WEBVTT
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in this problem who and re Pramuk tries occur as
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represented by the vector our tea with respect to its
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arc length. So go ahead and write out our
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function are if t here e to the power of
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to tee times co sign of duty. The second
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component function is just to and the third is e
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to the power of duty Sign, uh, two
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teeth. So we know that we need Teoh.
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We want to use the expression the equations on the
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right hand side and the top right corner to be
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able Teoh find s in the first place. So
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that way we can re premature eyes The curve in
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terms of it, remember s is the arc length
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. So first we want to be able to take
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the derivative of the vector are so we're gonna look
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at our prime of tea and we have to use
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our product rule to find the derivative of the first
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and third component functions so derivative of our first component
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function is going to be two times either the power
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of to t Times co sign of two t my
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s two times e to the power of T T
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time sign of two t derivative of the second component
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function is just going to be zero and derivative of
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the third component function is going to be two times
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e to the power of to t sign of two
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T plus two times e to the power of to
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t Times CO sign of two team. And if
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we want, we can make this just a little
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bit nicer to look at. I would like to
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factor out uh, the common term two times either
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the power of to t from my first and third
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component functions. So we're left with that term Times
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co sign of two T minus sign of two t
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second component function is still zero and factoring two times
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even to the power of to t we're left with
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sign of to t plus co sign off duty.
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I'm gonna go ahead and switch the order of those
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since its addition and the order doesn't matter just so
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that way. It's consistent with our with the order
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of sign and co sign in the first component function
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that will become a little bit easier later on toe
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simplify things. All right, so now that we
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have the derivative. We need to find the magnitude
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of that resulting vector, which is gonna be the
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square root of each of our component functions squared and
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added together. So for first component function squared,
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we have four times either the power of fourty times
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co sign of two T minus sign of two t
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not quantity squared. Plus C R squared is just
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zero and then plus third component functions squared so four
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times e to the power of 40 times co sign
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of two t plus sign of two t all squared
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, and we can simplify that further. So I
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noticed the common term four times each of the power
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of fourty. I'm gonna factor those out of both
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of them. And in fact, I'm gonna pull
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them right out of the square root sign. So
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we have two times e to the power of to
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t and then under the square root sign. We
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have co sign of two T minus sign of two
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t quantity square plus CO sign of two T plus
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sign of two t all squared. So, Teoh
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, simplify this down a little bit. What I'm
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gonna do is represent this as a simpler expression just
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for the sake of being able to simplify things faster
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. So I'm gonna let co sign of two TB
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a and sign of two t equal. Be so
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off to the side. I'm gonna write with that
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expression under the square root sign would be if I
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express it in terms of A and B. So
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what we would have is a minus B squared,
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plus a plus B squared. If I expand,
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boost both of those terms. I end up with
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a squared minus two, a B plus B squared
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for the first term and a squared plus two a
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B plus B squared for the second term. If
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I add my like terms together my negative to a
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B and my plot positive to a B and to
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be zero and then I'm left with two times a
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a squared plus two times B squared or two times
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a a squared plus B squared. Now we can
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go back to our original expression and right our value
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under the square root sign, as in terms of
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the simplified expression that we have in bread over on
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the right, so we would have is two times
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e to the power of to t times the square
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root of two and substituting co sign of two t
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and sign of two t back in for A and
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B we would have co sign of to t squared
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or co sine squared of two t plus I sine
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squared of two t And that becomes really nice because
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by that the factory and identity one of our trig
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identities. We know that this expression is equal to
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one. So what we're left with then is two
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times the square root of two times E to the
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power of two teeth, and that is the magnitude
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of the derivative of our. So what we can
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do then is we can use the expression s or
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s of t equal to the inner rule in the
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top right corner in that form. So we know
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that we're integrating based on the problem statement from zero
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in the increasing directions there from zero to t.
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And then I'm going to write my, uh,
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expression that I just found the magnitude in terms of
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you just a dummy variable to take the place.
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So that way we don't get it confused with the
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tea, Uh, in our bounds of integration,
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I'm gonna integrate that with respect to you. It's
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what I end up with is square it of to
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times e to the power of to you going from
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zero to t If I use the fundamental theorem of
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calculus to evaluate that I end up with squared of
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to times eat of the two teeth minus the square
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root of two times E to the 20 r or
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just zero We know that is equal toe one.
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So we end up with squared of to times E
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to the power of two key minus squared too.
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So we want to now solved for key because we
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want to be able to write the original parameter ization
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of the curve in terms of s now. So
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we have to substitute an expression that's equal to T
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. So what? Weaken Dio? We can look
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at the fact that right now we're at s is
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equal. Teoh swear it of two and on a
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factor that square to two out through 22 times e
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to the power of to t minus one. So
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I want to solve that four t So I'm gonna
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divide both sides by squared of to and that I'm
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gonna add one to both sides and then if we
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take the natural long of both sides of this equation
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we have on the right hand side is the natural
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log which is a log a rhythm based e of
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e to the power of to t. So we
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end up with not your log of e to the
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power of two teeth which we know simplifies to to
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t We're almost there now we just have to divide
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both sides by two and what we get is that
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T is equal to 1/2 times natural log of s
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over squared of two plus one. So now that
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we have an expression for tea, what we want
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to do is plug that back into our original privatization
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of of the Curve. So scroll down a bit
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. So the original way that we wrote our of
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tea was eat of the power of to t co
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sign of two t to e the power of to
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t sign duty. So now if we write this
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in terms of t of s right, we're thinking
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of the parameter t Now, as a function of
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the arc length s, we're gonna go ahead and
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we're gonna substitute that value in for tea now.
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So for first component function, we're going to simplify
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this quite a bit. We're gonna write it,
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um, as it would originally come. We have
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two times. T the 1/2 is gonna cancel out
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. So our numerator is just gonna be natch.
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Are exponents gonna be natural? Log s over squared
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of two plus one. That's gonna be times the
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co sign of two t again to and the 1/2
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are going to cancel out. So we end up
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with co sign of natural log of a super squared
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of two plus one That is our first component function
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. Second component function is just to, and that
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one does not change. And then, lastly,
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we have e Teoh the power of to t again
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R two and R 1/2 are gonna cancel out.
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So we're left with natural Log s over squared of
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to plus one time sign of to T or two
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times 1/2 which cancels out natural log s squared over
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a squared of two plus one. You got a
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lot going on there, but fortunately, some things
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were gonna simplify out for us. We have an
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e raised to the power of natural law, and
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we know that these two can essentially cancel out one
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another. So what we get is s over squared
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of two plus one times the co sign of the
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natural log of s over squared of two plus one
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. Choose our second component function on our last component
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function Again E and natural log cancel each other out
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. So we have s over squared of two plus
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one all times. Sign of natural log of s
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over square of two plus one. And that is
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the re prayer characterization of our curves Now, in
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terms of the Ark link.