00:01
In this question, we are asked to find the curvature at a given point.
00:04
Since it's a two -dimensional curve, we can find the curvature by the formula absolute value of x ' times y' ' minus y ' multiplied by x' ' divided by the sum of the squares of x ' and y ' raised to the three -halves power, where x and y are the coordinates of the vector r.
00:32
That means that x ' equals to 1 minus cos t, x' ' equals to sin t, y' ' equals to sin t, and y' ' equals to cos t.
00:56
Now we'll calculate each of the derivatives at the point 2pi over 3.
01:03
X ' of 2pi over 3 equals to 1 minus cos 2pi over 3.
01:13
2pi over 3 is 120 degrees, and cos equals to negative one -half there.
01:19
So we'll get 1 minus negative one -half.
01:23
This equals to three -halves.
01:26
X' ' of 2pi over 3 equals to sin 2pi over 3, equals to the square root of 3 over 2...