U1S7V06 Derivative of sine at 0.txt

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Recall that we're trying to compute
the derivative of sine x and we were
able to reduce this problem to finding
the derivative of sine x at 0 and the derivative of cosine x
at 0.
Let's go ahead and start with the derivative of sine
x at 0, which is given by this limit.

Now, we don't have a lot of tools
for computing limits like these, but we
do have an understanding of sine x geometrically.
So we're going to go ahead and give a geometric proof.
To help us, I'm going to make a slight change in notation.
Instead of writing delta x, I'm going
to replace this with theta because it's easier
to think about theta as an angle instead of delta x.
So literally, I've just replaced delta x with theta everywhere.
So now, we're trying to compute the limit as theta goes
to 0 of sine theta over theta.
Let's get started and draw our picture.
I'm going to start with the unit circle,
and I'm going to draw in an angle theta.
Then this vertical distance here is sine theta.
And I want to compare this to this arc length right here.
So what is this arc length?
Now, this is where it becomes incredibly important
that theta is being measured in units of radians.
We can't use degrees for this argument to work.
So if we're measuring our angle in radians,
then the total circumference of the circle
is 2pi, which is also equal to the number of radians
in the circle.
So this arc length is equal to the number of radians swept out
by that angle.
So this arc length is exactly equal to theta.
So now, we have a picture that we can understand.
We're trying to understand this limit sine theta over theta,
which is the ratio of this vertical line to this arc
length as theta approaches 0.
To help us understand this, we're
going to start making the angle theta smaller.
Of course, as I make the angle smaller,
it's harder to see what's going on.
So as I shrink the angle, I'm also
going to blow up the picture so that we
can see what's happening.
And what you'll notice is that, as theta
gets smaller and smaller, this vertical distance
becomes more and more indistinguishable
from this curved segment.
What this is telling us is exactly that,
as theta approaches 0, this limit is approaching 1.
The motto is "curved lines are essentially straight when
you zoom in far enough," and this
is an idea that we've already been
looking at with the derivative.
OK.
So now, we're done.
We've figured it out.
It's 1.
OK.
But I want to come back, and I want
to talk about why it's so important that we're
doing this in radians and why this doesn't make sense
otherwise.
So remember that this derivative is actually
the slope of the tangent line of sine theta at 0.
So if I draw this in radians, then there
are sort of 2pi radians in a circle.
And so you can sort of draw this out.
This is a rough sketch of what sine looks like.
So now, what this is saying is that the slope
of this tangent line at 0 is equal to 1.
Now, I told you this argument doesn't make sense
if we do this in degrees, and let's look
at what happens if we try to draw this picture.
Well, first of all, there are 360 degrees.
So if I want to try to make this picture to scale,
it's going to be really long.
OK.
So 360.
Here's 180, and then 1 is really tiny.
So if I draw this picture, you see
that the slope of this tangent line
is definitely not going to be equal to 1.
It's not going to make sense.
So the derivative of sine theta at 0 is only equal to 1
if theta is being measured in radians.
Otherwise, this is definitely not the case.