00:01
In the problem, the given curve is r of t being equal to 6t i cap added to 3 sine t j cap plus 3 cos t k cap.
00:15
We have to find the curvature and for that we are trying to find the value of r prime t, that is the first order derivative and r double prime t.
00:25
If we try to write it in the vector form, the given curve can be written 6t 3 sine t and 3 cos t.
00:34
We have taken the coefficients only.
00:38
So the first order differentiation r prime t will be 6 3 cos t minus 3 sine t and the second order derivative it becomes 0 minus 3 sine t minus 3 cos t.
00:59
Now we have to find the curvature.
01:02
If we take the curvature, the theorem becomes modulus value for r prime t cross product r double prime t divided by modulus of r prime t whole cube.
01:19
If we take the cross product, it can be written in the determinant form.
01:24
So i j k for r prime t it will be 6 3 cos t minus 3 sine t and for r double prime t it will be 0 minus 3 sine t minus 3 cos t.
01:45
If we simplify this for i it will be multiplication of 3 cos t and minus 3 cos t that will be minus 9 cos square t minus of multiplication of minus 3 sine t and minus 3 sine t.
02:02
That will give us minus 9 sine square t.
02:08
If we take the j it will be multiplication of 0 with minus 3 sine t that will be 0 minus of multiplication of 6 and minus 3 cos t that will give us 18 cos t.
02:20
Now we take k that will give us minus 18 sine t and multiplication of 0 with 3 cos t...