Problem 192 - Calculus Volume 3

Question

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

Step-by-Step Solution

Answer

Tangential component of acceleration is \( \frac{8 + 18t^2}{\sqrt{8 + 9t^2}} \); normal component needs further simplification.

Step 1: Find the velocity vector

To find the velocity vector, differentiate the position vector \( \mathbf{r}(t) = t^2 \mathbf{i} + t^2 \mathbf{j} + t^3 \mathbf{k} \) with respect to \( t \). The velocity vector \( \mathbf{v}(t) \) is \( \mathbf{v}(t) = \frac{d}{dt}(t^2 \mathbf{i} + t^2 \mathbf{j} + t^3 \mathbf{k}) = 2t \mathbf{i} + 2t \mathbf{j} + 3t^2 \mathbf{k} \).

Step 2: Find the acceleration vector

Differentiate the velocity vector \( \mathbf{v}(t) = 2t \mathbf{i} + 2t \mathbf{j} + 3t^2 \mathbf{k} \) with respect to \( t \). The acceleration vector \( \mathbf{a}(t) \) is \( \mathbf{a}(t) = \frac{d}{dt}(2t \mathbf{i} + 2t \mathbf{j} + 3t^2 \mathbf{k}) = 2 \mathbf{i} + 2 \mathbf{j} + 6t \mathbf{k} \).

Step 3: Find the magnitude of the velocity vector

The magnitude of the velocity vector \( \mathbf{v}(t) = 2t \mathbf{i} + 2t \mathbf{j} + 3t^2 \mathbf{k} \) is \( \| \mathbf{v}(t) \| = \sqrt{(2t)^2 + (2t)^2 + (3t^2)^2} = \sqrt{4t^2 + 4t^2 + 9t^4} = \sqrt{8t^2 + 9t^4} = t \sqrt{8 + 9t^2} \).

Step 4: Determine the tangential component of acceleration

The tangential component of the acceleration \( a_T \) is given by the formula \( a_T = \frac{\mathbf{a}(t) \cdot \mathbf{v}(t)}{\|\mathbf{v}(t)\|} \). First, calculate the dot product \( \mathbf{a}(t) \cdot \mathbf{v}(t) = (2 \mathbf{i} + 2 \mathbf{j} + 6t \mathbf{k}) \cdot (2t \mathbf{i} + 2t \mathbf{j} + 3t^2 \mathbf{k}) = 4t + 4t + 18t^3 = 8t + 18t^3 \). The tangential component is then \( a_T = \frac{8t + 18t^3}{t\sqrt{8 + 9t^2}} = \frac{8 + 18t^2}{\sqrt{8 + 9t^2}} \).

Step 5: Determine the normal component of acceleration

The normal component of the acceleration \( a_N \) is given by \( a_N = \sqrt{\| \mathbf{a}(t) \|^2 - a_T^2} \). First, calculate \( \| \mathbf{a}(t) \|^2 = (2 \mathbf{i} + 2 \mathbf{j} + 6t \mathbf{k}) \cdot (2 \mathbf{i} + 2 \mathbf{j} + 6t \mathbf{k}) = 4 + 4 + 36t^2 = 8 + 36t^2 \). So, the normal component is \( a_N = \sqrt{(8 + 36t^2) - \left(\frac{8 + 18t^2}{\sqrt{8 + 9t^2}}\right)^2} \). Simplify to find \( a_N \).

Key Concepts

Velocity Vector

A velocity vector is a fundamental concept in physics that describes the speed and direction of an objectt's movement.

• It is derived by differentiating a position function with respect to time.

In the context of the exercise, we have the position vector \[ \mathbf{r}(t) = t^2 \mathbf{i} + t^2 \mathbf{j} + t^3 \mathbf{k} \].By differentiating each component of this vector with respect to time,we find the velocity vector \( \mathbf{v}(t) = 2t \mathbf{i} + 2t \mathbf{j} + 3t^2 \mathbf{k} \).This differentiation gives us the rate of change of the position vector,indicating how fast and in what direction each component of the position changes as time progresses.Velocity is crucial since it provides more insights than speedby including direction, thus offering a complete description of motion.

