Question

Given a tangent vector on an oriented curve, how do you find the unit tangent vector?

Short answer

Answer: The steps to find the unit tangent vector for an oriented curve are as follows: 1. Find the tangent vector by differentiating the components of the position vector function with respect to the curve parameter t. 2. Calculate the magnitude of the tangent vector using the formula: ||v|| = √(vx^2 + vy^2 + vz^2). 3. Normalize the tangent vector by dividing each component by its magnitude, which results in obtaining the unit tangent vector with a length of 1 and the same direction as the original tangent vector.

Step-by-step solution

Find the tangent vector

To find the tangent vector, we first need to find the derivative of our curve's position vector function. The position vector function typically looks like \(\vec{r}(t) = x(t) \hat{i} + y(t) \hat{j} + z(t) \hat{k}\). To find the tangent vector, which is represented by \({\vec{T}(t)}\), we need to differentiate each component of the position vector function with respect to the curve parameter \(t\). This simply means that \(\vec{T}(t) = \frac{d}{dt}(x(t) \hat{i} + y(t) \hat{j} + z(t) \hat{k})\).

Find the magnitude of the tangent vector

Now that we have the tangent vector, we need to find its magnitude, which is denoted by \(||\vec{T}(t)||\). The magnitude (or length) of a vector can be found using the following formula: \(||\vec{v}|| = \sqrt{v_x^2 + v_y^2 + v_z^2}\), where \(v_x\), \(v_y\), and \(v_z\) are the components of the vector. In our case, the magnitude of the tangent vector is given by \(||\vec{T}(t)|| = \sqrt{(\frac{dx}{dt})^2 + (\frac{dy}{dt})^2 + (\frac{dz}{dt})^2}\).

Normalize the tangent vector

Now that we know the magnitude of our tangent vector, we can obtain the unit tangent vector, denoted as \(\hat{T}(t)\), by dividing each component of the tangent vector by its magnitude. This normalizes the vector, ensuring that it maintains the same direction as the tangent vector but now has a length of 1. So, the unit tangent vector is given by: \(\hat{T}(t) = \frac{1}{||\vec{T}(t)||}\vec{T}(t) = \frac{1}{\sqrt{(\frac{dx}{dt})^2 + (\frac{dy}{dt})^2 + (\frac{dz}{dt})^2}}(\frac{dx}{dt}\hat{i} + \frac{dy}{dt}\hat{j} + \frac{dz}{dt}\hat{k})\). That's it! You have now found the unit tangent vector for an oriented curve, given its position vector function.

Key concepts

Position Vector Function

A position vector function is a way to describe the position of a point in space, usually in the context of motion along a curve. This function is typically denoted as \( \vec{r}(t) \). It uses the parameter \( t \), which often represents time or any other continuous variable. The position vector function is usually expressed in terms of its components along the standard Cartesian coordinate axes as:

• \( x(t) \) along the \( \hat{i} \) direction,
• \( y(t) \) along the \( \hat{j} \) direction, and
• \( z(t) \) along the \( \hat{k} \) direction.

Thus, the general form of the position vector function can be written as \( \vec{r}(t) = x(t) \hat{i} + y(t) \hat{j} + z(t) \hat{k} \).
This function gives us a vector pointing from the origin to the point \((x(t), y(t), z(t))\) at any given \( t \). This is crucial for describing curves in space and allows us to compute derivatives required to find tangent vectors.

Tangent Vector

The tangent vector is crucial in understanding the direction in which a point on a curve is moving. It provides a vector that is tangent to the curve at a particular point. To find the tangent vector, we differentiate the position vector function \( \vec{r}(t) \) with respect to the curve parameter \( t \). The resulting vector, \( \vec{T}(t) \), indicates the direction of the curve at that specific point.
Differentiating each component of the position vector gives us:

• \( \frac{dx}{dt} \hat{i} \) as the rate of change in the \( i \) direction,
• \( \frac{dy}{dt} \hat{j} \) as the rate of change in the \( j \) direction, and
• \( \frac{dz}{dt} \hat{k} \) as the rate of change in the \( k \) direction.

Together, they form the tangent vector: \( \vec{T}(t) = \frac{dx}{dt} \hat{i} + \frac{dy}{dt} \hat{j} + \frac{dz}{dt} \hat{k} \).This vector tells us how the curve is oriented at a point and is a precursor to finding the unit tangent vector.

Curve Parameter

The curve parameter, typically \( t \), serves as a variable that allows us to trace out the path of a curve in space. Think of a curve parameter as a tool that can adjust the values of \( x(t) \), \( y(t) \), and \( z(t) \).
As \( t \) varies, the position vector sweeps out the path of the curve, giving us a detailed way to represent and analyze the movement or shape of the curve across different points.
In many problems, especially in physics and engineering, \( t \) often represents time. Thus, the position vector function not only represents spatial paths but also the trajectory of moving objects over time. Using \( t \) as a parameter helps simplify the study of motion and curve properties.

Magnitude of a Vector

The magnitude of a vector, often represented as \( ||\vec{v}|| \), is a measure of the vector's length. For a vector \( \vec{v} = v_x \hat{i} + v_y \hat{j} + v_z \hat{k} \), its magnitude determines the distance from the origin to the point \((v_x, v_y, v_z)\).
To find the magnitude, we use the formula: \[||\vec{v}|| = \sqrt{v_x^2 + v_y^2 + v_z^2}\]This is similar to calculating the Euclidean distance in a three-dimensional space.
For tangent vectors, finding the magnitude \( ||\vec{T}(t)|| \) is a critical step to normalizing the vector. Normalization transforms the tangent vector into a unit vector, retaining direction but ensuring its magnitude is just 1.
The computation of the magnitude is vital in various applications, including physics and computer graphics, where understanding vector lengths helps in modeling real-world scenarios accurately.