Problem 25 - Calculus Early Transcendentals: Pearson New International Edition

Question

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

Step-by-Step Solution

Answer

\( \mathbf{v} = -\mathbf{j} + \mathbf{k}, \mathbf{a} = \mathbf{i}, s = \sqrt{2} \) at \( t = \pi \).

Step 1: Differentiate to Find Velocity

To find the velocity vector \( \mathbf{v} \), differentiate the position vector \( \mathbf{r}(t) = \cos t \mathbf{i} + \sin t \mathbf{j} + t \mathbf{k} \) with respect to \( t \). This gives us:\[ \mathbf{v}(t) = \frac{d}{dt}(\cos t) \mathbf{i} + \frac{d}{dt}(\sin t) \mathbf{j} + \frac{d}{dt}(t) \mathbf{k} = -\sin t \mathbf{i} + \cos t \mathbf{j} + \mathbf{k} \]

Step 2: Evaluate Velocity at Indicated Time

Substitute \( t = \pi \) into the velocity vector expression:\[ \mathbf{v}(\pi) = -\sin(\pi) \mathbf{i} + \cos(\pi) \mathbf{j} + \mathbf{k} = 0 \mathbf{i} - \mathbf{j} + \mathbf{k} = -\mathbf{j} + \mathbf{k} \]

Step 3: Differentiate Velocity to Find Acceleration

To find the acceleration vector \( \mathbf{a} \), differentiate the velocity vector \( \mathbf{v}(t) = -\sin t \mathbf{i} + \cos t \mathbf{j} + \mathbf{k} \) with respect to \( t \). This gives:\[ \mathbf{a}(t) = \frac{d}{dt}(-\sin t) \mathbf{i} + \frac{d}{dt}(\cos t) \mathbf{j} + \frac{d}{dt}(\mathbf{k}) = -\cos t \mathbf{i} - \sin t \mathbf{j} + 0 \mathbf{k} = -\cos t \mathbf{i} - \sin t \mathbf{j} \]

Step 4: Evaluate Acceleration at Indicated Time

Substitute \( t = \pi \) into the acceleration vector expression:\[ \mathbf{a}(\pi) = -\cos(\pi) \mathbf{i} - \sin(\pi) \mathbf{j} = \mathbf{i} \]

Step 5: Calculate Speed

The speed \( s \) is the magnitude of the velocity vector \( \mathbf{v} \). First, calculate:\[ ||\mathbf{v}(\pi)|| = \sqrt{(0)^2 + (-1)^2 + (1)^2} = \sqrt{0 + 1 + 1} = \sqrt{2} \]

Step 6: Compile Results

At \( t = \pi \), the velocity \( \mathbf{v} = -\mathbf{j} + \mathbf{k} \), acceleration \( \mathbf{a} = \mathbf{i} \), and speed \( s = \sqrt{2} \).

Key Concepts

Velocity and Acceleration

In vector calculus, understanding velocity and acceleration is crucial, especially when dealing with motion in three-dimensional space. Velocity is the rate of change of the position vector over time. It can be seen as the derivative of the position vector. For any given position function, the velocity vector tells you the direction and speed at which a point is moving at any specific time. In our exercise, the position vector is given as \( \(\mathbf{r}(t)=\cos t \mathbf{i}+\sin t \mathbf{j}+t \mathbf{k} \)\). When we differentiate this with respect to \( t \), we get the velocity vector:\[ \mathbf{v}(t) = -\sin t \mathbf{i} + \cos t \mathbf{j} + \mathbf{k} \] Accurate calculations and comprehension of the velocity vector allow us to predict a point's future position and motion path. Acceleration, on the other hand, is the rate of change of the velocity vector over time. By differentiating the velocity vector, we derive the acceleration vector, which informs us about changes in speed and direction:\[ \mathbf{a}(t) = -\cos t \mathbf{i} - \sin t \mathbf{j} \] Acceleration is particularly important because it helps to understand the forces acting on the object. Calculating these vectors gives us more detail and insights into the movement dynamics.

Differentiation

Differentiation is a fundamental concept in calculus, used to compute the rate at which a function is changing at any given point. It forms the core process of understanding motion via derivatives of position and velocity vectors. In the given exercise, we start by differentiating the position vector \( \(\mathbf{r}(t)=\cos t \mathbf{i}+\sin t \mathbf{j}+t \mathbf{k} \)\). This process yields the velocity vector \( \mathbf{v}(t) = -\sin t \mathbf{i} + \cos t \mathbf{j} + \mathbf{k} \), representing how fast and in what direction the point is moving.Here's what happens during this differentiation:

• \(\frac{d}{dt}(\cos t) \rightarrow -\sin t\): Differentiating cosine yields negative sine.
• \(\frac{d}{dt}(\sin t) \rightarrow \cos t\): Differentiating sine yields cosine.
• \(\frac{d}{dt}(t) \rightarrow 1\): Differentiating t with respect to t yields 1.

Following this, differentiating the velocity vector provides the acceleration vector \( \mathbf{a}(t) = -\cos t \mathbf{i} - \sin t \mathbf{j} \). By understanding the derivatives, especially in terms of physical movement, we gain deeper insight into how dynamic systems evolve over time.

Magnitude of a Vector

The magnitude of a vector, often referred to as the vector's "length", gives us its size or extent but not its direction. In the context of physics, the magnitude of velocity is equivalent to the speed of an object at a given moment. To find the speed, you calculate the magnitude of the velocity vector. For our exercise, we find the velocity vector at \( t = \pi \), which is \( -\mathbf{j} + \mathbf{k} \), and then calculate its magnitude:\[ ||\mathbf{v}(\pi)|| = \sqrt{(-1)^2 + (1)^2} = \sqrt{2} \]Here’s how the magnitude is determined step by step:

• Square each component of the vector.
• Sum these squared values.
• Take the square root of this sum to get the magnitude.

The computation shows that the speed at \( t = \pi \) is \( \sqrt{2} \). This scalar quantity provides a direct sense of how fast the point is moving, without considering direction, which is key in understanding the object's kinetic properties.