Derivative as instantaneous rate of change
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Approximating instantaneous rate of change with average rate of change
The table gives a position s of a motorcyclist for t between 0 and 3, including 0 and 3. This is just saying that t is part of the interval, or t in the interval between 0 and 3. And we see that right over here, where the distance traveled s is measured in meters, and t is time in seconds. Assume the motorcyclist is accelerating during a three-second period. And they give us the information. This is time between 0 seconds and 3 seconds. And here we have the corresponding distance that you could view as a function of time. The average velocity for t between 1.5 and 2, so t between 1.5 and 2 is 23 meters per second. So what they did over here is they figured out, well, what is delta s over delta t in this interval? And they figured out that it was 23 meters per second. And you can verify that. Your change in s looks like it's 12.5. Your change in time is point-- or actually, this looks like it's 11.5. Yeah, 11.5. Your change in time is 0.5. 11.5 divided by 0.5 is 23. So that makes sense. And then they tell us the average velocity for t between 2 and 2.5. So change in our distance over change in time, they say is 31.8 meters per second. And then they say, estimate the instantaneous velocity at t equals 2 seconds and use this value to write the equation of a line tangent to s of t at the time t equals 2. So we can try to approximate. We can approximate the slope of the tangent line right over here, right when t equals 2 seconds, by taking the average of the slopes of the tangent lines between 1.5 and 2, and 2 and 2.5. So essentially, to approximate the slope of the tangent line, we're going to take the average of these two rates of change right over here, the average of these two slopes. So let's do that. So the average is going to be 23 plus 31.8 over 2. And let's see, what is that equal to? That is equal to 54.8/2. And what is that equal to? Let's see. 54 divided by 2 is 27. So it's 27.4. So we can use that as our approximation for the instantaneous rate of change for the slope of the tangent line. And now we have to actually figure out what that equation actually is. They don't just want the slope. So this is the slope right over here. And they say that they want it in point-slope form. And they remind us that t is the independent variable. So when you're putting something in point-slope form, it really just comes out of the definition of a line. A line always has a constant slope. So let's just imagine taking a random point on that line, t-- let me write it this way-- t comma capital S-- a random point on the line, on the tangent line here. Well, the slope between that and this point is always going to be constant. So what's the slope between this point and this point? Well, your change in S is going to be S minus 2 over your change in t, which is t minus-- oh sorry, your S is S. This is confusing sometimes. S minus 30.2 over your change in t, t minus 2, is equal to your slope. Anywhere along that tangent line, you're going to have that slope, 27.4. And then you multiply both sides by t minus 2, and you've put it in point-slope form. So this is the same thing as S minus 30.2 is equal to 27.4 times t minus 2.