WEBVTT
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PROFESSOR: Hi.
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Welcome back to recitation.
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In last lecture we talked
about finding the derivatives
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of trigonometric functions.
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In particular, the sine function
and the cosine function.
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So today let's do an example
of putting that into practice.
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So here's a function:
h of x equal to sine
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of x plus square root
of 3 times cosine of x.
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And I'm asking you to
find which values of x
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have the property that
the derivative of h of x
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is equal to 0.
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So why don't you take
a minute to think
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about that, work it out on
your own, pause the video,
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and we'll come back and
we'll work it out together.
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All right.
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So you've hopefully had a chance
to look over this problem,
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try it out for yourself.
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Now let's see how
to go about it.
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So we have the
function h of x-- it's
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equal to sine x plus
square root of 3 cosine x--
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and we want to know when its
derivative is equal to 0.
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So in order to
answer that question
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we should figure out what
its derivative actually
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is and try and write down a
formula for its derivative.
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So in this case
that's not that bad.
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If we take a
derivative of h, well h
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is a sum of two
functions-- sine x
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and square root of 3 cosine x.
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And we know that the
derivative of a sum
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is just the sum of
the derivatives.
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So we have the h prime
of x is equal to d
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over dx of sine x plus d
over dx of square root of 3
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times cosine x.
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Now we learned last
time in lecture
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that the derivative of
sine x is cosine of x,
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and we learned the derivative
of cosine of x is minus sine x.
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So here we have a
constant multiple,
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but by the constant multiple
rule that just gets pulled out.
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So this is equal to cosine
x minus square root of 3
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times sine x.
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So this is h prime of x and now
we want to solve the equation h
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prime of x equals 0.
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So we want to find
those values of x
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such that cosine x minus square
root of 3 sine x is equal to 0.
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Now there are a couple
different ways to go about this.
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I think my preferred way is I
would add the square root of 3
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sine x to one side, and then
I want to get my x's together,
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so I would divide by cosine x.
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So that gives me--
so on the left side
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I'll be left with cosine
x divided by cosine x,
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so that's just 1.
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And on the right side I'll
have square root of 3 times
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sine x over cosine x.
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So that's just square
root of 3 times tan x.
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Or, and I can rewrite
this as tan of x
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is equal to 1 divided
by square root of 3.
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Now to find x here, either you
can remember your special trig
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angles and know which
values of x make this work.
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Or you could apply the
arc tangent function here.
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So in either case,
the simplest solution
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here is x equals pi over 6.
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So if you like, you can draw
a little right triangle.
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You know, if this is x, if tan
x is 1 over square root of 3,
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we should have this side
being 1 and this side
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being square root of 3.
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And in-- OK, in that case,
in this right triangle
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the hypotenuse would be 2.
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And so then you, you
know, would recognize
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this is a 30 degree angle,
or pi over 6 radian angle.
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But one thing to
remember is that tangent
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of x is a periodic
function with period pi,
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so not only is pi
over 6 a solution,
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but pi over 6 plus
pi is a solution.
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So that's 7*pi over 6,
or pi over 6 plus 2*pi,
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which is 13*pi over 6,
or pi over 6 minus pi,
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which is minus 5*pi
over 6, et cetera.
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So there are actually
infinitely many solutions.
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They're given by pi over 6 plus
an arbitrary multiple of pi.
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So then we're done.
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So I do want to mention,
though, that there's
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another approach to
this question, which
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is, we can start by multiplying
this expression for h of x.
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So if you look at
h of x-- that's
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sine of x plus square
root of 3 times cosine x--
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it resembles closely one of
your trigonometric identities
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that you know about.
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So in particular, to make
it resemble it even more
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I can multiply and divide by 2.
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So I can rewrite h
of x equals 2 times
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1/2 sine x plus square root
of 3 over 2 times cosine x.
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And now 1/2 is equal
to cosine of pi over 3.
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And square root of 3 over 2
is equal to sine of pi over 3.
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So I can rewrite
this as 2 times--
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what did I say-- I said
cosine pi over 3 sine x
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plus sine pi over 3 cosine x.
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And this is exactly what you get
when you do the angle addition
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formula for sine.
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This is the expanded
out form, and so we
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can apply it in reverse
and get that this
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is equal to 2 times sine
of x plus pi over 3.
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So, so far we haven't
done any calculus.
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We've just done-- so in this
solution, our first solution
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we did some calculus
first and then
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some algebra and trigonometry.
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So, so far we've just done
some algebra and trigonometry.
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Now the points where h
prime of x is equal to 0
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are the points where the
graph of this function
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has a horizontal tangent line.
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So either you can compute its
derivative using your rules
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or by the definition.
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Or you can just say, oh, we
know what this graph looks like.
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So I've sort of drawn
a schematic up here.
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So this is a graph, this graph
is OK, so this is the graph,
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y equals 2 sine of
x plus pi over 3.
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It's what you get if you take
the graph y equals sine x,
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and you shift it
left by pi over 3
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and you scale it up
by a factor of 2.
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So this here is at x
equals minus pi over 3.
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This root is x
equals 2*pi over 3.
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And the points
we're interested in
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are the points where there's a
horizontal tangent line, where
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the derivative is 0.
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And so there's one of these
right at this value, which
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is pi over 6.
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And then the second one is
this, is that minimum there.
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So that happens at x
equals 7*pi over 6.
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pi over 6 because
for the usual sine
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function it happens
at pi over 2,
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but we shifted everything
left by pi over 3.
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And so pi over 2 minus
pi over 3 is pi over 6.
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And here for this, for
just y equals sine x,
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this minimum would
happen at 3*pi over 2,
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but we've shifted it
left by pi over 3.
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And so on. you
know, every, there's
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another trough over here,
and another peak over there,
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and so on.
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So that's the second way you
can do this question using
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this cute trig identity here.
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And that's that.