Problem 22 - Calculus Early Transcendentals: Pearson New International Edition

Question

Find the velocity \(\mathbf{v}\), acceleration \(\mathbf{a}\), and speed \(s\) at the indicated time \(t=t_{1}\). $$ \mathbf{r}(t)=t^{6} \mathbf{i}+\left(6 t^{2}-5\right)^{6} \mathbf{j}+t \mathbf{k} ; t_{1}=1 $$

Step-by-Step Solution

Answer

Velocity is \( 6\mathbf{i} + 72\mathbf{j} + 1\mathbf{k} \), acceleration is \( 30\mathbf{i} + 4320\mathbf{j} + 0\mathbf{k} \), and speed is \( \sqrt{5221} \).

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

The velocity \( \mathbf{v} \) is the first derivative of the position function \( \mathbf{r}(t) \) with respect to time \( t \). Differentiate each component of \( \mathbf{r}(t) = t^6 \mathbf{i} + (6t^2 - 5)^6 \mathbf{j} + t \mathbf{k} \) to get:\[ \mathbf{v}(t) = \frac{d}{dt}(t^6) \mathbf{i} + \frac{d}{dt}((6t^2 - 5)^6) \mathbf{j} + \frac{d}{dt}(t) \mathbf{k} \]Calculating each derivative, we find:- \( \frac{d}{dt}(t^6) = 6t^5 \)- \( \frac{d}{dt}((6t^2 - 5)^6) = 6((6t^2 - 5)^5) \cdot \frac{d}{dt}(6t^2 - 5) = 6((6t^2 - 5)^5)(12t) \)- \( \frac{d}{dt}(t) = 1 \)Thus, the velocity vector is:\[ \mathbf{v}(t) = 6t^5 \mathbf{i} + 72t(6t^2 - 5)^5 \mathbf{j} + 1 \mathbf{k} \]

Step 2: Evaluate \( \mathbf{v} \) at \( t = 1 \)

Substitute \( t = 1 \) into the velocity expression:\[ \mathbf{v}(1) = 6(1)^5 \mathbf{i} + 72(1)(6(1)^2 - 5)^5 \mathbf{j} + 1 \mathbf{k} \]Calculate each term:- \( 6(1)^5 = 6 \)- \( 6 \times 1^2 - 5 = 1 \), and \( 1^5 = 1 \), so \( 72 \times 1 \times 1 = 72 \)Thus, \( \mathbf{v}(1) = 6\mathbf{i} + 72\mathbf{j} + 1\mathbf{k} \)

Step 3: Find the acceleration \( \mathbf{a} \)

The acceleration \( \mathbf{a} \) is the derivative of the velocity function \( \mathbf{v}(t) \) with respect to time \( t \). Differentiate each component of \( \mathbf{v}(t) = 6t^5 \mathbf{i} + 72t(6t^2 - 5)^5 \mathbf{j} + 1 \mathbf{k} \):\[ \mathbf{a}(t) = \frac{d}{dt}(6t^5) \mathbf{i} + \frac{d}{dt}(72t(6t^2 - 5)^5) \mathbf{j} + \frac{d}{dt}(1) \mathbf{k} \]Calculate each derivative:- \( \frac{d}{dt}(6t^5) = 30t^4 \)- For the \( \mathbf{j} \) component, use the product rule: \( \frac{d}{dt}(uv) = u'v + uv' \) : - Let \( u = 72t \) and \( v = (6t^2 - 5)^5 \) - \( u' = 72 \), \( v' = 5(6t^2 - 5)^4 \cdot 12t = 60t(6t^2 - 5)^4 \) - So, \( \frac{d}{dt}(72t(6t^2 - 5)^5) = 72(6t^2 - 5)^5 + 72t(60t(6t^2 - 5)^4)\)- \( \frac{d}{dt}(1) = 0 \)Thus, the acceleration vector is complicated but can be further simplified if needed.

Step 4: Evaluate \( \mathbf{a} \) at \( t = 1 \)

Substitute \( t = 1 \) into the acceleration expression:- For \( 30t^4 \), at \( t = 1 \) it becomes \( 30 \).- For the \( \mathbf{j} \) component \( 72(1)(1)^5 + 72(1)(60(1)(1)^4) \), which simplifies to \( 4320 \).Thus, \( \mathbf{a}(1) = 30\mathbf{i} + 4320\mathbf{j} + 0\mathbf{k} \).

