Curvature of a Curve: Calculus 3 project by Ryan Taylor
Overview
This Project has been completed as part of a standard 10 weeks Calculus 3 asynchronous online course with optional WebEx sessions during Summer 2021 Semester at MassBay Community College, Wellesley Hills, MA.
Curvature of a Curve
What is Curvature
Driving down the road you see a bend in the road, and you turn your wheel a little bit to make that bend. A little bit farther down the road, you see another bend, but this one is very tight, and you will have to turn very quickly, so you turn your steering wheel a lot more and you make the bend. The amount that turned your steering wheel is a good representation of curvature.
Curvature measures how quickly a line is changing direction as it is travelled, so that tight turn where you had to turn your steering wheel a lot, had a large value for curvature while the first curve has a much smaller value. Another way to define this is that it is the inverse of the radius of the circle that would be created to fit that curve, the larger the radius, the smaller the curve.
Figure 1: Created with Solidworks
In this example, the radius of the circle that would hug the curve at that point is .5, since the curvature is the inverse of radius, the curvature would be 2 or 1/.5
Figure 2: Created with Solidworks
In this example, the radius of the circle is greater, which means the curvature is going to be smaller which means this is a more gentle curve. If the line were a road, the driver would have to turn the steering wheel a lot less to stay on the road for this curve rather than the first curve shown. This curvature value would be ~.5714 or 1/1.75.
The next challenge to overcome with curvature, is finding the curvature value when the radius of the curve is not given. That is what we will figure out next.
Equations
There are two equations to find curvature, they are as follows.
\(\displaystyle{k}=\frac{{\left\|{\vec{{T}}{\left({t}\right)}'}\right\|}}{{\left\|{\vec{{r}}{\left({t}\right)'}}\right\|}} ~~ \text{ or }~~ {k}=\frac{{\left\|{\vec{{r}}{\left({t}\right)}'\times\vec{{r}}{\left({t}\right)}{''}}\right\|}}{{\left\|{\vec{{r}}{\left({t}\right)}'}\right\|}^{3}} ~~ \text{ and }~~ \displaystyle\vec{{T}}{\left({t}\right)}=\frac{{\vec{{r}}{\left({t}\right)}'}}{{\left\|{\vec{{r}}{\left({t}\right)}'}\right\|}} \)
To see how these equations come from, watch the the following 3 videos. They provide an excellent explanation of the curvature formulas.
Let's use these formulas in a couple examples.
Example 1
First let's find the curvature of \(\displaystyle{\vec r}{\left({t}\right)}=<{4} \sin{{\left({t}\right)}},-{4} \cos{{\left({t}\right)}}>\)
Start by finding \(\vec {r}~'(t)\)
\(\vec {r}~'(t)=<4\cos(t),4\sin(t)>\)
This will allow us to find \(\vec{T}(t)\)
\(\displaystyle\vec{{T}}{\left({t}\right)}=\frac{{<{4} \cos{{\left({t}\right)}},{4} \sin{{\left({t}\right)}}>}}{\sqrt{{{16}{{\cos}^{2}{\left({t}\right)}}+{16}{{\sin}^{2}{\left({t}\right)}}}}}=\frac{{<{4} \cos{{\left({t}\right)}},{4} \sin{{\left({t}\right)}}>}}{{{4}\sqrt{{{{\cos}^{2}{\left({t}\right)}}+{{\sin}^{2}{\left({t}\right)}}}}}}=\frac{{<{4} \cos{{\left({t}\right)}},{4} \sin{{\left({t}\right)}}>}}{{4}}=<\cos(t),\sin(t)>\)
Next find \(\vec{T}~'(t)\)
\(\displaystyle\vec{{{T}}}~'{\left({t}\right)}=<- \sin{{\left({t}\right)}}, \cos{{\left({t}\right)}}>\)
Now that we have \(\vec{T}~'(t)\) we need to find \(||\vec{T}~'(t)||\) and \(||\vec{r}~'(t)||\).
\(\displaystyle{\left\|{\vec{{T}}{\left({t}\right)}'}\right\|}=\sqrt{{{\left(- \sin{{\left({t}\right)}}\right)}^{2}+{\left( \cos{{\left({t}\right)}}\right)}^{2}}}={1}\)
\(\displaystyle{\left\|{\vec{{r}}{\left({t}\right)}'}\right\|}=\sqrt{{{\left({4} \cos{{\left({t}\right)}}\right)}^{2}+{\left({4} \cos{{\left({t}\right)}}\right)}^{2}}}=\sqrt{{{16}{{\cos}^{2}{\left({t}\right)}}+{16}{{\sin}^{2}{\left({t}\right)}}}}={4}\sqrt{{{{\cos}^{2}{\left({t}\right)}}+{{\sin}^{2}{\left({t}\right)}}}}={4}\times{1}={4}\)
This will allow us to substitute these values into the first curvature equation.
