Problem 41 - Calculus III

Question

For each of the following problems, 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

The tangential component is \( a_T = \frac{8 + 18t^2}{\sqrt{8 + 9t^2}} \) and the normal component is \( a_N = \sqrt{\frac{(8 + 36t^2)(8 + 9t^2) - (8 + 18t^2)^2}{8 + 9t^2}} \).

Step 1: Find Velocity

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

Step 2: Find Speed

The speed is the magnitude of the velocity vector. Calculate this as: \[ v = \|\mathbf{v}(t)\| = \sqrt{(2t)^2 + (2t)^2 + (3t^2)^2} = \sqrt{4t^2 + 4t^2 + 9t^4} = \sqrt{8t^2 + 9t^4} = \sqrt{t^2(8 + 9t^2)} = t\sqrt{8 + 9t^2} \].

Step 3: Find Acceleration

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

Step 4: Find Tangential Component of Acceleration

The tangential component of acceleration, \( a_T \), is given by the derivative of speed with respect to time or the projection of acceleration onto the velocity direction: \[ a_T = \frac{\mathbf{v}(t) \cdot \mathbf{a}(t)}{\|\mathbf{v}(t)\|} \]. Calculate: \( \mathbf{v}(t) \cdot \mathbf{a}(t) = (2t \mathbf{i} + 2t \mathbf{j} + 3t^2 \mathbf{k}) \cdot (2 \mathbf{i} + 2 \mathbf{j} + 6t \mathbf{k}) = 4t + 4t + 18t^3 = 8t + 18t^3 \). Thus, \[ a_T = \frac{8t + 18t^3}{t\sqrt{8 + 9t^2}} = \frac{8 + 18t^2}{\sqrt{8 + 9t^2}} \].

Step 5: Find Normal Component of Acceleration

The normal component of acceleration, \( a_N \), can be found using: \[ a_N = \sqrt{\|\mathbf{a}(t)\|^2 - a_T^2} \]. First, find the magnitude of acceleration: \( \|\mathbf{a}(t)\| = \sqrt{2^2 + 2^2 + (6t)^2} = \sqrt{4 + 4 + 36t^2} = \sqrt{8 + 36t^2} \). Then, calculate \( a_N = \sqrt{8 + 36t^2 - \left(\frac{8 + 18t^2}{\sqrt{8 + 9t^2}}\right)^2} \), which simplifies to: \[ a_N = \sqrt{\frac{(8 + 36t^2)(8 + 9t^2) - (8 + 18t^2)^2}{8 + 9t^2}} \].

Key Concepts

Differentiation of Vectors

Differentiation of vectors is a crucial concept in understanding motion along a trajectory. Imagine you have a position vector, which describes where an object is at any moment in time. By differentiating this vector with respect to time, you can find out how fast and in which direction it is moving. This process involves taking the derivative of each component of the vector separately. For instance, given a position vector \( \mathbf{r}(t) = t^2 \mathbf{i} + t^2 \mathbf{j} + t^3 \mathbf{k} \), the velocity vector is found by differentiating each component to get \( \mathbf{v}(t) = 2t \mathbf{i} + 2t \mathbf{j} + 3t^2 \mathbf{k} \).

• Helps us understand how the position of an object changes.
• Essential for calculating velocity and further, acceleration.

Remember, differentiating a vector is like finding slopes of curves, which tells us the rate of change at any point in time.

Velocity Vector

The velocity vector provides a clear picture of how an object's position changes over time. It is obtained by differentiating the position vector. For instance, the velocity vector \( \mathbf{v}(t) = 2t \mathbf{i} + 2t \mathbf{j} + 3t^2 \mathbf{k} \) gives the speed and direction as functions of time.

• Magnitude: Speed, which is a scalar of how fast the object is moving.
• Direction: Given by the directional components \( \mathbf{i}, \mathbf{j}, \mathbf{k} \).

The velocity vector is foundational for any further motion analysis, as seen when calculating acceleration as the rate of change of velocity.

Acceleration Vector

Once you grasp the velocity vector, the next step is understanding acceleration. The acceleration vector pictures how the velocity itself changes over time. Found by differentiating the velocity vector, it's a rate of a rate. For the example given, \( \mathbf{a}(t) = 2 \mathbf{i} + 2 \mathbf{j} + 6t \mathbf{k} \) illustrates how acceleration affects each directional component.

• Analytical Purpose: Helps derive both tangential and normal components of acceleration.
• Components: Constant terms indicate steady acceleration, whereas time-dependent terms show changing acceleration over time.

By analyzing these components, you can determine how an object speeds up or slows down, and in which directions these changes occur.

Vector Magnitude Calculation

The magnitude of a vector is a measure of its length without considering direction, often referred to as its 'norm.' It's pivotal for understanding the scale of vectors like velocity and acceleration. The calculation involves the Pythagorean theorem extended to multiple dimensions. For a vector \( \mathbf{v}(t) = 2t \mathbf{i} + 2t \mathbf{j} + 3t^2 \mathbf{k} \), the magnitude \( v = \|\mathbf{v}(t)\| = \sqrt{(2t)^2 + (2t)^2 + (3t^2)^2} \) gives us the speed at any time \( t \).

• Critical for finding the scalar quantity associated with velocity, known as speed.
• Also important for computing components of acceleration.

Ensure you compute magnitudes correctly, as they are foundational for subsequent calculations involving any motion along a path.