Main content

## Derivative as instantaneous rate of change

Current time:0:00Total duration:4:15

# Approximating instantaneous rate of change with average rate of change

## Video transcript

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.