Question

Find the velocity \(\mathbf{v}\), acceleration \(\mathbf{a}\), and speed \(s\) at the indicated time \(t=t_{1}\). $$ \mathbf{r}(t)=\sin 2 t \mathbf{i}+\cos 3 t \mathbf{j}+\cos 4 t \mathbf{k} ; t_{1}=\frac{\pi}{2} $$

Short answer

Velocity: \(-2 \mathbf{i} + 3 \mathbf{j}\), Acceleration: \(-16 \mathbf{k}\), Speed: \(\sqrt{13}\)

Step-by-step solution

Differentiate position vector to get velocity

The position vector is \( \mathbf{r}(t) = \sin 2t \mathbf{i} + \cos 3t \mathbf{j} + \cos 4t \mathbf{k} \). To find the velocity vector \( \mathbf{v}(t) \), we differentiate each component with respect to \( t \):\[ \mathbf{v}(t) = \frac{d}{dt}(\sin 2t) \mathbf{i} + \frac{d}{dt}(\cos 3t) \mathbf{j} + \frac{d}{dt}(\cos 4t) \mathbf{k} \] Apply the chain rule: \[ \mathbf{v}(t) = 2 \cos 2t \mathbf{i} - 3 \sin 3t \mathbf{j} - 4 \sin 4t \mathbf{k} \]

Evaluate velocity at the indicated time

Substitute \( t_1 = \frac{\pi}{2} \) into the velocity expression: \[ \mathbf{v}\left(\frac{\pi}{2}\right) = 2 \cos\left(2 \cdot \frac{\pi}{2}\right) \mathbf{i} - 3 \sin\left(3 \cdot \frac{\pi}{2}\right) \mathbf{j} - 4 \sin\left(4 \cdot \frac{\pi}{2}\right) \mathbf{k} \] Simplify each term: \[ \mathbf{v}\left(\frac{\pi}{2}\right) = 2 \cdot (-1) \mathbf{i} - 3 \cdot (-1) \mathbf{j} - 4 \cdot 0 \mathbf{k} \] \[ \mathbf{v}\left(\frac{\pi}{2}\right) = -2 \mathbf{i} + 3 \mathbf{j} \]

Differentiate velocity vector for acceleration

Differentiate each component of the velocity \( \mathbf{v}(t) = 2 \cos 2t \mathbf{i} - 3 \sin 3t \mathbf{j} - 4 \sin 4t \mathbf{k} \) with respect to \( t \) to get the acceleration vector \( \mathbf{a}(t) \):\[ \mathbf{a}(t) = \frac{d}{dt}(2 \cos 2t) \mathbf{i} + \frac{d}{dt}(-3 \sin 3t) \mathbf{j} + \frac{d}{dt}(-4 \sin 4t) \mathbf{k} \] \[ \mathbf{a}(t) = -4 \sin 2t \mathbf{i} - 9 \cos 3t \mathbf{j} - 16 \cos 4t \mathbf{k} \]

Evaluate acceleration at the indicated time

Substitute \( t_1 = \frac{\pi}{2} \) into the acceleration expression: \[ \mathbf{a}\left(\frac{\pi}{2}\right) = -4 \sin\left(2 \cdot \frac{\pi}{2}\right) \mathbf{i} - 9 \cos\left(3 \cdot \frac{\pi}{2}\right) \mathbf{j} - 16 \cos\left(4 \cdot \frac{\pi}{2}\right) \mathbf{k} \] Simplify each term: \[ \mathbf{a}\left(\frac{\pi}{2}\right) = -4 \cdot 0 \mathbf{i} - 9 \cdot 0 \mathbf{j} - 16 \cdot 1 \mathbf{k} \] \[ \mathbf{a}\left(\frac{\pi}{2}\right) = -16 \mathbf{k} \]

Compute speed from velocity vector

The speed \( s \) is the magnitude of the velocity vector. Compute \( \|\mathbf{v}\left(\frac{\pi}{2}\right)\| \):\[ s = \sqrt{(-2)^2 + 3^2 + 0^2} = \sqrt{4 + 9} = \sqrt{13} \]

Key concepts

Velocity

In vector calculus, velocity refers to the rate of change of position with time. It is a vector quantity, meaning it has both magnitude and direction. To find velocity, you need to differentiate the position vector with respect to time. The position vector in the problem is given as \( \mathbf{r}(t) = \sin 2t \mathbf{i} + \cos 3t \mathbf{j} + \cos 4t \mathbf{k} \). By applying differentiation, we arrive at the velocity vector \( \mathbf{v}(t) = 2 \cos 2t \mathbf{i} - 3 \sin 3t \mathbf{j} - 4 \sin 4t \mathbf{k} \). This expression tells us how fast and in which direction the object is moving at any time \( t \). Evaluating this at \( t = \frac{\pi}{2} \), we get a specific velocity vector portraying the motion of the object at that instant.

Acceleration

Acceleration is another crucial concept in vector calculus, representing the rate of change of velocity with time. Like velocity, it is also a vector quantity, having direction as well as magnitude. Acceleration can be found by differentiating the velocity vector. In this problem, the velocity found earlier is \( \mathbf{v}(t) = 2 \cos 2t \mathbf{i} - 3 \sin 3t \mathbf{j} - 4 \sin 4t \mathbf{k} \). Differentiating this with respect to \( t \) yields the acceleration vector \( \mathbf{a}(t) = -4 \sin 2t \mathbf{i} - 9 \cos 3t \mathbf{j} - 16 \cos 4t \mathbf{k} \). Evaluating this at \( t = \frac{\pi}{2} \) gives the precise acceleration of the object at that time.

Speed

Speed is the magnitude of velocity, telling us how fast an object is moving regardless of its direction. It is a scalar quantity, which means it has only magnitude and no direction. To compute speed from a velocity vector, we calculate the magnitude of the velocity vector:

• Magnitude is the square root of the sum of the squares of the vector's components.

For the velocity vector \( \mathbf{v}\left(\frac{\pi}{2}\right) = -2 \mathbf{i} + 3 \mathbf{j} \), the speed \( s \) is calculated using \[ s = \sqrt{(-2)^2 + 3^2 + 0^2} = \sqrt{13} \].Thus, the object's speed at \( t = \frac{\pi}{2} \) is \( \sqrt{13} \).

Differentiation

Differentiation is a key mathematical method in calculus, used to compute the rate at which a function is changing at any point. In the context of vector calculus, differentiation helps determine both the velocity and acceleration from a given position vector. For example, differentiating the position vector \( \mathbf{r}(t) \) yields the velocity vector, and further differentiating the velocity vector gives the acceleration. Moreover, differentiation involves applying rules such as the chain rule, which was used here to manage the trigonometric functions like \( \sin \) and \( \cos \). Understanding differentiation allows you to move from position to velocity and finally to acceleration, providing a complete picture of motion.

Position Vector

The position vector is a fundamental concept in describing motion in vector calculus. It describes the position of a point in space relative to a given origin, represented as a vector. In this problem, the position vector is given as \( \mathbf{r}(t) = \sin 2t \mathbf{i} + \cos 3t \mathbf{j} + \cos 4t \mathbf{k} \). Each of its components represents a coordinate function in different directions: \( \mathbf{i}, \mathbf{j}, \) and \( \mathbf{k} \). When you differentiate the position vector concerning time, you derive the velocity, which shows how the position is changing, and further differentiation gives the acceleration, showing how the velocity evolves. The position vector lies at the heart of kinematics analysis, helping to determine various just by differentiation.