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14.4 Length of Curves

Given a vector function \(\vec{r}(t)\) for \(a \le t \le b\), we want to find the length of the curve.

Figure: A curve in 2D space starting at t=a and ending at t=b, with the question 'length = ?' written below it.

Note: if \(h\) is small, then \(|\vec{r}(t+h) - \vec{r}(t)| \approx \text{actual length from } t \text{ to } t+h\).

How to calculate \(|\vec{r}(t+h) - \vec{r}(t)|\) to estimate a small segment of length? Once done, accumulate by integration.

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Derivation of Arc Length

Recall the definition of the derivative:

\[ \vec{r}'(t) = \lim_{h \to 0} \frac{\vec{r}(t+h) - \vec{r}(t)}{h} \]

This means when \(h\) is small:

\[ \vec{r}'(t) \approx \frac{\vec{r}(t+h) - \vec{r}(t)}{h} \]

So,

\[ \vec{r}(t+h) - \vec{r}(t) \approx h \vec{r}'(t) \] \[ |\vec{r}(t+h) - \vec{r}(t)| \approx h |\vec{r}'(t)| \]

So, one small segment has length given by \(|\vec{r}'(t)| h\).

The total length \(L\) is the sum of all segments:

\[ L = \sum_{i=1}^{n} |\vec{r}'(t)| h \approx \sum_{i=1}^{n} |\vec{r}'(t)| \Delta t \]

Where \(h\) is the change in \(t\), called \(\Delta t\). Now let \(n \to \infty\), so \(\sum \to \int\) and \(\Delta t \to dt\).

Therefore, the exact length \(L\) for \(a \le t \le b\) is:

\[ L = \int_{a}^{b} |\vec{r}'(t)| dt \]

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Example

Given the vector function:

\[ \vec{r}(t) = \langle \cos t, \sin t, t \rangle, \quad \frac{\pi}{2} \le t \le 2\pi \]

This represents a helix traveling on the surface of a cylinder.

Figure: A 3D coordinate system showing a helical path winding around a cylinder, with points marked at t = pi/2 and t = 2pi.

\[ L = \int_{a}^{b} |\vec{r}'(t)| \, dt \]

Calculating the derivative and its magnitude:

\[ \vec{r}' = \langle -\sin t, \cos t, 1 \rangle \] \[ |\vec{r}'| = \sqrt{\sin^2 t + \cos^2 t + 1} = \sqrt{2} \]

Evaluating the integral:

\[ \int_{\pi/2}^{2\pi} \sqrt{2} \, dt = \sqrt{2} t \Big|_{\pi/2}^{2\pi} = \sqrt{2} \left( 2\pi - \frac{\pi}{2} \right) = \boxed{\frac{3\pi}{2} \sqrt{2}} \]

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Arc Length as a Function of t

What if we want the length as a function of t? Let S(t) = ? for a ≤ t ≤ something else.

We modify the arc length formula:

\[ L = \int_{a}^{b} |\vec{r}'(t)| \, dt \]

Remove b and replace it with t:

\[ S(t) = \int_{a}^{t} |\vec{r}'(u)| \, du \]

Note: u is called a "dummy variable". We don't want t to be the variable when we integrate it to t.

Try it on the helix

Given \(\vec{r}(t) = \langle \cos t, \sin t, t \rangle\) and |\(\vec{r}'| = \sqrt{2}\):

\[ S(t) = \int_{\pi/2}^{t} \sqrt{2} \, du = \sqrt{2} u \Big|_{\pi/2}^{t} = \boxed{\sqrt{2} \left( t - \frac{\pi}{2} \right)} \]

At \(t = 2\pi\):

\[ S(2\pi) = \sqrt{2} \left( 2\pi - \frac{\pi}{2} \right) = \frac{3\pi}{2} \sqrt{2} \quad \text{(same as before)} \]

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\[ S(t) = \int_{a}^{t} |\vec{r}'(u)| du \]

gives relationship between distance and time.

In \(\vec{r}(t) = \langle \cos t, \sin t, t \rangle\) gives us the position at a given time \(t\).

but we know \(S(t)\) so we can change the parameter from time \(t\) to distance \(S\) to get \(\vec{r}(S)\) → location having traveled distance of \(S\).

back to helix : \(\vec{r}(t) = \langle \cos t, \sin t, t \rangle\), \(\frac{\pi}{2} \le t \le 2\pi\)

we found \(S(t) = \sqrt{2}(t - \pi/2)\) from \(t = \pi/2\) to some other \(t\)

\(S = \sqrt{2}(t - \pi/2) \rightarrow t = \frac{S}{\sqrt{2}} + \frac{\pi}{2}\)

Sub in \(\vec{r}(t)\):

\[ \vec{r}(S) = \langle \cos(\frac{S}{\sqrt{2}} + \frac{\pi}{2}), \sin(\frac{S}{\sqrt{2}} + \frac{\pi}{2}), \frac{S}{\sqrt{2}} + \frac{\pi}{2} \rangle \]

"using arc length as parameter"

\(\frac{\pi}{2} \le t \le 2\pi \rightarrow \text{length at } t=\pi/2 \le S \le \text{length at } t=2\pi\)

\(0 \le S \le \frac{3\pi}{2}\sqrt{2}\)

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14.5 Curvature

curvature is a measure of how "curvy" a curve is.

Straight line has curvature of 0.

given \(\vec{r}(t)\), how to find curvature \(\kappa(t)\).

recall \(\vec{T}\) is the unit tangent vector:

\[ \vec{T} = \frac{\vec{r}'(t)}{|\vec{r}'(t)|} \]

the unit tangent vector \(\vec{T}\) changes more rapidly on more curvy portion.

so we need to track the change in \(\vec{T}\) as we move along:

\[ \left| \frac{d\vec{T}}{ds} \right| = \kappa \]

where \(s\) is length

Figure: Three line segments showing increasing curvature: a straight line, a gentle curve, and a sharper curve, each with tangent vectors.

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but \(\frac{d\vec{T}}{ds}\) is not always most practical, since \(\vec{r}\) is usually a function of \(t\) (because \(\vec{r}(t)\) is more often given as a function of \(t\))

→ find alternative formula that deals with \(t\)

\(s(t) = \int_{a}^{t} |\vec{r}'(u)| du\)

we get \(\frac{d}{dt} s(t) = \frac{d}{dt} \int_{a}^{t} |\vec{r}'(u)| du\)

\(\frac{ds}{dt} = |\vec{r}'(t)|\) from the Fundamental Theorem of Calculus (part 1)

\(\kappa = |\frac{d\vec{T}}{ds}| = \frac{|d\vec{T}|}{|ds|} = \frac{|\frac{d\vec{T}}{dt}|}{|\frac{ds}{dt}|} = \frac{|\vec{T}'(t)|}{|\vec{r}'(t)|} = \kappa\)

other forms (vector) : \(\kappa = \frac{|\vec{r}'' \times \vec{r}'|}{|\vec{r}'|^3}\)