# 7 Geometry

**(Relevant Section of the Textbook: 13.3 Arc Length and Curvature)**

## 7.1 Arc Length

**Definition 7.1 (Infinitesimal Arclength)** If \(\vec{r}(t)\) is a parametric curve, its infinitesimal arclength is measured by \[ds=|\vec{r}^\prime(t)|\,dt\]

This makes sense: after all the derivative \(\vec{r}^\prime(t)\) is the *velocity*, \(\|\vec{r}^\prime(t)\|\) is the *speed*, and \(dt\) is an infinitesimal length of time. Thus, the product \(\|\vec{r}^\prime(t)\|dt\) is an infinitesimal bit of *distance* - a small length along the curve. To take this infinitesimal information and get something useful out - we need to *integrate* along the curve.

**Definition 7.2 (Arclength)** If \(\vec{r}(t)\) is a parametric curve, its length between \(t=a\) and \(t=b\) is given by \[L=\int_a^b ds=\int_a^b |\vec{r}^\prime(t)|dt\]

**Example 7.1 (Arclength of a Helix)** Find the arclength of \(\vec{r}(t)=(\cos(t),\sin(t),t)\) from \(t=0\) to \(t=2\pi\). First, we need to find the velocity \(\vec{r}^\prime\): \[\vec{r}^\prime(t)=\langle -\sin(t),\cos(t),1\rangle\] Next, we need to take this velocity and find the speed: \[\|\vec{r}^\prime(t)\|=\sqrt{(-\sin t)^2+(\cos t)^2+1}=\sqrt{2}\] Finding arclenght is just integrating this over the domain: \[\int_0^{2\pi}\|vec{r}^\prime(t)\|dt=\int_0^{2\pi}\sqrt{2}dt=2\pi\sqrt{2}\]

Often arclength integrals can be challenging to do, because of the square root. But with some algebra and integration tricks, alot can be learned.

From this idea, we can define the *arclength function* whcih measures the length of a curve \(\vec{r}(t)\) from a starting point \(t=a\):

**Definition 7.3 (The Arclength Function)** If \(\vec{r}(t)\) is a parametric curve, for any given starting point \(t=a\) we may define the *arclength function* which measures the length of curve between \(a\) and \(t\): \[s(t)=\int_a^t |\vec{r}^\prime(u)|du\] (Note we have changed the variable of integration so that \(t\) is not used in two different contexts)

What is the arclength function for the helix in our earlier example, starting from \(t=0\)? Since \(\|\vec{r}^\prime(t)\|=\sqrt{2}\), we see that

\[s(t)=\int_0^t \sqrt{2}dt=\sqrt{2}t\]

This tells us that after \(t\) seconds, we have traced out \(\sqrt{2}t\) units of arclength. How could we reparameterize this curve so that its arclength function is just \(s(t)=t\) (tracing out \(t\) units of arc in \(t\) units of time)?

**Definition 7.4 (Unit Speed Curve)** A curve \(\vec{c}(t)\) is *unit speed* if \(\|\vec{c}^\prime(t)\|=1\) for all times \(t\). This means that it after \(t\) seconds, the curve has traversed \(t\) units of length. For this reason, we also call unit speed curves *arclength parameterized curves*.

In our example, to make the helix unit speed we need to slow it down by a factor of \(\sqrt{2}\): that is, we need \(\vec{r}(t/\sqrt{2})\):

\[\vec{r}\left(\frac{t}{\sqrt{2}}\right)=\left(\cos\frac{t}{\sqrt{2}},\sin\frac{t}{\sqrt{2}},\frac{t}{\sqrt{2}}\right)\]

## 7.2 Curvature

Besides the length of a curve, one of the most powerful things calculus allows us to do is rigorously study its *curvature*. How can we quantify the fact that some curves bend gently and others turn sharply in space? One means of trying to do this is by looking at the tangent vectors to the curve, and trying to determing how quickly they are changing.

Of course, there’s a complication to this: a tangent vector can change in *length* without changing in direction. This doesn’t mean that a curve is *curving*, but rather that the particle tracing it out is *accelerating*.

To remove this worry, we define the *unit tangent vector* to a curve. Just divide the derivative by its magnitude!

**Definition 7.5 (Unit Tangent Vector)** The unit tangent vector to the curve \(\vec{r}(t)\) is the vector of length \(1\) which is parallel to \(\vec{r}^\prime(t)\): \[\vec{T}(t)=\frac{\vec{r}^\prime(t)}{|\vec{r}^\prime(t)|}\]

This allows a clean definition of curvature: it is how much the unit tangent vector turns per arclength.

