WEBVTT
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JOEL LEWIS: Hi.
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Welcome to recitation.
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In lecture you've learned
how to compute derivatives
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of polynomials,
and you've learned
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the relationship between
derivatives and tangents lines.
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So let's do a quick example that
puts those two ideas together.
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So here I have a
question on the board:
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compute the tangent
line to the curve y
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equals x cubed minus
x at the point (2, 6).
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So why don't you
take a minute, work
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on that yourself, pause
the video, we'll come back
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and we'll do it together.
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All right.
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Welcome back.
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So we have this function,
y equals x cubed minus x.
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Let's just draw a
quick sketch of it.
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So looks to me like it has
zeros at 0, 1, minus 1.
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And it sort of does something
like this in between.
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Very rough sketch there.
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And way over-- well, OK.
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We'll call that
the point (2, 6).
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Feel a little sketchy,
but all right.
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OK.
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So we want to know
what the tangent line
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to the curve at that point is.
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So in order to do that, we need
to know what its derivative is,
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and then that'll
give us the slope.
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And then with the
slope, we have the slope
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and we have a point,
so we can slap
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that into, say, your
point-slope formula for a line.
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So, all right, so the derivative
of this function is y prime.
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So, OK, so here we have
a sum of two things,
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and they're both powers of x.
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And so we learned
our rules for a power
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of x that the
derivative of x to the n
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is n times x to the n minus 1.
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And so we also learned
that the derivative
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of a sum of two things is
the sum of the derivatives.
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So in this case, so the
derivative of x cubed
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minus x is 3 x squared minus 1.
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OK.
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So this is the slope of
the function in terms of x.
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But in order to compute
the tangent line,
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we need the slope at the
particular point in question.
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Right?
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This is really important.
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So we aren't going to use 3 x
squared plus 1 as our slope.
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Right?
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We want the slope at
the point x equals 2.
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Right?
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OK.
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So what we want for the
slope of the tangent line
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is y prime of 2.
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Right?
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We want it at this point
when x is equal to 2.
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So that's equal to, well, 3
times 2 squared is 12, minus 1
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is 11.
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So this is the slope, this is
the slope of the tangent line.
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I just want to say this
one more time for emphasis.
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This is a really
common mistake that we
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see on lots of homework and
exams when teaching calculus.
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You have to remember
that when you compute
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the slope of the tangent line,
you compute the derivative
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and then you need to
plug in the value of x
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at the point in question.
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Or the value of x and y.
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You know, you need to plug
in the values of the point
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that you have.
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So here, the derivative
is 3 x squared minus 1,
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so the slope at the
point (2, 6) is 11.
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It's just a number.
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The slope at that point
is that particular number.
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OK, so now, to compute
the tangent line,
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we have a point, (2, 6),
and we have a slope, 11.
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So we can plug into
point-slope form.
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So the equation of the-- on
the-- tangent line is y minus 6
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is equal to 11 times x minus 2.
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Right?
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So it's y minus y_0 is equal
to the slope times x minus x_0.
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If you like, some
people prefer to write
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their equations of their
lines in slope-intercept form.
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So if you wanted to do
that, you could just
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multiply through by 11 and then
bring the constants together.
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So we could say, or
y equals 11x-- well,
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we get minus 22
plus 6 is minus 16.
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So either of those is
a perfectly good answer
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to the question.
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So that's that.