Problem 19 - Calculus Early Transcendentals: Pearson New International Edition

Question

In Problems 19-30, find the velocity \(\mathbf{v}\), acceleration \(\mathbf{a}\), and speed \(s\) at the indicated time \(t=t_{1}\). \mathbf{r}(t)=4 t \mathbf{i}+5\left(t^{2}-1\right) \mathbf{j}+2 t \mathbf{k} ; t_{1}=1

Step-by-Step Solution

Answer

Velocity is \( 4\mathbf{i} + 10\mathbf{j} + 2\mathbf{k} \), acceleration is \( 10\mathbf{j} \), and speed is \( 2\sqrt{30} \).

Step 1: Find the velocity vector \( \mathbf{v}(t) \)

Velocity is the derivative of the position vector \( \mathbf{r}(t) \) with respect to time \( t \). Compute the derivative:\[ \mathbf{v}(t) = \frac{d}{dt} \left( 4t\mathbf{i} + 5(t^2 - 1)\mathbf{j} + 2t\mathbf{k} \right) = 4\mathbf{i} + 10t\mathbf{j} + 2\mathbf{k} \]

Step 2: Find the acceleration vector \( \mathbf{a}(t) \)

Acceleration is the derivative of the velocity vector \( \mathbf{v}(t) \). Compute the derivative:\[ \mathbf{a}(t) = \frac{d}{dt} \left( 4\mathbf{i} + 10t\mathbf{j} + 2\mathbf{k} \right) = 10\mathbf{j} \]

Step 3: Evaluate the velocity vector at \( t = t_1 \)

Insert \( t_1 = 1 \) into the velocity vector \( \mathbf{v}(t) \):\[ \mathbf{v}(1) = 4\mathbf{i} + 10(1)\mathbf{j} + 2\mathbf{k} = 4\mathbf{i} + 10\mathbf{j} + 2\mathbf{k} \]

Step 4: Evaluate the acceleration vector at \( t = t_1 \)

Insert \( t_1 = 1 \) into the acceleration vector \( \mathbf{a}(t) \):\[ \mathbf{a}(1) = 10\mathbf{j} \]

Step 5: Calculate the speed \( s(t) \) at \( t = t_1 \)

The speed is the magnitude of the velocity vector \( \mathbf{v}(t) \). Compute it for \( t = 1 \):\[ s(1) = \| \mathbf{v}(1) \| = \sqrt{(4)^2 + (10)^2 + (2)^2} = \sqrt{16 + 100 + 4} = \sqrt{120} = 2\sqrt{30} \]

Key Concepts

Velocity calculation

Velocity is all about how fast something is moving and in what direction. In our problem, we start with a position vector \( \mathbf{r}(t) = 4t\mathbf{i} + 5(t^2 - 1)\mathbf{j} + 2t\mathbf{k} \). The first step to finding the velocity vector, \( \mathbf{v}(t) \), is to calculate the derivative of this position vector with respect to time, \( t \). By taking this derivative, we get:

• The \( \mathbf{i} \) component is derived as \( \frac{d}{dt}(4t) = 4 \).
• The \( \mathbf{j} \) component is \( \frac{d}{dt}(5(t^2 - 1)) = 10t \).
• The \( \mathbf{k} \) component becomes \( \frac{d}{dt}(2t) = 2 \).

These steps help us form the velocity vector: \( \mathbf{v}(t) = 4\mathbf{i} + 10t\mathbf{j} + 2\mathbf{k} \).
A key takeaway is that velocity doesn't just describe how fast, but also in which direction the object is heading. Evaluating this derivative gives a clear picture of the dynamic behavior of the object.

Acceleration determination

Acceleration shows how quickly the velocity is changing, essentially describing how fast speed itself is increasing or decreasing. Given the velocity vector \( \mathbf{v}(t) = 4\mathbf{i} + 10t\mathbf{j} + 2\mathbf{k} \), we determine the acceleration vector \( \mathbf{a}(t) \) by differentiating the velocity vector with respect to time.
Focusing on each component, we find:

• The \( \mathbf{i} \) component, being a constant \( 4 \), has a derivative of 0.
• The \( \mathbf{j} \) component evaluates to \( \frac{d}{dt}(10t) = 10 \).
• The \( \mathbf{k} \) component, a constant \( 2 \), also derives to 0.

Consequently, the acceleration vector simplifies to \( \mathbf{a}(t) = 10\mathbf{j} \), indicating an acceleration solely in the \( \mathbf{j} \) direction.
Understanding acceleration is pivotal as it identifies how quickly velocity changes, aiding in comprehending forces acting on an object.

Speed evaluation

Speed quantifies how fast an object moves irrespective of direction, distinct from velocity. To calculate speed, we need the magnitude of the velocity vector. For our velocity vector \( \mathbf{v}(1) = 4\mathbf{i} + 10\mathbf{j} + 2\mathbf{k} \), the speed at \( t = 1 \) is determined by the formula for the magnitude:
\[ s(1) = \sqrt{(4)^2 + (10)^2 + (2)^2} = \sqrt{16 + 100 + 4} = \sqrt{120} = 2\sqrt{30} \]This calculation involves squaring each component of the velocity vector, summing these squares, and taking the square root of the result.
Summing up, speed is the absolute value of how fast something is moving, providing insight into motion without regard for directionality.