Concepts in Curvature

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Step 2: Find the Unit Tangent Vector

The Unit Tangent Vector (TT) is the tangent of the curve with a uniform length of one. To find TT, we must first find dsdt\frac{ds}{dt}, which is the rate of change of a point traveling along the curve with respect to time. In other words, it's the tangent to the curve (you could also think of it as "velocity").

Given that s(t)s(t) is defined by the following components,

x(t)=atbsin(t)y(t)=abcos(t)x(t) = at-b\sin(t) \\ y(t) = a-b\cos(t)

We can find dsdt\frac{ds}{dt} by taking the derivative of each component.

dsdt=dxdt,dydt=abcos(t),bsin(t)\frac{ds}{dt} = \left\langle \frac{dx}{dt}, \frac{dy}{dt} \right\rangle = \left\langle a-b\cos(t), b\sin(t) \right\rangle

Now that we have dsdt\frac{ds}{dt}, we can find TT by dividing dsdt\frac{ds}{dt} by its magnitude, which gives us a vector with a length of one.

T=dsdtdsdt=abcos(t),bsin(t)(abcos(t))2+(bsin(t))2T = \frac{\frac{ds}{dt}}{\lVert\frac{ds}{dt}\rVert} = \frac{\left\langle a-b\cos(t), b\sin(t) \right\rangle}{\sqrt{(a-b\cos(t))^2+(b\sin(t))^2}}

The specific math is getting a bit messy at this point, but the important part is that you're able to understand what dsdt\frac{ds}{dt} and TT are. Try flipping between the visualizations of both of these vectors to get a better understanding of what they represent. Once you're ready, move on to the next step.

dsdt\frac{ds}{dt}TT

t=0.000t = 0.000ds/dt=1.000\left\lVert ds/dt \right\rVert= 1.000T=1.000\left\lVert T \right\rVert= 1.000