Problem 49 - Calculus III

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 \(\theta\).

Step 1: Find the Velocity Vector

Differentiate the position vector \( \mathbf{r}(t) = t \cos(t) \mathbf{i} + t \sin(t) \mathbf{j} + 3t \mathbf{k} \) with respect to \(t\) to find the velocity vector \( \mathbf{v}(t) \).\[ \mathbf{v}(t) = \frac{d}{dt} (t \cos(t)) \mathbf{i} + \frac{d}{dt} (t \sin(t)) \mathbf{j} + \frac{d}{dt} (3t) \mathbf{k} \]Using the product rule:\[ \frac{d}{dt} (t \cos(t)) = \cos(t) - t \sin(t) \]\[ \frac{d}{dt} (t \sin(t)) = \sin(t) + t \cos(t) \]Thus, the velocity vector is:\[ \mathbf{v}(t) = (\cos(t) - t \sin(t)) \mathbf{i} + (\sin(t) + t \cos(t)) \mathbf{j} + 3 \mathbf{k} \]

Step 2: Find the Acceleration Vector

Differentiate the velocity vector from Step 1 to find the acceleration vector \( \mathbf{a}(t) \).\[ \mathbf{a}(t) = \frac{d}{dt} (\cos(t) - t \sin(t)) \mathbf{i} + \frac{d}{dt} (\sin(t) + t \cos(t)) \mathbf{j} + \frac{d}{dt} (3) \mathbf{k} \]Calculate each component:\[ \frac{d}{dt} (\cos(t) - t \sin(t)) = -\sin(t) - \sin(t) - t \cos(t) = -2\sin(t) - t \cos(t) \]\[ \frac{d}{dt} (\sin(t) + t \cos(t)) = \cos(t) + \cos(t) - t \sin(t) = 2\cos(t) - t \sin(t) \]So, the acceleration vector is:\[ \mathbf{a}(t) = (-2\sin(t) - t \cos(t)) \mathbf{i} + (2\cos(t) - t \sin(t)) \mathbf{j} \]

Step 3: Evaluate Vectors at \(t = 1.5\)

Substitute \( t = 1.5 \) into the velocity and acceleration vectors from Steps 1 and 2:Velocity vector:\[ \mathbf{v}(1.5) = (\cos(1.5) - 1.5 \sin(1.5)) \mathbf{i} + (\sin(1.5) + 1.5 \cos(1.5)) \mathbf{j} + 3 \mathbf{k} \]Acceleration vector:\[ \mathbf{a}(1.5) = (-2\sin(1.5) - 1.5 \cos(1.5)) \mathbf{i} + (2\cos(1.5) - 1.5 \sin(1.5)) \mathbf{j} \]

Step 4: Compute the Dot Product of Velocity and Acceleration

Calculate the dot product \( \mathbf{v}(1.5) \cdot \mathbf{a}(1.5) \) to utilize it in finding the angle.\[ \mathbf{v}(1.5) \cdot \mathbf{a}(1.5) = [(\cos(1.5) - 1.5 \sin(1.5))(-2\sin(1.5) - 1.5 \cos(1.5)) + (\sin(1.5) + 1.5 \cos(1.5))(2\cos(1.5) - 1.5 \sin(1.5)) + 0] \]

Step 5: Magnitudes of Velocity and Acceleration

Compute the magnitudes of \( \mathbf{v}(1.5) \) and \( \mathbf{a}(1.5) \).Magnitude of \( \mathbf{v}(1.5) \):\[ ||\mathbf{v}(1.5)|| = \sqrt{(\cos(1.5) - 1.5 \sin(1.5))^2 + (\sin(1.5) + 1.5 \cos(1.5))^2 + 3^2} \]Magnitude of \( \mathbf{a}(1.5) \):\[ ||\mathbf{a}(1.5)|| = \sqrt{(-2\sin(1.5) - 1.5 \cos(1.5))^2 + (2\cos(1.5) - 1.5 \sin(1.5))^2 } \]

Step 6: Find the Angle between the Vectors

Use the formula for the angle \( \theta \) between two vectors: \[ \cos(\theta) = \frac{\mathbf{v}(1.5) \cdot \mathbf{a}(1.5)}{||\mathbf{v}(1.5)|| \times ||\mathbf{a}(1.5)||} \]Compute \( \theta = \cos^{-1} \left(\frac{\mathbf{v}(1.5) \cdot \mathbf{a}(1.5)}{||\mathbf{v}(1.5)|| \times ||\mathbf{a}(1.5)||}\right) \) using calculated values.

