00:01
In this problem, we are given a parametric curve, from which you want to find, is curvature at point p.
00:06
So here we have the x component of our curve equal to t minus t squared, and the one component of our curve equal to 1 minus t cube, for which we want to find as curvature at the point x equal to 0 and y is equal to 1.
00:23
So here we have our curve written in parametric format, but let's rewrite it in vector form.
00:30
So our curve, r of t, will have x component t minus t squared and y component 1 minus t cube.
00:44
Let's evaluate its first and second derivative.
00:49
Why? because we're going to need these two for the calculation of the curvature.
00:56
The first derivative of our prime, excuse me, the first derivative of our curve r will give us 1 minus 2t.
01:03
In the s component and minus 3 t square in the y component.
01:10
The front chain is expression again, we'll find r prime prime of t, which will give us minus 2 in the x component and minus 6t in the y component.
01:28
The curvature, kappa, is defined as the magnitude of the cross product between r prime of t and r prime prime of t.
01:45
Divided by the magnitude cube of r prime of t.
01:57
So we've already evaluated r prime and r prime prime prime.
02:01
What's next is to evaluate the value of these two vector positions at the point p, because we were precisely looking for the curvature at this point p.
02:15
But to do so, we need to find a parametric variable t that will satisfy the given x and y component, meaning that we want to find t such that on our curve x is equal to 0 and y is equal to 1.
02:39
Meaning that we want to simultaneously solve the following two equations...