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Video transcript

- [Voiceover] Johanna jogs along a straight path. For zero is less than or equal to t is less than or equal to 40, Johanna's velocity is given by a differentiable function v. Selected values of v of t, where t is measured in minutes and v of t is measured in meters per minute, are given in the table above. We see that right over there. Use the data in the table to estimate the value of not v of 16 but v prime of 16. And we see here, they don't even give us v of 16, so how do we think about v prime of 16. Well, just remind ourselves, this is the rate of change of v with respect to time when time is equal to 16. So, we can estimate it, and that's the key word here, estimate. We can estimate v prime of 16 by thinking about what is our change in velocity over our change in time around 16. And when we look at it over here, they don't give us v of 16, but they give us v of 12. They give us v of 20. So, let's figure out our rate of change between 12, t equals 12, and t equals 20. So, our change in velocity, that's going to be v of 20, minus v of 12. And then our change in time is going to be 20 minus 12. And so, this is going to be equal to v of 20 is 240. 240 v of 12. We see right there is 200. And so, this is going to be 40 over eight, which is equal to five. And we would be done. For good measure, it's good to put the units there. So, this is our rate. This is how fast the velocity is changing with respect to time. So, the units are gonna be meters per minute per minute. So, we could write this as meters per minute squared, per minute, meters per minute squared. And we're done. Now, if you want to get a little bit more of a visual understanding of this, and what I'm about to do, you would not actually have to do on the actual exam. Well, let's just try to graph. Let's graph these points here. Let me give myself some space to do it. So, if we were, if we tried to graph it, so I'll just do a very rough graph here. So, let's say this is y is equal to v of t. And we see that v of t goes as low as -220. It goes as high as 240. So, let me give, so I want to draw the horizontal axis some place around here. T is positive. It would look something like that. And so, let's just make, let's make this, let's make that 200 and, let's make that 300. And so, then this would be 200 and 100. 200 and 100 there. And then that would be -100. And then that would be -200. And then that would be -300. And we see on the t axis, our highest value is 40. We go between zero and 40. And so, these obviously aren't at the same scale. But this is going to be zero. If we put 40 here, and then if we put 20 in-between. And so, this would be 10. Let me do a little bit to the right. That would be 10. And then, that would be 30. And so, what points do they give us? So, they give us, I'll do these in orange. Zero, comma, zero. So, that is right over there. They give us when time is 12, our velocity is 200. So, when the time is 12, which is right over there, our velocity is going to be 200. So, that's that point. When our time is 20, our velocity is going to be 240. So, when our time is 20, our velocity is 240, which is gonna be right over there. And then, when our time is 24, our velocity is -220. So, she switched directions. So, when it's 24. So, 24 is gonna be roughly over here. It's gonna be -220. So, -220 might be right over there. And then, finally, when time is 40, her velocity is 150, positive 150. So, at 40, it's positive 150. And so, these are just sample points from her velocity function. But what we wanted to do is we wanted to find in this problem, we want to say, okay, when t is equal to 16, when t is equal to 16, what is the rate of change? And we don't know much about, we don't know what v of 16 is. But what we could do is, and this is essentially what we did in this problem. We could say, alright, well, we can approximate with the function might do by roughly drawing a line here. So, if you draw a line there, and you say, alright, well, v of 16, or v prime of 16, I should say. So, v prime of 16 is going to be approximately the slope is going to be approximately the slope of this line. That's going to be our best job based on the data that they have given us of estimating the value of v prime of 16. So, we literally just did change in v, which is that one, delta v over change in t over delta t to get the slope of this line, which was our best approximation for the derivative when t is equal to 16.
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