Question

Prove that the line \(\mathbf{r}(t)=\left\langle x_{0}+a t, y_{0}+b t, z_{0}+c t\right\rangle\) is parameterized by arc length, provided \(a^{2}+b^{2}+c^{2}=1\)

Short answer

Question: Prove that the parameterized line \(\mathbf{r}(t)=\left\langle x_{0}+at, y_{0}+bt, z_{0}+ct\right\rangle\) is parameterized by arc length, given that \(a^2 + b^2 + c^2 = 1\). Answer: To prove that the given line is parameterized by arc length, we showed that the magnitude of the tangent vector (derivative vector) is equal to 1. We found the derivative of \(\mathbf{r}(t)\) as \(\mathbf{r}'(t)=\left\langle a, b, c\right\rangle\). Its magnitude is \(\sqrt{a^2 + b^2 + c^2}\), which, given the condition \(a^2 + b^2 + c^2 = 1\), is equal to 1. Thus, the given line is parameterized by arc length.

Step-by-step solution

Find the derivative of the vector function

To find the tangent vector (velocity vector) of the vector function \(\mathbf{r}(t)=\left\langle x_{0}+at, y_{0}+bt, z_{0}+ct\right\rangle\), take the derivative of the function with respect to \(t\). The derivative is given by: $$ \mathbf{r}'(t)=\frac{d}{dt}\left\langle x_{0}+at, y_{0}+bt, z_{0}+ct\right\rangle = \left\langle a, b, c\right\rangle $$

Find the magnitude of the derivative vector

Now that we have found the derivative of \(\mathbf{r}(t)\), let's find its magnitude. Magnitude of a vector \(\mathbf{v} = \left\langle a, b, c\right\rangle\) is given by \(\sqrt{a^2+b^2+c^2}\). So, the magnitude of the derivative vector \(\mathbf{r}'(t)\) is: $$ \left\| \mathbf{r}'(t) \right\| = \sqrt{a^2 + b^2 + c^2} $$

Prove that the magnitude of the derivative vector is equal to 1

We want to show that the magnitude of the derivative vector is equal to 1. Let's look at the magnitude of the derivative vector we found in Step 2: $$ \left\| \mathbf{r}'(t) \right\| = \sqrt{a^2 + b^2 + c^2} $$ We are given that \(a^2 + b^2 + c^2 = 1\). Therefore, the magnitude of the derivative vector becomes: $$ \left\| \mathbf{r}'(t) \right\| = \sqrt{1} =1 $$ Since the magnitude of the tangent vector (derivative vector) is equal to 1, we can conclude that the given line is indeed parameterized by arc length.

Key concepts

Vector Function Derivative

When dealing with a vector function, just like in calculus with scalar functions, we can find what's known as the derivative. In the context of a vector function, the derivative reflects how the vector function changes with respect to some variable, typically time. Imagine you're driving a car – the derivative of your position vector would give you the velocity vector, indicating not only how fast you're going but also in what direction.

We take the derivative of each component of the vector function with respect to the variable. For example, if we have a vector function \( \mathbf{r}(t)=\left\langle x_0+at, y_0+bt, z_0+ct\right\rangle \), its derivative \( \mathbf{r}'(t) \) is obtained by differentiating each component separately: \( \mathbf{r}'(t)=\left\langle a, b, c\right\rangle \). Essentially, the process of taking the derivative 'peels away' any constant factors and reduces the function down to its most variable parts, streamlining it to the vectors that best describe the instant change.

Magnitude of a Vector

The magnitude of a vector represents its length. In physical terms, it's the 'strength' or 'intensity' of the vector. Mathematically, it’s the distance from the tail to the head of the vector, considering all dimensions it spans across. Magnitude is always a non-negative value (since it's derived from a square root) and is calculated using the Pythagorean theorem extended into our vector's dimensions. For a 3D vector \( \mathbf{v} = \left\langle a, b, c\right\rangle \), the magnitude is \( \| \mathbf{v} \| = \sqrt{a^2 + b^2 + c^2} \).

To give you a concrete example, if the coefficients of the vector \( \mathbf{r}'(t) = \left\langle a, b, c\right\rangle \) add up to 1 (that is, \( a^2 + b^2 + c^2 = 1 \)), the magnitude is \( \sqrt{1} = 1 \), meaning the vector has a 'unit length'. This notion of unit vectors is crucial in fields like physics and computer graphics where normalization (scaling a vector to unit length) is a common requirement for simplifying calculations.

Tangent Vector

In the world of curves and surfaces, the tangent vector holds a special place as it touches upon the idea of directionality along a path. A tangent vector at a point on a curve indicates the direction in which the curve is heading at that exact point. If you imagine walking along a path, the tangent vector would represent the direction of your feet pointing forward – it's the 'instantaneous' direction of the motion.

Getting back to the exercise, the derivative of our line \( \mathbf{r}(t) \) gives us the tangent vector \( \mathbf{r}'(t) \) which, due to having a constant magnitude of 1, means that our line moves at a constant speed in a steady direction. This signifies a uniform motion, and hence the line parameterized by \( t \) has been efficiently described by its arc length - a concept intimately tied with the curve's geometry, useful for simplifying complex calculus and physics problems involving motion and curves.