Problem 21 - Calculus Early Transcendentals: Pearson New International Edition

Question

Find the velocity \(\mathbf{v}\), acceleration \(\mathbf{a}\), and speed \(s\) at the indicated time \(t=t_{1}\). $$ \mathbf{r}(t)=(1 / t) \mathbf{i}+\left(t^{2}-1\right)^{-1} \mathbf{j}+t^{5} \mathbf{k} ; t_{1}=2 $$

Step-by-Step Solution

Answer

Velocity: \( \mathbf{v}(2) = -\frac{1}{4} \mathbf{i} - \frac{4}{9} \mathbf{j} + 80 \mathbf{k} \). Acceleration: \( \mathbf{a}(2) = \frac{1}{4} \mathbf{i} + \frac{10}{27} \mathbf{j} + 160 \mathbf{k} \). Speed: \( s \approx 80.001 \).

Step 1: Differentiate for Velocity

To find the velocity vector \( \mathbf{v}(t) \), we need to differentiate the position vector \( \mathbf{r}(t) \) with respect to time \( t \). The position vector is given by \( \mathbf{r}(t) = \frac{1}{t} \mathbf{i} + (t^2 - 1)^{-1} \mathbf{j} + t^5 \mathbf{k} \). \( \frac{d}{dt} \left(\frac{1}{t}\right) = -\frac{1}{t^2} \)\( \frac{d}{dt} \left((t^2 - 1)^{-1}\right) = \frac{-2t}{(t^2 - 1)^2} \)\( \frac{d}{dt} \left(t^5\right) = 5t^4 \)Thus, \( \mathbf{v}(t) = -\frac{1}{t^2} \mathbf{i} + \frac{-2t}{(t^2 - 1)^2} \mathbf{j} + 5t^4 \mathbf{k} \).

Step 2: Evaluate Velocity at t=2

Substitute \( t = 2 \) into \( \mathbf{v}(t) \) to find the velocity at \( t_1 = 2 \).\( \mathbf{v}(2) = -\frac{1}{2^2} \mathbf{i} + \frac{-2 \times 2}{(2^2 - 1)^2} \mathbf{j} + 5 \times 2^4 \mathbf{k} \)= \(-\frac{1}{4} \mathbf{i} - \frac{4}{9} \mathbf{j} + 80 \mathbf{k} \).

Step 3: Differentiate Velocity for Acceleration

To find the acceleration vector \( \mathbf{a}(t) \), differentiate the velocity vector \( \mathbf{v}(t) \) with respect to time \( t \).\( \frac{d}{dt} \left(-\frac{1}{t^2}\right) = \frac{2}{t^3} \)\( \frac{d}{dt} \left(\frac{-2t}{(t^2 - 1)^2}\right) = \frac{2(t^2 + 1)}{(t^2 - 1)^3} \)\( \frac{d}{dt} \left(5t^4\right) = 20t^3 \)Thus, \( \mathbf{a}(t) = \frac{2}{t^3} \mathbf{i} + \frac{2(t^2 + 1)}{(t^2 - 1)^3} \mathbf{j} + 20t^3 \mathbf{k} \).

Step 4: Evaluate Acceleration at t=2

Substitute \( t = 2 \) into \( \mathbf{a}(t) \) to determine acceleration at \( t_1 = 2 \).\( \mathbf{a}(2) = \frac{2}{2^3} \mathbf{i} + \frac{2(2^2 + 1)}{(2^2 - 1)^3} \mathbf{j} + 20 \times 2^3 \mathbf{k} \)= \( \frac{1}{4} \mathbf{i} + \frac{10}{27} \mathbf{j} + 160 \mathbf{k} \).

