The last video we talked about
what instantaneous velocity meant but I didn’t tell you how to
actually compute it. I mean I gave you a limit but
there was no particular way to compute that limit. In this video we’re actually
going to compute a limit. So here’s the problem.
We’ve got a particle that’s moving and we have a formula for where
it is at any given time. It’s position is 4t^2 + 3. We’re measuring t in seconds
and s in feet. So what’s the velocity
at t=2? Straight forward question.
Let’s figure it out. So here’s a sketch of its position
as a function of time. It’s useful to draw a graph.
Graph’s don’t replace everything but they’re useful. And what we need to do is
we need to figure out what is the average velocity between
time 2 and some other time t. So at time t,
our position is 4t^2 + 3. I guess it’s t, that. Okay? But its height is 4t^2+3.
And so what we need to do is we need to figure out the average velocity
between time 2 and time t. And then we’re going to take a limit as
t gets closer and closer and closer to 2. Now in our last example,
we considered times that were earlier than 2 o’clock.
Here we’re just going to consider any ol’ T.
I drew it as if t is bigger than 2 but when we take a limit,
we want to take a limit both as t approaches 2 from above and
as t approaches 2 from below. We have to consider all
possible values of t. So since the position is 4t^2 – 3, and at time 2 we were at position 19 –
that’s 4 * 2^2 +3, the change in position is the new
position minus the old, so that’s 4t^2 + 3 – 19
so that’s 4t^2 – 16 And the amount of time it took
was t-2. So our average velocity is
4t^2-16/(t-2) and graphically, this is the slope of a line between those two points. This slope is (4t^2 – 16) / (t-2) okay?
So the slope of the secant line is the average velocity.
But we want the instantaneous velocity. So to take the instantaneous velocity
we have to take a limit so the instantaneous rate of change, IROC
stands for instantaneous rate of change. The instantaneous rate of change
is the limit as t approaches 2 of the average rate of change.
We factor out a 4 and 4t^2 – 16 becomes 4(t^2-4) and then we factor that as
t-2 * t+2 and you notice that you can’t plug in
t=2 or else you get 0 / 0 but for any value of t other than 2,
you can cancel the factors of t-2 and you get the limit of
4 (t+2) and that’s 4*4 which is 16. Since we’re measuring a velocity,
the change in the number of feet per number of seconds, it’s
actually 16 feet per second. That is our instantaneous
velocity at t=2. It’s also the slope of the line
that’s tangent to the curve here. There’s a line that’s tangent here
and that tangent line has a slope of 16.