00:01
Hello, welcome to this lesson.
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In this lesson we'll find the parametric equation of the tangent line to the kef.
00:07
So the kef is rt which is equal to 2 ln t plus 1.
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Then we have t equals 2t plus 2.
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Then we have e to the power t.
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So we are evaluating this or we are looking at the tangent line at 0, 2 and 1.
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The first thing i will do is that we find the directional vector.
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The vector would evaluate it at t equals to 0, finding the first derivative with respect to t.
00:49
So here we have 2 ln t plus 1 and we are taking the derivative of this.
00:57
So we'll have 2 on t plus 1.
01:02
The next thing we are looking at 2, we are looking at t equals 2t plus 2.
01:10
Taking the derivative of this.
01:12
So first of all we hold t and differentiate cos 2t and we'll have negative, so negative t, all right and this is multiplying sine.
01:31
Actually let me bring the 2.
01:38
So i'm holding the t and differentiating cos 2t.
01:41
So i have negative 2 sine 2t.
01:47
Then again i'll hold the cos 2t and differentiate t.
01:51
So this is plus cos 2t, all right and the last component we are differentiating e to the power t.
02:06
So that is e to the power t.
02:10
So let this be vain and let's evaluate it at 0.
02:16
So we have r prime of 0.
02:19
If we put 0 there we have 2 on 0 plus 1, all right.
02:25
We have 0, this is 2t sine 2 times 0, okay, plus cos 2 times 0, all right and this is e to the power 0...