U1S7V09 Recitation video.txt

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Hi.
Welcome back to recitation.
In last lecture we talked about finding the derivatives
of trigonometric functions-- in particular, the sine function
and the cosine function.
So today let's do an example of putting that into practice.
So here's a function h of x equal to sine
of x plus square root of 3 times cosine of x.
And I'm asking you to find which values of x
have the property that the derivative of h of x
is equal to 0.
So why don't you take a minute to think
about that, work it out on your own, pause the video,
and we'll come back and we'll work it out together.

All right so you've hopefully had a chance
to look over this problem, try it out for yourself.
Now let's see how to go about it.
So we have the function h of x that's
equal to sine x plus square root of 3 cosine x.
And we want to know when its derivative's equal to 0.
So in order to answer that question,
we should figure out what its derivative actually
is and try and write down a formula for its derivative.
So in this case, that's not that bad.
If we take a derivative of h, well, h
is a sum of two functions-- sine x
and square root of 3 cosine x.
And we know that the derivative of a sum
is just the sum of the derivatives.
So we have the h prime of x is equal to d over dx
sine x plus d over dx of square root of 3 times cosine x.
Now, we learned last time in lecture
that the derivative of sine x is cosine of x.
And we learned the derivative of cosine of x is minus sine x.
So here we have a constant multiple,
but by the constant multiple rule that just gets pulled out.
So this is equal to cosine x minus square root of 3
times sine x.
So this is h prime of x.
And now we want to solve the equation h prime of x equals 0.
So we want to find those values of x
such that cosine x minus square root of 3 sine x is equal to 0.
Now, there are a couple of different ways
to go about this.
I think my preferred way is I would add the square root of 3
sine x to one side, and then I want to get my x's together,
so I would divide by cosine x.
So that gives me--
so on the left side I'll be left with cosine x divided
by cosine x, so that's just 1.
And on the right side I'll have square root of 3 times
sine x over cosine x--
so that's just square root of 3 times tan x.
And I can rewrite this as tan of x
is equal to 1 divided by square root of 3.
Now, to find x here either you can remember your special trig
angles and know which values of x make this work
or you could apply the arctangent function here.
So in either case, the simplest solution
here is x equals pi over 6.
So if you like, you can draw a little right triangle.

If this is x--
if tan x is 1 over square root of 3,
we should have this side being 1 and this side
being square root of 3, and--
OK, in that case in this right triangle
the hypotenuse would be 2, and so then you
would recognize this is a 30 degree angle or pi
over 6 radian angle.
But one thing to remember is that tangent
of x is a periodic function with period pi.
So not only is pi over 6 a solution,
but pi over 6 plus pi is a solution.
So that's 7 pi over 6 or pi over 6
plus 2 pi, which is 13 pi over six or pi over 6
minus pi, which is minus 5 pi over 6, et cetera.
So there are actually infinitely many solutions.
They're given by pi over 6 plus an arbitrary multiple of pi.
So then we're done.
So I do want to mention, though, that there's
a there's another approach to this question, which
is we can start by multiplying this expression for h of x.
So if you look at h of x-- that's sign of x plus square
root of 3 times cosine x--
it resembles closely one of your trigonometric identities
that you know about.
So in particular, to make it resemble it even more,
I can multiply and divide by 2.
So I can rewrite h of x equals 2 times
1/2 sine x plus square root of 3 over 2 times cosine x.
And now 1/2 is equal to cosine of pi over 3,
and square root of 3 over 2 is equal to sine of pi over 3.
So I can rewrite this as 2 times cosine pi
over 3 sine x plus sine pi over 3 cosine x.
And this is exactly what you get when you do the angle addition
formula for sine.
This is the expanded out form, and so we
can apply it in reverse and get that this
is equal to 2 times sine of x plus pi over 3.
So so far we haven't done any calculus.
We've just done-- so in this solution, our first solution,
we did some calculus first, and then
some algebra and trigonometry.
So so far we've just done some algebra and trigonometry.
Now, the points where h prime of x is equal to 0
are the points where the graph of this function
has a horizontal tangent line.
So either you can compute its derivative using your rules
or by the definition, or you can just
say, oh, we know what this graph looks like.
So I've drawn a schematic up here.
So this is a graph.
This graph is y equals 2 sine of x plus pi over 3--
it's what you get if you take the graph y equals sine x
and you shift it left by pi over 3
and you scale it up by a factor of 2.
So this here is at x equals minus pi over 3.
This route is x equals 2 pi over 3.
And the points we're interested in
are the points where there's a horizontal tangent line, where
the derivative is 0.
And so there's one of these right at this value, which
is pi over 6.
And the second one is this is that minimum there,
so that happens at x equals 7 pi over 6--
pi over 6 because for the usual sine
function it happens at pi over 2,
but we've shifted everything left by pi over 3,
and so pi over 2 minus pi over 3 is pi over 6.
And here for y equals sine x, this minimum
would happen at 3 pi over 2, but we've
shifted it left by pi over 3.
And so on.
There's another trough over here,
and another peak over there, and so on.
So that's the second way you can do this question, using
this cute trig identity here.
And that's that.