Problem 123 - Calculus Volume 3

Question

Find \(\mathbf{T}(t)\) for the curve \(\mathbf{r}(t)=\left(t^{3}-4 t\right) \mathbf{i}+\left(5 t^{2}-2\right) \mathbf{j}\)

Step-by-Step Solution

Answer

\( \mathbf{T}(t) = \frac{(3t^2 - 4) \mathbf{i} + (10t) \mathbf{j}}{\sqrt{9t^4 + 76t^2 + 16}} \)

Step 1: Differentiate the curve

To find the tangent vector \( \mathbf{T}(t) \), we first need to find the derivative of the curve \( \mathbf{r}(t) \). Differentiate each component of \( \mathbf{r}(t) = \left(t^{3}-4t\right) \mathbf{i} + \left(5t^{2}-2\right) \mathbf{j} \): \[\frac{d}{dt} \mathbf{r}(t) = \left(3t^2 - 4\right) \mathbf{i} + \left(10t\right) \mathbf{j}. \] This derivative is the velocity vector \( \mathbf{v}(t) \).

Step 2: Compute the magnitude of the velocity vector

Next, we calculate the magnitude of the velocity vector \( \mathbf{v}(t) = (3t^2 - 4) \mathbf{i} + (10t) \mathbf{j} \). The magnitude is: \[\| \mathbf{v}(t) \| = \sqrt{(3t^2 - 4)^2 + (10t)^2}.\] Simplify the expression under the square root to get: \[\sqrt{9t^4 - 24t^2 + 16 + 100t^2} = \sqrt{9t^4 + 76t^2 + 16}.\]

Step 3: Normalize the velocity vector

The unit tangent vector \( \mathbf{T}(t) \) is found by dividing the velocity vector by its magnitude. Thus: \[\mathbf{T}(t) = \frac{(3t^2 - 4) \mathbf{i} + (10t) \mathbf{j}}{\sqrt{9t^4 + 76t^2 + 16}}.\] This expression gives us the unit tangent vector at any point \( t \) on the curve.

Key Concepts

Unit Tangent Vector

The unit tangent vector, often denoted as \( \mathbf{T}(t) \), is a vector that points in the direction of the curve at a specific point \( t \). It is derived from the velocity vector, which is the tangent vector to the curve, by normalizing it. Normalizing involves scaling the vector to have a unit length of 1. This process keeps the direction of the tangent vector unchanged but ensures the magnitude is 1, making it easier to analyze the curve's direction at that point.

For our specific curve function \( \mathbf{r}(t) \), the unit tangent vector provides insight into how the path behaves as \( t \) changes. It allows us to understand the trajectory's direction independent of the speed at which the point moves along the path. Once we compute \( \mathbf{v}(t) \), the velocity vector, we normalize it to find \( \mathbf{T}(t) \), presenting a simplified and precise way to grasp the orientation of the curve.

Velocity Vector

The velocity vector \( \mathbf{v}(t) \) is a crucial element when studying the motion along a curve. It describes the rate of change of position with respect to time. Essentially, this vector gives us both the speed and direction of a point moving along the curve at any instant.

In mathematical terms, for a given vector function \( \mathbf{r}(t) \), the velocity vector is the derivative of \( \mathbf{r}(t) \). The computation of the velocity vector involves differentiating each component of the vector function with respect to \( t \). In our example of \( \mathbf{r}(t) = (t^{3}-4t) \mathbf{i} + (5t^2-2) \mathbf{j} \), the derivative yields \( (3t^2 - 4) \mathbf{i} + (10t) \mathbf{j} \).

This result gives us the velocity vector, indicating both the speed and the direction of movement along the curve at any parameter \( t \).

Vector Differentiation

Vector differentiation is the process of taking derivatives of vector functions, which is essential in finding tangent vectors and analyzing their properties. It involves differentiating each component of the vector function separately.

When applied to a vector function such as \( \mathbf{r}(t) = (t^{3}-4t)\mathbf{i} + (5t^2-2)\mathbf{j} \), vector differentiation requires applying the rules of differentiation to each component. For our vector, differentiating \( (t^{3}-4t) \mathbf{i} \) gives \( (3t^{2}-4) \mathbf{i} \), and differentiating \( (5t^{2}-2) \mathbf{j} \) yields \( 10t \mathbf{j} \).

This process results in the velocity vector, \( \mathbf{v}(t) = (3t^{2}-4)\mathbf{i} + (10t)\mathbf{j} \), which describes how the vector function changes with respect to time. Understanding vector differentiation is fundamental to comprehending more complex calculus of vector functions.

Curve Normalization

Curve normalization plays a vital role in simplifying vector functions for analysis by ensuring vectors have a unit length. This is particularly useful when dealing with tangent vectors, as it allows us to focus on direction without being influenced by magnitude.

The process of normalization involves dividing a vector by its magnitude. For the velocity vector \( \mathbf{v}(t) \), computed previously, we find the magnitude by applying the Pythagorean theorem:

• Calculate the square of each component: \((3t^2 - 4)^2\) and \((10t)^2\).
• Sum these squares and take the square root.

This yields the magnitude \( \| \mathbf{v}(t) \| = \sqrt{9t^4 + 76t^2 + 16} \).

To achieve a unit tangent vector \( \mathbf{T}(t) \), we divide each component of \( \mathbf{v}(t) \) by this magnitude, resulting in a directionally equivalent vector of length 1. This normalized vector provides a standardized approach to understanding curves, especially when comparing multiple curves or analyzing variations along a single curve.