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.