Step 5: Find the speed \( s \)

The speed \( s \) is the magnitude of the velocity \( \mathbf{v} \). Use the formula for magnitude of a vector: \( |\mathbf{v}| = \sqrt{v_x^2 + v_y^2 + v_z^2} \).Substitute the components from \( \mathbf{v}(1)=6\mathbf{i}+72\mathbf{j}+1\mathbf{k} \):\[ s = \sqrt{6^2 + 72^2 + 1^2} \]Calculate:- \( 6^2 = 36 \)- \( 72^2 = 5184 \)- \( 1^2 = 1 \)The speed is \( s = \sqrt{36 + 5184 + 1} = \sqrt{5221} \).

Step 6: Compute the final answers

We have found the velocity, acceleration, and speed at \( t = 1 \):- \( \mathbf{v}(1) = 6\mathbf{i} + 72\mathbf{j} + 1\mathbf{k} \)- \( \mathbf{a}(1) = 30\mathbf{i} + 4320\mathbf{j} + 0\mathbf{k} \)- \( s = \sqrt{5221} \)

Key Concepts

Velocity and Acceleration

In vector calculus, velocity and acceleration help us understand motion in terms of direction and rate of change. Velocity is the rate of change of position with respect to time. It is represented as a vector since it has both magnitude and direction. For any position function like \( \mathbf{r}(t) \), velocity \( \mathbf{v}(t) \) can be found by differentiating the position function with respect to time (t).
The velocity vector gives insight into how fast an object moves and in which direction. Once you have the velocity, you can find acceleration by differentiating the velocity again.
Acceleration tells us how the velocity of an object is changing with time. It is also a vector and can provide information about changes in speed or direction over time.

• Velocity: first derivative of position function.
• Acceleration: first derivative of velocity function.
• Both are vectors indicating direction and magnitude.

Differentiation

Differentiation is a key concept in calculus that deals with finding derivatives. Derivatives measure how a function changes as its input changes and are central to calculating velocity and acceleration.
When you differentiate a position function \( \mathbf{r}(t) \), you find its velocity \( \mathbf{v}(t) \). Similarly, differentiating the velocity \( \mathbf{v}(t) \) provides the acceleration \( \mathbf{a}(t) \).
One often uses differentiation rules such as the power rule, product rule, and chain rule to find these derivatives.

• Power rule: \( \frac{d}{dt}(t^n) = nt^{n-1} \)
• Product rule: \( \frac{d}{dt}(uv) = u'v + uv' \)
• Chain rule: Used for composing two functions, especially in complex differentiation tasks.

Understanding these rules helps calculate how each component of a vector function will change, offering insight into physical phenomena described by the function.

Magnitude of Velocity

The magnitude of velocity, often known as speed, tells us how fast something is moving regardless of direction. It's crucial for understanding the dynamics of motion in parametric equations.
Given a velocity vector \( \mathbf{v}(t) = v_x \mathbf{i} + v_y \mathbf{j} + v_z \mathbf{k} \), the magnitude of velocity \( s \) is found using the formula:\[s = |\mathbf{v}| = \sqrt{v_x^2 + v_y^2 + v_z^2}\]This is the distance that would be traveled by the object if it continued at the same velocity indefinitely.

• Focuses on how fast, not in which direction.
• Uses Pythagorean theorem to calculate overall speed.

Once you have the magnitude, you can better analyze the effect of different forces or slopes along the path of movement.

Parametric Equations

Parametric equations come in handy in vector calculus to describe paths in space. Instead of using \( y = f(x) \) forms, we use separate functions for each coordinate as a function of a parameter, typically time \( t \).
For example, \( \mathbf{r}(t) = t^6 \mathbf{i} + (6t^2 - 5)^6 \mathbf{j} + t \mathbf{k} \) describes a path in three-dimensional space, where each component \( (x, y, z) \) is a function of time. Parametric equations allow greater flexibility and precision when describing complex paths and motions.

• Each coordinate can be a function of time.
• Useful to represent complex curves and motions.
• Enable analysis of the vector's path over time.

This approach is useful for modeling paths in physics, animation, and various engineering fields.