00:01
Hi there, so for this problem, we are given this expression for the curve and we need to determine the tangent and normal and the binormal vectors.
00:11
So first of all, we are going to obtain the tangent vector.
00:19
And the tangent vector is just the derivative of the curve divided by the magnitude of that derivative of the curve.
00:29
Now, let's determine first the derivative of this curve.
00:34
That will give us then 10 times the sine of five times the time.
00:42
We need to include a minus sign in there and we will have 10 times the cosine of five times the time and three for the last component.
00:54
Now, the magnitude of this will be just simply the square root of the sum of the squares of these components.
01:06
So we will obtain then that that is just simply 100 plus because that is 10 to the square and remember that the sine plus the cosine square is one and then we will have three to the square, which is nine.
01:22
So this will give us the square root of 109.
01:25
Okay, so the tangent vector, we just remember that we want this at the time equals to zero.
01:34
So we just need to evaluate this up zero.
01:37
So that will give us that the tangent vector at zero is just zero for the first component because the sine of zero is zero.
01:45
10 for the next component because the cosine of zero is one and three for the last component.
01:50
Oh, of course we need to multiply this by one divided by the square root of 109.
02:02
So that's the solution for the first question.
02:05
Now for the next question is about the normal vector and the normal vector at any given time.
02:13
So let's determine first this at any given time is just the tangent vector, the derivative of the tangent vector divided by the magnitude of that vector.
02:24
Okay, so let's first determine the derivative of the tangent vector.
02:30
That will be the derivative of this one in here.
02:34
But of course we need to divide this by 109.
02:37
So that will be give us one divided by 109.
02:40
Then this will be the derivative in here.
02:44
So that will be then five times 10.
02:47
So that will give us minus 50 times the cosine of five times the time.
02:52
Then the next component is just 50, well, minus 50 because that will give us the sine of five times the time.
03:02
And the last component is just zero.
03:05
Now we determine the magnitude of this vector.
03:11
So that will be the square root of then 50 to the square divided by 109...