Derivative as instantaneous rate of change
Tangent slope as instantaneous rate of change
The data for three points on a smooth function f is given in the table. So let's actually graph these, just so that we can visualize it a little bit. So our vertical axis, let's just call that y, is equal to f of x. And then we have our horizontal axis. We have our horizontal axis, this is our x-axis. And we're going to skip some space here, just because we immediately go to very high values of f of x. And I'll even skip some space here, since we already go up to 6.5 all the way up to 7.5. So let me just show that I'm skipping some space here. And we're going to start. Let's say this is 108, 108, 109, 110. And let's say that this right over here is 6.5, that is 7, and then that is 7.5. And we can plot these points. 6.5 comma 108.25, that'll put us right over there. 7 comma 109.45, that puts us right around there. And then 7.5, 110.15 puts us right around here. And these are just three points on a smooth function f. So we can visualize what that smooth function f might look like. It might look something like this. So maybe it looks something like this. So our smooth function f might look something-- who knows what it looks like after this. So that's, this right over here is, this is the graph of y is equal to f of x, Now let's try to answer their questions. What is the average rate of change of f with respect to x as x goes from 6.5 to 7, 7 to 7.5, and 6.5 to 7.5? So let's do them one at time. 6.5 to 7, our change in x here, our change in x is 7 minus 6.5, which is equal to 0.5. And our change in y here is equal to-- let's see. We end up at 109.45 minus 108.25, which is 1.2. So our average rate of change over this interval is our change in y over our change in x, or 1.2 divided by 0.5. Let me write this down. Our change in y over our change in x is equal to 1.2 over 0.5, which is equal to 2.4. So this is equal to 2.4. Now let's do the next interval. Our change in x, once again, is 7.5 minus 6.5. Change in x is still 0.5. That's hard to read, let me write that a little bit neater. So our change in x is still 0.5, 7 to 7.5. And our change in y, let's see. We end up at 110.15. We started at 109.45. So our change is 0.7. So it's 0.7. So our change in y over our change in x is equal to 0.7/0.5, which is equal to 1.4. And so you can see that the slope of the secant line-- this essentially the slope of the secant line between these two points. So this one, this one right over here-- let me try my best to draw it. This one right over here. You can see it's steeper than the second one, than this secant line right over here, than this secant line right over there. And now let's find the slope-- the average rate of change. I should say, or the slope of the secant. The average rate of change over this interval, which is the same thing as the slope of the secant line between that point and that point. So let's think about our delta x. And maybe I'll do it up here. So our delta x, we're going from 6.5 to 7.5. So our delta x is equal to 1 and what is our delta y? So our delta y, our change in y, is equal to, let's see. We end up at 110.15. We started at 108.25. 110.15 minus 108.25, we increase by 1.9. So our change in y over change in x is equal to 1.9 over 1, which is equal to 1.9. Fair enough. And that's the slope of this secant line between this point and this point right over here. Slope of the secant line. It would look something like that. You see that it's slightly less steep than the magenta secant, and it's slightly more steep than the orange secant. It's kind of in between the two. Now they ask us this final question. Use the average rate of change of f on the larger interval from here to here-- which we already figured out, that's 1.9-- as an approximation for the slope of the line tangent to f at x equals 7. So we're trying to approximate the slope of the line tangent to f at f equals 7. So the line tangent might look something like this. And we see, at least visually, it looks like that little blue line we drew. This one right over here does look like it's-- at least the way I've drawn it-- it looks like it's a pretty good approximation for the slope up there. So we'll use this as the slope, as an approximation for the slope of the tangent line to f at x equals 7. They say, write an equation for the line tangent f at 709.45 using point slope form. So it's going to be a line where we're going to use this as an approximation for slope. And it's going to contain this line. It's tangent. It touches to the curve at this point. And so point slope form, just as a reminder, it's just another way of saying that every point on this line, every point on this line, x comma y-- If we were to find the change in x and change in y relative to this point right over here, it's always going to have a constant slope. So one way you could say is if you take an arbitrary point x, y on this line, you could say your change in y. So y minus 109.45 over your change in x, so we know that the 0.7 is there. Or the 0.7 is-- when x is equal to 7, f of x is 109.45. So we're referencing the same point. When our change in x is x minus 7, our change in y is y minus 109.45. And this is always going to be constant for any x, y that I pick on this line there. And we're using this as an approximation for what that's going to be. That's going to be equal to 1.9. So if you want to put it in point slope form, you just multiply both sides times x minus 7. And you get y minus 109.45 is equal to 1.9 times x minus 7. And we are done. That is an approximation. We've approximated the slope. And if this was the slope, this would be the equation of the tangent line in point slope form.