U1S2V11 Falling Pumpkin Example revisited.txt

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Let's look at the pumpkin example
we've considered before where we go up to the top of a 100 meter
building and toss a pumpkin.
Please do not try this at home.
We have a formula for the height of the pumpkin measured
in meters in terms of the time measured in seconds given
by the polynomial f of t is equal to 100
plus 20t minus 5t squared.
Now, we're interested in the tangent line to this function,
so let's draw a graph.
The vertical axis f of t is measured
in meters and the horizontal axis t is measured in seconds.
Here is 100 meters.
The height of the building.
And each tick mark along the vertical axis
is measured in units of 20 meters.
Each horizontal tick mark is one second.
Let's plot a few points.
And here is our graph.
We actually computed the derivative of the function
before, and we found that f prime of 1
is equal to 10 meters per second, f prime of 2
is equal to 0 meters per second, and f prime of 3
is equal to negative 10 meters per second.
Let's go ahead and compare this to the slope
of the tangent line at each of these points.
At time t equals 1, this is the tangent line.
What is the slope of this line?
Well, the rise is about 20 meters over a run of 2 seconds.
So the slope is 20 meters over 2 seconds or 10 meters
per second, which definitely agrees
with what we computed before.
Let's erase this tangent line and draw in the tangent line
at t equals 2.
This is a horizontal line, so the slope
is 0, which agrees with the computation
of 0 meters per second that we found before.
Let's erase this and draw in the tangent line at t equals 3.
What is the slope?
Well, this line has a rise of about negative 20
meters over a run of two seconds,
giving us negative 20 meters over 2 seconds
or negative 10 meters per second.
So what we've seen in this worked
example is that by estimating the value of the slope
of the tangent line at a point we were actually
able to get a pretty good handle on the value of the derivative.
So now it's your turn.
I want you to estimate the value of the derivative
by estimating the value of the slope of the tangent line
at various points of a function along a graph.
So good luck and we'll see you in a little bit.