**Definition 7.6 (Curvature of a Curve)** The curvature of a curve is \[\kappa = \left|\frac{d\vec{T}}{ds}\right|
=\left|\vec{T^\prime}(t)\right|\Bigg/\left|\vec{r}^\prime(t)\right|
\]

Where the second equality is derived via the chain rule: \[\left|\frac{d\vec{T}}{ds}\right|=\left|\frac{d\vec{T}}{dt}\frac{dt}{ds}\right|=\left|\frac{\frac{d\vec{T}}{dt}}{\frac{ds}{dt}}\right| =\frac{\left|\vec{T}^\prime\right|}{|\vec{r}^\prime|} \]

This formula is difficult to apply in genreral, as the unit tangent vector \(T\) might have a pretty scary looking formula, and so taking its derivative can be a lot of work.

Doing some calculus we can get a simpler formula:

**Theorem 7.1 (Curvature of a Curve)** The curvature of \(\vec{r}(t)\) is given by \[\kappa(t)=\frac{|\vec{r}^\prime(t)\times\vec{r}^{\prime\prime}(t)|}{|\vec{r}^\prime(t)|^3}\]

This is something that’s relatively easy to compute (though perhaps tedious) from any parameterization: you just need to find the first and second derivatives, take a cross product, and then plug into the formula!

But, if we further restrict ourselves to the case that \(\vec{r}(t)=(t,f(t))\) traces the graph of a function, we can simplify this calculation even more:

**Theorem 7.2 (Curvature of a Graph)** If \(y=f(x)\) is a function, the curvature of its graph is \[\kappa(x)=\frac{|f^{\prime\prime}(x)|}{|1+f^\prime(x)|^{3/2}}\]

The below graphing calculator lets you entere a function \(\kappa(s)\) that specifies the curvature of a curve, and then it computes a curve which has that curvature! Try even just the case \(k(s)=s\) and think about the result - what sort of curve do you expect to see if the curvature grows linearly along the length of the curve?

## 7.3 Framing a Curve

The unit tangent vector provides us with a very useful “pointer” - always oriented directly along a curve. But in any serious application of parametric curves, we need more information: we would like a whole \(x,y,z\) coordinate frame at each point of the curve.

To start, we’ll look for one vector which is orthogonal to, or *normal to* our curve.

How can we find one? Well, the unit tangent is of constant length (its the *unit tangent*, after all). We can use the *product rule for dot products* to understand \(T^\prime\):

\[(T\cdot T)=1 \implies (T\cdot T)^\prime = 0\] \[(T\cdot T)^\prime = T^\prime \cdot T + T\cdot T^\prime = 2 T\cdot T^\prime\]

Thus, we see that \(2T\cdot T^\prime =0\), so \(T\) is orthogonal to \(T^\prime\)!. To find a unit vector orthogonal to \(T\), we just need to normalize \(T^\prime\).

**Definition 7.7 (Normal Vector)** \[\vec{N}(t)=\frac{\vec{T}^\prime(t)}{|\vec{T}^\prime(t)|}\]

Given these two, its easy to find a third unit vector: just take the cross product of \(T\) and \(N\)! The result is called the *binormal* as its a second normal vector to the curve.

**Definition 7.8 (Binormal Vector)** \[\vec{B}(t)=\vec{T}(t)\times\vec{N}(t)\]

Together these three vectors provide a coordinate system at each point along the curve: \(T\) measures distance in the tangent direction to the curve, \(N\) measures distance in the direction the curve is bending fastest, and \(B\) is orthogonal to both. This collection of vectors is called the *Frenet Frame* and is heavily used in computations in physics, engineering, and computer graphics.

**Definition 7.9 (Frenet Frame)** \[\vec{T}(t)=\frac{\vec{r}^\prime(t)}{|\vec{r}^\prime(t)}\hspace{0.5cm}\vec{N}(t)=\frac{\vec{T}^\prime(t)}{|\vec{T}^\prime(t)|}\hspace{0.5cm}\vec{B}(t)=\vec{T}(t)\times\vec{N}(t)\]

## 7.4 Videos

Here are some useful videos reviewing the sort of examples that we have covered in class:

#### 7.4.0.1 Arclength of Curves

Parameterizing a curve with respect to arclength (a unit speed parameterization)

#### 7.4.0.2 Unit Tangents and Normals

The calculus Blue series on Tangent, Normal and Curvature:

The binormal vector and the Frenet Frame:

#### 7.4.0.3 Curvature

An application of this: finding the point on a curve where it is maximally curved (say, you wanted to find the sharpest bend of a roller coaster)