Question

Zero curvature 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

Question: Prove that the curve given by the position vector 𝑟(𝑡) = ⟨𝑎 + 𝑏𝑡^𝑝, 𝑐 + 𝑑𝑡^𝑝, 𝑒 + 𝑓𝑡^𝑝⟩ has zero curvature. Solution: To prove that the given curve has zero curvature, we found the tangent vector, the normal vector, and calculated the curvature. The magnitude of the cross product of the tangent and normal vectors turned out to be 0, resulting in a curvature of 0. Therefore, the curve has zero curvature.

Step-by-step solution

Calculate the tangent vector (𝑟'(𝑡))

Differentiate each component of the position vector 𝑟(𝑡) with respect to 𝑡: $$ \mathbf{r}'(t) = \left\langle \frac{d}{dt}(a+bt^p), \frac{d}{dt}(c+dt^p), \frac{d}{dt}(e+ft^p) \right\rangle = \left\langle pb t^{p-1}, pd t^{p-1}, pf t^{p-1} \right\rangle $$

Calculate the normal vector (𝑟''(𝑡))

Now, we need to differentiate the tangent vector (𝑟'(𝑡)) with respect to 𝑡 to find the normal vector: $$ \mathbf{r}''(t) = \left\langle \frac{d}{dt}(pb t^{p-1}), \frac{d}{dt}(pd t^{p-1}), \frac{d}{dt}(pf t^{p-1}) \right\rangle = \left\langle p(p-1)bt^{p-2}, p(p-1)dt^{p-2}, p(p-1)ft^{p-2}\right\rangle $$

Calculate the curvature (𝜅)

To find the curvature, we need to find the magnitude of the cross product of the tangent and normal vectors (𝑟'(𝑡) × 𝑟''(𝑡)) and divide it by the cube of the magnitude of the tangent vector (|𝑟'(𝑡)|³): $$ \kappa(t) = \frac{|\mathbf{r}'(t) \times \mathbf{r}''(t)|}{|\mathbf{r}'(t)|^3} $$ Calculate the cross product of 𝑟'(𝑡) and 𝑟''(𝑡): $$ \mathbf{r}'(t) \times \mathbf{r}''(t) = \left\langle (pd t^{p-1})(p(p-1)ft^{p-2}) - (pf t^{p-1})(p(p-1)dt^{p-2}), (pf t^{p-1})(p(p-1)bt^{p-2}) - (pb t^{p-1})(p(p-1)ft^{p-2}) , (pb t^{p-1})(p(p-1)dt^{p-2}) - (pd t^{p-1})(p(p-1)bt^{p-2}) \right\rangle $$ The cross product simplifies to: $$ \mathbf{r}'(t) \times \mathbf{r}''(t) = \left\langle 0, 0, 0 \right\rangle $$ Now we need to find the magnitude of the tangent vector 𝑟'(𝑡): $$ |\mathbf{r}'(t)|=\sqrt{(pb t^{p-1})^2+(pd t^{p-1})^2+(pf t^{p-1})^2}=t^{p-1}\sqrt{p^2(b^2+d^2+f^2)} $$ Now, plug the cross product, and the tangent vector magnitude into the formula for the curvature: $$ \kappa(t) = \frac{|0|}{(t^{p-1}\sqrt{p^2(b^2+d^2+f^2)})^3} $$

Show that the curvature is zero

The magnitude of the cross product of the tangent and normal vectors is 0, so the curvature is: $$ \kappa(t) = \frac{0}{(t^{p-1}\sqrt{p^2(b^2+d^2+f^2)})^3} = 0 $$ Since the curvature (𝜅) is 0, we have proven that the given curve has zero curvature.

Key concepts

Tangent Vector

The tangent vector of a curve is essential for understanding its direction at any given point. To find this vector, we differentiate the position vector with respect to the parameter, usually denoted as time or simply "t."
This process gives us the rate of change of the position vector, which effectively tells us the direction in which the curve passes through a point.

For the curve \(\mathbf{r}(t) = \langle a + b t^{p}, c + d t^{p}, e + f t^{p} \rangle\), the tangent vector is derived by differentiating each component:

• The derivative of the \(x\)-component is \(pb t^{p-1}\).
• The derivative of the \(y\)-component is \(pd t^{p-1}\).
• The derivative of the \(z\)-component is \(pf t^{p-1}\).

Thus, the tangent vector is \(\mathbf{r}'(t) = \langle pb t^{p-1}, pd t^{p-1}, pf t^{p-1} \rangle\). This vector points in the direction the curve is moving at any point \(t\).
It provides a straightforward linear approximation of the curve's path near that point, crucial for analyzing the curve's properties.

Normal Vector

The normal vector plays a key role in differentiating a curve from a straight line. It specifies the direction in which you would "veer off" if you continued along the curve but slightly away from the tangent line.
To find the normal vector, we differentiate the tangent vector we previously found. This requires taking each component of the tangent vector \(\mathbf{r}'(t)\) and differentiating it again with respect to \(t\).

For our curve's tangent vector \(\mathbf{r}'(t) = \langle pb t^{p-1}, pd t^{p-1}, pf t^{p-1} \rangle\), differentiating gives:

• First component: \(p(p-1) b t^{p-2}\)
• Second component: \(p(p-1) d t^{p-2}\)
• Third component: \(p(p-1) f t^{p-2}\)

Thus, the normal vector is \(\mathbf{r}''(t) = \langle p(p-1) b t^{p-2}, p(p-1) d t^{p-2}, p(p-1) f t^{p-2} \rangle\).
This vector is perpendicular to the tangent vector, giving us insight into how sharply or gently the curve bends.

Position Vector

The position vector offers insight into where exactly a point on a curve is located within space.
For a parametric curve defined as \(\mathbf{r}(t) = \langle a + b t^{p}, c + d t^{p}, e + f t^{p} \rangle\), it represents the precise coordinates of the curve at any parameter \(t\).

This vector helps us understand:

• The starting point of the curve when \(t = 0\), simplifying to \(\langle a, c, e \rangle\).
• How the curve evolves as \(t\) changes, dictated by the rate of change components \(b t^{p}, d t^{p}, \) and \(f t^{p}\).

By focusing on the position vector, we align ourselves with the specific trajectory that the curve follows in 3D space. It offers a map that is crucial for plotting the points of the curve and understanding its overall shape.

Curvature Formula

The curvature formula tells us how sharply a curve bends at any particular point. The curvature, denoted as \(\kappa(t)\), is calculated using the magnitude of the cross product of the tangent vector \(\mathbf{r}'(t)\) and the normal vector \(\mathbf{r}''(t)\), divided by the cube of the magnitude of the tangent vector.

For our curve:

• Calculate the cross product \(\mathbf{r}'(t) \times \mathbf{r}''(t)\), which results in \(\langle 0, 0, 0 \rangle\).
• This result simplifies the curvature \(\kappa(t) = \frac{|0|}{(t^{p-1}\sqrt{p^2(b^2+d^2+f^2)})^3} = 0\).

Since the cross product is zero, it indicates there is no bending or twisting, hence zero curvature.
Essentially, this means our curve behaves more like a straight line, even if it may not appear to at first glance. This understanding of curvature is foundational in determining the "flatness" or "straightness" of curves in mathematical space.