Step 5: Find Speed at t=2

Speed is the magnitude of the velocity vector \( \mathbf{v}(2) \). Use the formula for magnitude:\[ s = \sqrt{\left(-\frac{1}{4}\right)^2 + \left(-\frac{4}{9}\right)^2 + 80^2} \]Calculate:\[ s = \sqrt{\frac{1}{16} + \frac{16}{81} + 6400} = \sqrt{6400.2222} \approx 80.001 \]

Key Concepts

Differentiation

Differentiation is a fundamental concept in calculus. It involves finding the rate of change of a function with respect to a variable. In our exercise, we start by differentiating the position vector \(\mathbf{r}(t)\). This vector describes the position in 3D space as a function of time \(t\). By differentiating it, we can determine how fast the position is changing. This results in the velocity vector \(\mathbf{v}(t)\).

For example, to find the velocity, we take the derivative of each component of \(\mathbf{r}(t)\). The calculus rules for differentiating each part are applied:

• \(\frac{d}{dt} \left(\frac{1}{t}\right) = -\frac{1}{t^2}\).
• \(\frac{d}{dt} \left((t^2 - 1)^{-1}\right) = \frac{-2t}{(t^2 - 1)^2}\).
• \(\frac{d}{dt} \left(t^5\right) = 5t^4\).

The outcome is the velocity vector \(\mathbf{v}(t) = -\frac{1}{t^2} \mathbf{i} + \frac{-2t}{(t^2 - 1)^2} \mathbf{j} + 5t^4 \mathbf{k}\). Differentiation provides a vital tool for analyzing how quantities evolve over time.

Velocity

Velocity is an important vector in mathematics and physics that describes both the speed and direction of an object's movement. In the exercise, we determine this from the position vector by differentiation.

The velocity vector \(\mathbf{v}(t)\) provides insight into how the object moves through space over time. At time \(t = 2\), we substitute into our velocity equation to calculate the specific velocity: \(\mathbf{v}(2) = -\frac{1}{4} \mathbf{i} - \frac{4}{9} \mathbf{j} + 80 \mathbf{k}\).

This result tells us the speed and direction along each axis at that precise moment. The negative values in the \(\mathbf{i}\) and \(\mathbf{j}\) components indicate a movement in the opposite direction along those axes, while the large positive \(\mathbf{k}\) component suggests significant movement along the third axis.

Acceleration

Acceleration represents the rate at which velocity changes with time. To determine the acceleration vector \(\mathbf{a}(t)\), we differentiate the velocity vector \(\mathbf{v}(t)\).

The differences in acceleration highlight how the velocity vector evolves over time. This is crucial for understanding dynamics in mechanics and illustrates how quickly an object changes its speed.

Using differentiation, we find:

• The derivative \(\frac{d}{dt} \left(-\frac{1}{t^2}\right) = \frac{2}{t^3}\).
• \(\frac{d}{dt} \left(\frac{-2t}{(t^2 - 1)^2}\right) = \frac{2(t^2 + 1)}{(t^2 - 1)^3}\).
• \(\frac{d}{dt} \left(5t^4\right) = 20t^3\).

At \(t = 2\), the acceleration vector evaluates to \(\mathbf{a}(2) = \frac{1}{4} \mathbf{i} + \frac{10}{27} \mathbf{j} + 160 \mathbf{k}\). This shows the instantaneous change in velocity, indicating how quickly the object is speeding up or slowing down in the direction of each axis.

Speed

Speed is a scalar that measures how fast an object is moving regardless of direction. It is distinct from velocity because it lacks directional information.

To find the speed at a given instance, we compute the magnitude of the velocity vector \(\mathbf{v}(t)\). The magnitude formula in three dimensions \(\sqrt{v_x^2 + v_y^2 + v_z^2}\) is used to ensure the calculation considers all movement dimensions.

For our exercise at \(t = 2\), the speed \(s\) is calculated as:

• \[ s = \sqrt{\left(-\frac{1}{4}\right)^2 + \left(-\frac{4}{9}\right)^2 + 80^2} = \sqrt{6400.2222} \approx 80.001 \]

Thus, the speed is approximately 80.001 units, which gives us an idea of how fast the object travels without concern for its direction, only its fastness.