors

The angle between two vectors is determined by their dot product and magnitudes. The relationship is defined by:\[\cos(\theta) = \frac{\mathbf{v} \cdot \mathbf{a}}{\|\mathbf{v}\| \|\mathbf{a}\|}\]Calculating this assists in understanding the directional relationship between vectors at a specific time \( t \). If \( \theta \) is known, it indicates:

• Zero angle: vectors are in the same direction.
• \(90^\circ\) angle: vectors are perpendicular.
• Obtuse angle: vectors point more away from each other.

In our problem, using \( \cos^{-1} \) allows us to determine the precise angle at \( t = 1.5 \), encapsulating the interaction between velocity and acceleration as the object moves.