00:01
We will find the tangent vector, the normal vector and the binormal vector t and b for the curve r of t equal 4 cosine of 2t, 4 sine of 2t and 2t at the point t equals 0.
00:19
So we are going to find in general the tangent vector and normal vector and then to calculate the binormal vector we do it specifically at 0.
00:31
So first we find the derivative of the curve that is a vector formed by the derivative of each component of r of t.
00:49
That is 4 times derivative of cosine of 2t is negative sine of 2t times derivative of 2t.
00:57
Because 2t depends on t and is not only variable t.
01:02
So we say it is negative 8 sine of 2t.
01:09
I repeat that comes from the fact that we have constant 4 and then we find the derivative of cosine of 2t that is negative sine of 2t but we have to multiply that by the derivative of 2t which is 2.
01:25
So the 4 we had before and that 2 that comes from the derivative of 2t gives us the factor a.
01:34
And the derivative of the function we saw is negative sine of 2t.
01:39
Similar situation for the second component.
01:42
We have a constant 4 and then derivative of sine of 2t is cosine of 2t times derivative of 2t that is times 2.
01:52
We finally get 8 cosine of 2t.
01:58
And derivative of 2t is 2.
02:05
Now we calculate the norm of this vector.
02:13
The norm of this derivative is the square root of the sum of the squares of the components that is negative 8 sine of 2t squared plus 8 cosine of 2t squared plus 2 squared.
02:39
So this is the norm of the derivative of the curve.
02:43
So it is the square root of 64 sine squared of 2t plus 64 cosine squared of 2t plus 4.
02:59
Now you see the first two terms has a common factor 64 so we get 64 times sine squared of 2t plus cosine squared of 2t and that plus 4.
03:20
Now we remember that the sum of the squares of sine and cosine function at the same angle that is sine squared of an angle plus cosine squared of the same angle is equal to 1.
03:32
So this is the square root of 64 plus 4 because this expression inside parentheses is equal to 1 and that is the square root of 68 which is just 2 times square root of 17 because 17 times 4 is 68.
03:57
So the square root of 4 is 2 so that gets out of the square root and inside the square root we keep the value of 70.
04:10
And so this is the norm of the derivative of the curve vector.
04:25
And so the tangent vector is in general that is for any parameter t is t of t equal the derivative of the curve.
04:50
Let me arrange this a bit.
04:55
The derivative of the curve at t over the norm of that derivative vector.
05:08
And so tangent vector at t is 1 over the norm of the derivative vector that is 1 over 2 square root of 17.
05:23
I say 1 over that number because the norm of the vector is in the denominator so it's the same as we write 1 over the norm of the vector, derivative vector times derivative vector.
05:44
So 1 over the norm is 1 over 2 square root of 17 times the vector, derivative vector is this one right here, is negative 8 sine of 2t, 8 cosine of 2t and 2.
06:13
Now we multiply this scalar to each component of the vector and so we get that the tangent vector at t is vector negative 8 over 2 square root of 17 is 4 over square root of 17 sine of 2t.
06:40
Then 8 over 2 square root of 17 is 4 over square root of 17 cosine of 2t.
06:51
And 2 times 1 over 2 square root of 17 is 1 over square root of 17.
06:59
That's the tangent vector in general at any t and so the tangent vector at 0 is sine of 0 is 0 so we get 0 as the first component, the second one cosine of 0 is 1 so we get 4 over square root of 17.
07:22
And the third component is a constant component that does not depend on t...