00:01
We are provided with the position vector r of t it is equal to twice t plus 1, 4t minus 5, 6t plus 11.
00:11
Now we have to calculate the unit tangent vector and the curvature k.
00:18
For that first of all we will calculate the velocity vector v of t which is nothing but the d by dt of r of t and it is equal to 2, 4, 6 since derivative of this with respect to t is 2, derivative of this with respect to t is 4, derivative of 6t plus 11 with respect to t is 6.
00:43
So this is the required vector v of t.
00:48
Now magnitude of velocity is nothing but the speed it is equal to s of t and it is the mod of velocity vector v of t.
00:58
It is equal to under root of 2 square that is 4 plus 4 square is 16 plus 6 square is 36 and the value of this root it is 56, root 56 it is equal to 2 root 14.
01:14
So this is the value of s of t that is the magnitude of velocity.
01:19
Now we can calculate the unit tangent vector t which is equal to the velocity vector upon the speed that is vector upon its magnitude.
01:32
So vector is equal to 2, 4, 6 and its magnitude is 2 into under root 14.
01:42
So t is given as 1 upon root 14, 2 upon root 14, 3 upon root 14.
01:56
This is the required t which is known as the unit tangent vector...