00:01
Let's talk about this question.
00:02
We need to find the parametric equations of the tangent line to the curve with the parametric equations at a specific point.
00:09
So the point is 2, 4, comma, minus 1.
00:14
Let's write the r vector first that's given as t square plus 1.
00:19
That's x, that's x coordinate, y coordinate is 4 root t, and the z coordinate is e raised to t squared minus t.
00:26
So for that, we first need to find the derivative of the r vector.
00:30
So that's going to be the derivative of t square plus one, which is 2t.
00:33
This is going to be 4 over 2 root t.
00:36
And this is going to be e raise to t square minus t.
00:39
We're going to use the chain rule here.
00:41
And the differentiation of power is going to be 2t minus 1.
00:46
This can be read written as 2t times 4 over 2 is just 2.
00:50
So this is 2 over root 2.
00:52
2 over root 2.
00:53
And this is going to be 2t minus 1 times e raised to t square minus 1...