Question

Prove that the curve \(\mathbf{r}(t)=\left\langle a+b t^{p}, c+d t^{p}, e+f t^{p}\right\rangle\), where \(a, b, c, d, e,\) and \(f\) are real numbers and \(p\) is a positive integer, has zero curvature. Give an explanation.

Short answer

Answer: The curvature of the curve is zero.

Step-by-step solution

Find the first and second derivatives of the position vector

Given the curve \(\mathbf{r}(t) = \langle a + bt^p, c + dt^p, e + ft^p \rangle\), where \(a, b, c, d, e, f\) are real numbers and \(p\) is a positive integer. Compute the first and second derivatives \(\mathbf{r}'(t)\) and \(\mathbf{r}''(t)\): $$ \mathbf{r}'(t) = \langle pbt^{p-1}, pdt^{p-1}, pft^{p-1} \rangle \\ \mathbf{r}''(t) = \langle p(p-1)bt^{p-2}, p(p-1)dt^{p-2}, p(p-1)ft^{p-2} \rangle $$

Compute the tangent vector

The tangent vector \(\mathbf{T}(t)\) is the normalized first derivative of the position vector: $$ \mathbf{T}(t) = \frac{\mathbf{r}'(t)}{|\mathbf{r}'(t)|} = \frac{t^{p-1}\langle pb, pd, pf \rangle}{|t^{p-1}|\sqrt{(pb)^2 + (pd)^2 + (pf)^2}} $$ We can simplify the tangent vector as follows: $$ \mathbf{T}(t) = \frac{\langle pb, pd, pf \rangle}{\sqrt{(pb)^2 + (pd)^2 + (pf)^2}} $$

Find the derivative of the tangent vector

Compute the derivative of the tangent vector \(\mathbf{T}'(t)\): $$ \mathbf{T}'(t) = \frac{d}{dt} \frac{\langle pb, pd, pf \rangle}{\sqrt{(pb)^2 + (pd)^2 + (pf)^2}} = 0 $$ Calculating the derivative of a constant vector results in the zero vector, which implies that the tangent vector does not change.

Compute the curvature

The curvature \(\kappa(t)\) is the magnitude of the derivative of the tangent vector: $$ \kappa(t) = |\mathbf{T}'(t)| = |0| = 0 $$

Show that the curvature is zero

We have shown that the curvature \(\kappa(t)\) is equal to zero, which means the curve \(\mathbf{r}(t) = \langle a + bt^p, c + dt^p, e + ft^p \rangle\) has zero curvature.

Key concepts

Curvature of a Curve

Understanding the curvature of a curve is essential for studying the geometry and motion along a curve in space. The curvature, denoted as \(\kappa(t)\), measures how sharply a curve bends at a given point. The curvature is defined as the magnitude of the derivative of the unit tangent vector with respect to the curve’s parameter, usually time or arc length.

For the given exercise, proving that a curve has zero curvature means showing that the curve does not bend at any point, implying that it's straight in the geometric sense. It's important to recognize that curvature is a local property, which means it describes how the curve bends at a specific point, rather than over its entire length.

Tangent Vector

In vector calculus, a tangent vector is a vector that touches a curve at a point without crossing it. It points in the direction of the curve's immediate path as one moves along the curve. Mathematically, it's the vector obtained by normalizing the first derivative of the position vector \(\mathbf{r}(t)\), known as the velocity vector.

The normalization process involves dividing the velocity vector by its own magnitude, which yields a unit tangent vector \(\mathbf{T}(t)\). This unit vector is essential for computing the curvature since it tells us how the direction of the curve changes. In the step by step solution, the tangent vector was shown to be constant, and its derivative, therefore, was zero, leading to zero curvature.

Derivatives

Derivatives play a key role in understanding the behavior of functions, including those that describe curves in space.

Vector Calculus

Vector calculus expands upon the traditional calculus to deal with vector fields as well as functions of multiple variables.