Acceleration Vector

The acceleration vector represents the rate of change of the velocity vector,providing insights into how the velocity of an object is changing over time.To determine the acceleration vector,we differentiate the velocity vector with respect to time.

• The acceleration vector is typically given by \( \mathbf{a}(t) = \frac{d}{dt} \mathbf{v}(t) \).

In the exercise, starting from the velocity vector\( \mathbf{v}(t) = 2t \mathbf{i} + 2t \mathbf{j} + 3t^2 \mathbf{k} \),we differentiate to find the acceleration vector:\( \mathbf{a}(t) = 2 \mathbf{i} + 2 \mathbf{j} + 6t \mathbf{k} \).This vector indicates how each component of velocity is increasing or decreasing at a specific time.Acceleration is a key concept as it tells us whether and how an object is speeding up, slowing down, or changing direction,effectively describing dynamic changes in motion.

Magnitude of a Vector

The magnitude of a vector measures its size or length,providing a scalar quantity that reflects how large the vector is without considering its direction.To calculate the magnitude of a vector, use:\[ \|\mathbf{v}\| = \sqrt{v_1^2 + v_2^2 + v_3^2} \]where \(v_1, v_2,\) and \(v_3\) are the component functions of the vector.For the velocity vector in the exercise, \( \mathbf{v}(t) = 2t \mathbf{i} + 2t \mathbf{j} + 3t^2 \mathbf{k} \),the magnitude is:\[ \| \mathbf{v}(t) \| = \sqrt{(2t)^2 + (2t)^2 + (3t^2)^2} = \sqrt{8t^2 + 9t^4} = t \sqrt{8 + 9t^2} \].The magnitude provides a way to assess speed without directional bias,allowing us to compare the sizes of different vectors precisely.It is crucial for computing other quantities such as acceleration components.

Dot Product

The dot product is a mathematical operation that takes two equal-length sequences of numbersand returns a single number. This operation can help in determining the angle between two vectors,as well as playing a key role in finding components of accelerations.The dot product of two vectors \( \mathbf{a} \) and \( \mathbf{b} \) is calculated as:\[ \mathbf{a} \cdot \mathbf{b} = a_1b_1 + a_2b_2 + a_3b_3 \]In this exercise, we find the dot product of the acceleration \( \mathbf{a}(t) = 2 \mathbf{i} + 2 \mathbf{j} + 6t \mathbf{k} \)and the velocity \( \mathbf{v}(t) = 2t \mathbf{i} + 2t \mathbf{j} + 3t^2 \mathbf{k} \):\[ \mathbf{a}(t) \cdot \mathbf{v}(t) = 4t + 4t + 18t^3 = 8t + 18t^3 \].The resulting scalar is significant in physics for computing the tangential component of acceleration,indicating how much of the acceleration is going in the same direction as the velocity.

Vector Differentiation

Vector differentiation is a vital technique in calculus,used to calculate rates of change of vectors with respect to a variable,often time.It's just like differentiating scalar functions,but applied separately to each component of a vector.In this exercise, we first differentiate the position vector \( \mathbf{r}(t) \) to find the velocity vector.Later, we take the derivative of the velocity vector \( \mathbf{v}(t) \) to obtain the acceleration vector.

• Velocity: \( \mathbf{v}(t) = \frac{d}{dt}(\mathbf{r}(t)) \)

This process involves taking the derivative of each component:\[ \begin{align*} \frac{d}{dt}(t^2\mathbf{i}) & = 2t\mathbf{i} \ \frac{d}{dt}(t^2\mathbf{j}) & = 2t\mathbf{j} \ \frac{d}{dt}(t^3\mathbf{k}) & = 3t^2\mathbf{k} \\end{align*} \]Understanding vector differentiation enables us to analyze motion,providing insights into velocity and acceleration over time,and is foundational for mechanics and dynamic systems.