Key Concepts

Velocity Vector

A velocity vector describes how fast an object is moving and in what direction. In the context of a position function, it is derived from the first derivative of the position vector concerning time. To identify the velocity vector from a given position function, you differentiate each component of the position vector separately with respect to the parameter, usually time (\( t \)).

For example, given a position function \( \mathbf{r}(t) = t \cos(t) \mathbf{i} + t \sin(t) \mathbf{j} + 3t \mathbf{k} \), we differentiate each term:

• \( \frac{d}{dt}(t \cos(t)) = \cos(t) - t \sin(t) \)
• \( \frac{d}{dt}(t \sin(t)) = \sin(t) + t \cos(t) \)
• \( \frac{d}{dt}(3t) = 3 \)

Hence, the velocity vector is \( \mathbf{v}(t) = (\cos(t) - t \sin(t)) \mathbf{i} + (\sin(t) + t \cos(t)) \mathbf{j} + 3 \mathbf{k} \). This expression tells us both the direction and magnitude of movement.

Acceleration Vector

An acceleration vector indicates how the velocity vector of an object changes over time. It highlights the changes in speed and direction. The process to find the acceleration vector involves differentiating the velocity vector with respect to time.

Given a velocity vector from the differentiation in the earlier step, like \( \mathbf{v}(t) = (\cos(t) - t \sin(t)) \mathbf{i} + (\sin(t) + t \cos(t)) \mathbf{j} + 3 \mathbf{k} \), we differentiate this to find the acceleration:

• \( \frac{d}{dt}(\cos(t) - t \sin(t)) = -2\sin(t) - t \cos(t) \)
• \( \frac{d}{dt}(\sin(t) + t \cos(t)) = 2\cos(t) - t \sin(t) \)
• \( \frac{d}{dt}(3) = 0 \)

Therefore, the acceleration vector is \( \mathbf{a}(t) = (-2\sin(t) - t \cos(t)) \mathbf{i} + (2\cos(t) - t \sin(t)) \mathbf{j} \). This vector describes how the velocity of the object is changing over time.

Dot Product

The dot product between two vectors is crucial in finding the angle between them. It provides a scalar value that indicates how much one vector goes in the direction of the other. The dot product is computed using the formula:

• \( \mathbf{u} \cdot \mathbf{v} = u_1 v_1 + u_2 v_2 + u_3 v_3 \) for vectors \( \mathbf{u} = (u_1, u_2, u_3) \) and \( \mathbf{v} = (v_1, v_2, v_3) \).

This operation is particularly important when it comes to finding the angle between vectors because it directly relates to the cosine formula used later. For the exercise we discussed:

• Given \( \mathbf{v}(t) \) and \( \mathbf{a}(t) \), substituting \( t = 1.5 \) and calculating \( \mathbf{v}(1.5) \cdot \mathbf{a}(1.5) \) provides the crucial value needed for further calculations.

Product Rule

In calculus, the product rule is a technique used for differentiating functions that are products of two or more functions. It's essential for determining derivatives where terms involve products of functions of \( t \).

The rule is stated as follows: if \( u(t) \) and \( v(t) \) are functions of \( t \), the derivative of their product is:

• \( \frac{d}{dt}[u(t)v(t)] = u'(t)v(t) + u(t)v'(t) \)

In the example of the velocity vector, for instance, to differentiate \( t \cos(t) \), you apply:

• \( \frac{d}{dt}(t \cos(t)) = \frac{d}{dt}(t) \cdot \cos(t) + t \cdot \frac{d}{dt}(\cos(t)) = \cos(t) - t \sin(t) \)

The product rule provides a systematic way to deal with products of variables and is indispensable in the derivation process of both velocity and acceleration vectors.

Cosine Formula

The cosine formula, especially in the context of vectors, is used to find the angle between two vectors. It connects the dot product, magnitudes of vectors, and the cosine of the angle between them:

• \( \cos(\theta) = \frac{\mathbf{u} \cdot \mathbf{v}}{||\mathbf{u}|| \times ||\mathbf{v}||} \)

This is directly applied after calculating the dot product of the velocity and acceleration vectors, as well as their magnitudes. With these values, the cosine of the angle can be found, and thus the angle itself can be determined by solving for \( \theta \) as:

• \( \theta = \cos^{-1}\left(\frac{\mathbf{v}(1.5) \cdot \mathbf{a}(1.5)}{||\mathbf{v}(1.5)|| \times ||\mathbf{a}(1.5)||}\right) \)

Understanding each part of the process aids in comprehending the interaction between vectors and how the principles of trigonometry and algebra play a part in this context.