Main content

## Derivative as instantaneous rate of change

Current time:0:00Total duration:7:21

# Tangent slope as instantaneous rate of change

## Video transcript

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.