\(\displaystyle\kappa=\frac{{\left\|{\vec{{T}}{\left({t}\right)}'}\right\|}}{{\left\|{\vec{{r}}{\left({t}\right)}'}\right\|}}=\frac{1}{{4}}\)
The curvature of \(\displaystyle{\vec r}{\left({t}\right)}=<{4} \sin{{\left({t}\right)}},-{4} \cos{{\left({t}\right)}}>\) is \(\kappa= \dfrac{1}{4}\)1/4 or .25. The inverse of this is 4, which is the radius of the equation. To check this, we can look at a graph of \(\vec{r}(t)\).
Figure 3: Created with Matlab
Example 2
Let's find the curvature of \(\displaystyle \vec{{r}}{\left({t}\right)}=<\frac{{t}^{3}}{{3}},{4}{t},\frac{{t}^{5}}{{5}}>\)
To start, let's graph this equation.
Figure 4: Created with Matlab
3D Graph.fig
This time we are going to use the second formula for curvature. That equation is as follows:
\(\displaystyle\kappa=\frac{{\left\|{\vec{{r}}{\left({t}\right)}'\times\vec{{r}}{\left({t}\right)}{''}}\right\|}}{{\left\|{\vec{{r}}{\left({t}\right)}'}\right\|}^{3}}\)
To make things simpler, we are going to find \(\vec{r}~'(t)\) and \(\vec{r}~''(t)\), then find the cross product, and then put everything into the equation.
\(\displaystyle\vec{{r}}{\left({t}\right)}'=<{t}^{2},{4},{t}^{4}>\)
\(\displaystyle\vec{{r}}{\left({t}\right)}{''}=<{2}{t},{0},{4}{t}^{3}>\)
Next we will need to find the cross product.
\(\displaystyle\vec{{r}}{\left({t}\right)}'\times\vec{{r}}{\left({t}\right)}{''}={\left|\begin{matrix}\hat{{i}}&\hat{{j}}&\hat{{k}}\\{t}^{2}&{4}&{t}^{4}\\{2}{t}&{0}&{4}{t}^{3}\end{matrix}\right|}=<{16}{t}^{3},-{2}{t}^{5},-{8}{t}>\)
From here we will put the vectors we found into the curvature formula.
\(\displaystyle\kappa=\frac{{\left\|{\vec{{r}}{\left({t}\right)}'\times\vec{{r}}{\left({t}\right)}{''}}\right\|}}{{\left\|{\vec{{r}}{\left({t}\right)}'}\right\|}^{3}}=\frac{\sqrt{{{\left({16}{t}^{3}\right)}^{2}+{\left(-{2}{t}^{5}\right)}^{2}+{\left(-{8}{t}\right)}^{2}}}}{{\left(\sqrt{{{\left({t}^{2}\right)}^{2}+{\left({4}\right)}^{2}+{\left({t}^{4}\right)}^{2}}}\right)}^{3}}=\frac{\sqrt{{{256}{t}^{6}+{4}{t}^{10}+{64}{t}^{2}}}}{{\left(\sqrt{{{t}^{4}+{16}+{t}^{8}}}\right)}^{3}}\)
To find the curvature at a specific point, a \(t\) value can be inputted, for example, let's use \(t=10\).
\(\displaystyle\frac{\sqrt{{{256}{t}^{6}+{4}{t}^{10}+{64}{t}^{2}}}}{{\left(\sqrt{{{t}^{4}+{16}+{t}^{8}}}\right)}^{3}}={2.0061}\times{10}^{ -{{7}}}\)
Curvature
Being able to calculate curvature can open a lot of doors and can be very helpful, and thankfully, it is fairly simple to do. The two formulas will both give the same results, but in some situations one may be easier than the other. Knowing when to use which equation will save you a lot of time and so practiving with each can be quite helpful. Now you will be able to calculate the curvature of the road you are driving on, or the curvature of a graph you are looking at. I hope you have found this to be a helpful guide!
If you would like to learn more about calculus, I highly encourage you to visit the following wiki pages that I found very informative!
Gradient and Directional Derivatives
References