We want to quantify the "curviness" of a curve in a way that is independent of parameterization. In other words, our measure of curvature should be intrinsic to the plane curve itself and not the way we trace it out. [br][br]Measuring the change in the unit tangent was a good start, but we found that this change was at least in part dependent on the parameterization. To avoid this complication let's first require that any parameterization used to measure curvature be an [i]arc length[/i] parameterization - that is, we will start with a [i]unit-speed[/i] parameterization.[br][br]Under these circumstances now, we define curvature as follows:[br][br]Let [math]C[/math] be a curve and [math]P_0[/math] a point on [math]C[/math]. Then the curvature of [math]C[/math] at [math]P_0[/math] is computed by first finding an arc length parameterization of [math]C[/math]: [math]\vec{c}\left(s\right)[/math]. if [math]s_0[/math] is the real number so that [math]\vec{c}\left(s_0\right)=P_0[/math] then the [b][color=#ff0000]curvature[/color][/b] of [math]C[/math] at [math]P_0[/math] is given by:[br][math]\kappa\left(s_0\right)=\left|\left|\vec{T}\,'\left(s_0\right)\right|\right|[/math][br][br]The GeoGebra applet below is first and foremost probably trying to do too much. Sorry. In the top screen you'll see a curve - you can enter any parameterization here together with a domain of definition. The graph of curvature will be displayed below (the [math]y-[/math]axis tracks curvature while the [math]x-[/math]axis tracks the value of the parameter over the given domain of definition). You can reparameterize by changing the function [math]p\left(x\right)[/math] but because GeoGebra is computing curvature via an arclength parameterization (that I'm not showing you) the graph of curvature will not change. Animate will move a point along the curve as [math]t[/math] runs through the domain of definition and simultaneously run a corresponding point along the graph of curvature illustrating how the curvature of the image curve changes.[br][br]As you experiment think about the following questions:[br][list][*]Can you create a curve with constant curvature everywhere?[/*][*]Can you create a curve that has zero curvature somewhere? everywhere?[/*][*]What kinds of curves create a discontinuous curvature graph?[/*][/list][br]If it helps - here are some of the interesting curves we've encountered so far in this course:[br][list][*]Cycloid: [math]\vec{c}\left(t\right)=\left(t-\sin t,1-\cos t\right),t\in\left[0,4\pi\right][/math] [/*][*]Astroid: [math]\vec{c}\left(t\right)=\left(cos^3t,sin^3t\right),t\in\left[0,2\pi\right][/math] [/*][*]Figure eight: [math]\vec{c}\left(t\right)=\left(\sin t,\sin t\cos t\right),t\in\left[0,2\pi\right][/math] [/*][*]Ellipse: [math]\vec{c}\left(t\right)=\left(2\cos t,3\sin t\right),t\in\left[0,2\pi\right][/math][/*][/list]
We've defined curvature of a curve at a point as the length of [math]\vec{T}'\left(s\right)[/math] where it is understood we have an arc length parameterization. However finding an arc length parameterization is often tedious. There are other methods for computing curvature from an arbitrary parameterization. I will leave it as an exercise for you to show that these are all equivalent:[br][br]If [math]\vec{c}\left(t\right)[/math] is a twice-differentiable regular curve then the curvature of the image curve at a point [math]\vec{c}\left(t_0\right)[/math] can be computed via:[br][br][math]\kappa\left(t_0\right)=\frac{\left|\left|\vec{T}\,'\left(t_0\right)\right|\right|}{\left|\left|\vec{c}\,'\left(t_0\right)\right|\right|}[/math][br]or via[br][math]\kappa\left(t_0\right)=\frac{\left|\left|\vec{c}\,'\left(t_0\right)\times\vec{c}\,''\left(t_0\right)\right|\right|}{\left|\left|\vec{c}\,'\left(t_0\right)\right|\right|^3}[/math]
Show that the curvature of a circle of radius [math]R[/math] is constantly [math]\frac{1}{R}[/math] at every point.
A circle with radius [math]R[/math] and center [math]\left(h,k\right)[/math] has the arc length parameterization:[br][math]\vec{c}\left(s\right)=\left(R\cos\left(\frac{s}{R}\right)+h,R\sin\left(\frac{s}{R}\right)+k\right),s\in\left[0,2\pi\right][/math][br][br]We compute:[br][math]\vec{c}\,'\left(s\right)=\left(-\sin\left(\frac{s}{R}\right),\cos\left(\frac{s}{R}\right)\right)[/math][br]Since this is an arc length parameterization we have [math]\vec{T}\left(s\right)=\vec{c}\,'\left(s\right)[/math], so we compute:[br][math]\vec{T}\,'\left(s\right)=\left(-\frac{1}{R}\cos\left(\frac{s}{R}\right),-\frac{1}{R}\sin\left(\frac{s}{R}\right)\right)[/math][br]Finally [math]\kappa\left(s\right)=\left|\left|\vec{T}\,'\left(s\right)\right|\right|=\sqrt{\left(-\frac{1}{R}\cos\left(\frac{s}{R}\right)\right)^2+\left(-\frac{1}{R}\sin\left(\frac{s}{R}\right)\right)^2}=\sqrt{\left(-\frac{1}{R}\right)^2\left(\cos^2\left(\frac{s}{R}\right)+\sin^2\left(\frac{s}{R}\right)\right)}=\sqrt{\left(-\frac{1}{R}\right)^2}=\frac{1}{R}[/math][br]Since the curvature does not depend at all on the value of [math]s[/math] we know it is constant across the circle and moreover is the reciprocal of the radius. This should agree with your intuition - a circle with a very large radius (think of the equator of the earth) feels flatter (less curvy) than one with a very small radius (like a penny).
Show that the curvature of a line is always 0.
Let [math]P_1=\left(x_1,y_1\right)[/math] and [math]P_2=\left(x_2,y_2\right)[/math] be two points on the line. We have the following arc length parameterization:[br][br][math]\vec{c}\left(s\right)=\left(x_1+\frac{s}{\sqrt{\left(x_2-x_1\right)^2+\left(y_2-y_1\right)^2}}\left(x_2-x_1\right),y_1+\frac{s}{\sqrt{\left(x_2-x_1\right)^2+\left(y_2-y_1\right)^2}}\left(y_2-y_1\right)\right),s\in\mathbb{R}[/math][br][br]We compute:[br][math]\vec{c}\,'\left(s\right)=\left(\frac{x_2-x_1}{\sqrt{\left(x_2-x_1\right)^2+\left(y_2-y_1\right)^2}},\frac{y_2-y_1}{\sqrt{\left(x_2-x_1\right)^2+\left(y_2-y_1\right)^2}}\right)[/math][br]Again since we started with an arclength parameterization we know [math]\vec{c}'\left(s\right)=\vec{T}\left(s\right)[/math].[br]Note that [math]\vec{T}\left(s\right)[/math] does not depend on [math]s[/math], hence [math]\vec{T}\,'\left(s\right)=\vec{0}[/math].[br]Thus [math]\kappa\left(s\right)=\left|\left|\vec{0}\right|\right|=0[/math] always.