Problem 33 - Essential Calculus Early Transcendentals

Question

Find the tangential and normal components of the acceleration vector. $$\mathbf{r}(t)=\left(3 t-t^{3}\right) \mathbf{i}+3 t^{2} \mathbf{j}$$

Step-by-Step Solution

Answer

The tangential and normal components of the acceleration vector are derived from the vector's differentiation and subsequent manipulations.

Step 1: Find the Velocity Vector

The velocity vector \( \mathbf{v}(t) \) is the derivative of the position vector \( \mathbf{r}(t) \) with respect to time \( t \).\[ \mathbf{v}(t) = \frac{d}{dt}(3t - t^3) \mathbf{i} + \frac{d}{dt}(3t^2) \mathbf{j} \]Calculating the derivatives, we have:\[ \mathbf{v}(t) = (3 - 3t^2) \mathbf{i} + 6t \mathbf{j} \]

Step 2: Find the Acceleration Vector

The acceleration vector \( \mathbf{a}(t) \) is the derivative of the velocity vector \( \mathbf{v}(t) \).\[ \mathbf{a}(t) = \frac{d}{dt}(3 - 3t^2) \mathbf{i} + \frac{d}{dt}(6t) \mathbf{j} \]After differentiation, we get:\[ \mathbf{a}(t) = (-6t) \mathbf{i} + 6 \mathbf{j} \]

Step 3: Calculate Speed

The speed \( \|\mathbf{v}(t)\| \) is the magnitude of the velocity vector \( \mathbf{v}(t) \).\[ \|\mathbf{v}(t)\| = \sqrt{(3 - 3t^2)^2 + (6t)^2} \]Simplifying this expression gives:\[ \|\mathbf{v}(t)\| = \sqrt{9 - 18t^2 + 9t^4 + 36t^2} = \sqrt{9t^4 + 18t^2 + 9} \]

Step 4: Calculate the Tangential Component of Acceleration

The tangential component of acceleration \( a_T \) is given by the formula:\[ a_T = \frac{d}{dt} \| \mathbf{v}(t) \| \]First, find \( \frac{d}{dt} \| \mathbf{v}(t) \| \) by differentiating step 3.Replacing \( x = t^2 \), we rewrite the expression as:\[ \frac{d}{dt} \sqrt{9x^2 + 18x + 9} = \frac{1}{2\sqrt{9x^2 + 18x + 9}} \times (18x + 18) \times 2t \]Substitute back \( x = t^2 \) and rearrange:\[ a_T = \frac{18t^3 + 18t}{\sqrt{t^4 + 2t^2 + 1}} \]

Step 5: Calculate the Normal Component of Acceleration

The normal component of acceleration \( a_N \) is given by:\[ a_N = \sqrt{\| \mathbf{a}(t) \|^2 - a_T^2} \]First, calculate \( \| \mathbf{a}(t) \| \):\[ \| \mathbf{a}(t) \| = \sqrt{(-6t)^2 + 6^2} = \sqrt{36t^2 + 36} = \sqrt{36(t^2 + 1)} = 6\sqrt{t^2 + 1} \]Now substitute \( a_T \) from Step 4 to find \( a_N \).

Key Concepts

Velocity Vector

In physics, the concept of a velocity vector is essential for understanding motion. This vector describes both the speed and direction of an object's movement. For a given position vector \( \mathbf{r}(t) \), the velocity vector \( \mathbf{v}(t) \) is obtained by differentiating the position vector with respect to time \( t \). In our exercise, we start with the position vector \( \mathbf{r}(t) = (3t - t^3) \mathbf{i} + 3t^2 \mathbf{j} \). To find the velocity vector, we calculate the derivative:

• Differentiate \( 3t - t^3 \) with respect to \( t \), resulting in \( 3 - 3t^2 \) in the \( \mathbf{i} \) direction.
• Differentiate \( 3t^2 \), yielding \( 6t \) in the \( \mathbf{j} \) direction.

Therefore, the velocity vector \( \mathbf{v}(t) \) becomes: \[ \mathbf{v}(t) = (3 - 3t^2) \mathbf{i} + 6t \mathbf{j} \]This vector is crucial because it encapsulates how fast and in which direction the object is moving at any given time.

Acceleration Vector

The acceleration vector \( \mathbf{a}(t) \) tells us how the velocity of an object changes over time. It's essentially a derivative of the velocity vector. By finding this vector, we can describe how the motion itself is evolving.For our problem, we've determined the velocity vector \( \mathbf{v}(t) = (3 - 3t^2) \mathbf{i} + 6t \mathbf{j} \). We need to differentiate it to find the acceleration:

• The rate of change in the \( \mathbf{i} \) direction is found by differentiating \( 3 - 3t^2 \), which results in \( -6t \).
• Similarly, differentiating \( 6t \) gives us a constant \( 6 \) in the \( \mathbf{j} \) direction.

Thus, the acceleration vector becomes:\[ \mathbf{a}(t) = (-6t) \mathbf{i} + 6 \mathbf{j} \]This vector informs us about changes in speed and direction of the velocity vector. It's key for understanding how forces act on the object to influence its motion.

Magnitude of Velocity

The magnitude of the velocity vector, often referred to as speed, is vital to comprehend how fast an object is traveling regardless of direction. This is calculated by taking the length of the velocity vector \( \mathbf{v}(t) \).Given the velocity vector \( \mathbf{v}(t) = (3 - 3t^2) \mathbf{i} + 6t \mathbf{j} \), we find the magnitude \( \| \mathbf{v}(t) \| \) as follows:\[ \| \mathbf{v}(t) \| = \sqrt{(3 - 3t^2)^2 + (6t)^2} \]Simplifying this expression, we get:\[ \| \mathbf{v}(t) \| = \sqrt{9 - 18t^2 + 9t^4 + 36t^2} = \sqrt{9t^4 + 18t^2 + 9} \]By finding this magnitude, we isolate the speed, which measures how quickly the position of the object changes regardless of its travel direction.

Differentiation

Differentiation is a fundamental concept in calculus used to determine the rate of change of a function. In the context of this exercise, differentiation helps us find both the velocity and acceleration vectors from the position vector \( \mathbf{r}(t) \).Let's consider its role in our solution:

• We first differentiate the position vector to obtain the velocity vector. This step reveals how the position changes over time.
• Then, we differentiate the velocity vector to get the acceleration vector. Here, differentiation tells us how the velocity itself varies with time.

Thus, differentiation bridges the gap between successive rates of change, from position to velocity to acceleration. It's the key mathematical tool that links the movement of objects to time, showing not just where they are, but how fast they're moving, and how their speed itself is changing.