Section 2.3 Arc Length for Vector Functions

2.3 Arc Length for Vector Functions

In this section, we calculate the arc length of vector functions. There is not much difference between 2-dimensional space that we learn in calculus II. Recall, [latex]\overrightarrow{r(t)}=\lt f(t),g(t)\gt[/latex] is a curve in 2-dimensional space then the arc length of the curve from [latex]t=a[/latex] to [latex]t=b[/latex] is \[ L=\int_{a}^{b}\sqrt{(f'(t))^{2}+(g'(t))^{2}}dt.\]

Theorem: Arc-Length 

Given a smooth curve [latex]C[/latex] defined by the function [latex]\overrightarrow{r(t)}=\lt f(t),g(t),h(t)\gt[/latex] where [latex]t[/latex] lies within the interval [latex][a,b][/latex], the arc length of [latex]C[/latex] over the interval is  \[ L=\int_{a}^{b}||\overrightarrow{r'(t)}||dt=\int_{a}^{b}\sqrt{(f'(t))^{2}+(g'(t))^{2}+(h'(t))^{2}}dt.\]

Proof: Use in 3D, [latex]a^{2}+b^{2}+c^{2}=d^{2}[/latex].

Example 1: Find the length of the arc of the circular helix with vector equation [latex]\overrightarrow{r(t)}=\lt \text{cos}(t),\text{sin}(t),t\gt[/latex] from the point [latex](1,0,0)[/latex] to the point[latex](-1,0,\pi)[/latex].

Exercise 1: Find the length of the arc of the circular helix with vector equation [latex]\overrightarrow{r(t)}=\lt 2\text{sin}(t),2\text{cos}(t),t\gt[/latex] from the point [latex](0,2,0)[/latex] to the point[latex](2,0,\frac{\pi}{2})[/latex].

Example 2: Find the length of the arc length of the curve [latex]\overrightarrow{r(t)}=\lt \text{cos}(t),\text{sin}(t),\text{ln}(\text{sin}(t))\gt[/latex] for [latex]\frac{\pi}{4}\leq t\leq\frac{\pi}{2}[/latex].

Exercise 2: Find the length of the arc length of the curve [latex]\overrightarrow{r(t)}=\lt \text{ln}(\text{cos}(t)),\text{cos}(t),\text{sin}(t)\gt[/latex] for [latex]0\leq t\leq\frac{\pi}{3}[/latex].

Recall the Fundamental Theorem of Calculus: [latex]f(t)[/latex] is a continue function on [latex][a,\infty)[/latex] then  \[ g(t)=\int_{a}^{t}f(x)dx \]  is continuous and differential such that [latex]g'(t)=f(t)[/latex]. We have similar theorem for the arc length function.

Theorem: Arc-Length Function

Let [latex]\overrightarrow{r(t)}[/latex] describe a smooth curve for [latex]t\geq a[/latex]. Then the arc-length function is given by  \[ s(t)=\int_{a}^{t}||\overrightarrow{r'(x)}||dx.\] Furthermore, [latex]\frac{d}{dt}[s(t)]=||\overrightarrow{r'(t)}||\gt 0[/latex]. 

Example 3: Find the arc length of the circular helix with vector equation [latex]\overrightarrow{r(t)}=\lt \text{cos}(t),\text{sin}(t),t\gt[/latex] starting from [latex](1,0,0)[/latex]. Reparametrize the helix with respect to arc length. 

Exercise 3: Find the arc length function of the circular helix with vector equation [latex]\overrightarrow{r(t)}=\lt 2\text{sin}(t),2\text{cos}(t),t\gt[/latex] starting from [latex](0,2,0)[/latex]. Reparametrize the helix with respect to arc length.

Example 4: Find the arc length function of the circular helix with vector equation [latex]\overrightarrow{r(t)}=\lt 2\text{cos}(t),-2\text{cos}(t),2\sqrt{2}\text{sin}(t)\gt[/latex] starting from [latex](2,-2,0)[/latex]. Reparametrize the helix with respect to arc length. 

Exercise 4: Find the arc length function of the circular helix with vector equation [latex]\overrightarrow{r(t)}=\lt 3\text{sin}(t),-3\sqrt{2}\text{cos}(t),3\text{sin}(t)\gt[/latex] starting from [latex](0,-3\sqrt{2},0)[/latex]. Reparametrize the helix with respect to arc length.

The curvature of [latex]C[/latex] at a given point is a measure of how quickly the curve changes direction at that point. Specifically, we define it to be the magnitude of the rate of change of the unit tangent vector with respect to arc length.

Definition

Let [latex]C[/latex] be a smooth curve in the plane or in space given by [latex]\overrightarrow{r(s)}[/latex], where [latex]s[/latex] is the arc-length parameter. The curvature [latex]\mathit{k}[/latex] at [latex]s[/latex] is \[ k=||\frac{\overrightarrow{T(s)}}{ds}||=||\overrightarrow{T'(s)}||. \]

Theorem

Let [latex]C[/latex] be a smooth curve in the plane or in space given by [latex]\overrightarrow{r(t)}[/latex], where [latex]t[/latex] is the parameter. The curvature [latex]\mathit{k}[/latex] at [latex]t[/latex] is \[ k=\frac{||\overrightarrow{T'(t)}||}{||\overrightarrow{r'(t)}||}. \]

Example 5: Show that the curvature of a circle of radius [latex]a[/latex] is [latex]\frac{1}{a}[/latex].

Exercise 5: Find the curvature of [latex]\overrightarrow{r(t)}=\lt 3\text{sin}(t),3\text{cos}(t)\gt[/latex]. 

Example 6: Find the curvature of [latex]\overrightarrow{r(t)}=\lt 2\text{sin}(t),2\sqrt{2}\text{cos}(t),-2\text{sin}(t)\gt[/latex]

Exercise 6: Find the curvature of [latex]\overrightarrow{r(t)}=\lt -3\sqrt{2}\text{sin}(t),3\text{cos}(t),3\text{cos}(t)\gt[/latex]. 

Example 7: Set up an integration to find the arc length of curve of the intersection of two surfaces [latex]9y^{2}+z^{2}=9[/latex] and [latex]x+y+z=1[/latex].

Example 8: Suppose you start at the point [latex](1,0,0)[/latex] and move [latex]4[/latex] units along the curve [latex]x=\text{cos}(t)[/latex], [latex]y=\text{sin}(t)[/latex], [latex]z=2t[/latex] in the positive direction. Where are you now?

Group work:

1. Set up an integration to find the arc length of curve of the intersection of two surfaces [latex]x^{2}+4z^{2}=4[/latex] and [latex]x+y=1[/latex].

2. Suppose you start at the point [latex](0,0,4)[/latex] and move [latex]3[/latex] units along the curve [latex]x=4\text{sin}(t)[/latex], [latex]y=3t[/latex], [latex]z=4\text{cos}(t)[/latex] in the positive direction. Where are you now?

3. Find the arc length function of the curve [latex]\overrightarrow{r(t)}=\lt \sqrt{2}t,e^{t},e^{-t}\gt[/latex] over the interval [latex]0\leq t\leq1[/latex]. 

4. Set up an integration to find the arc length of curve of the intersection of two surfaces [latex]1=x^{2}+4y^{2}[/latex] and [latex]x-y+z=1[/latex].