00:01
We're given s of t, which equals 106 minus 100e to the negative point 2t.
00:06
And we're asked what is the rate of change at various points.
00:09
So let's go ahead and find s prime of t, which is the derivative, because the derivative is the same as the slope at a point.
00:18
The derivative of 106 is zero.
00:21
And so then we'll have minus 100, and then the derivative of e to the stuff is e to the stuff.
00:29
And the reason i didn't write the 100 yet is because we also have to multiply by 0 .2.
00:35
So we're going to actually multiply by negative 0 .2.
00:38
So the doubled negative will cancel out.
00:40
So we get 0 .2 times 100, which is 20.
00:43
So s prime of t is 20 e to the negative 0 .2t.
00:48
So if we want to plug in at 1 or 5, we just plug it in.
00:52
So s prime of 1 and s prime of 5, i will just plug in 20 e.
01:00
To the negative 0 .2 times 1...