WEBVTT
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all right. In this problem, we have a
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curve represented by the vector ar 15 with three component
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functions three sine of T four t and three co
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sign of tea. And what we want to think
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about is starting at the 30.3 and moving five units
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along this curve, and we want to figure out
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where we are after we've moved five units. So
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what points do we end up with after moving five
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units along the curve? So we do want to
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think about our formula for the length of the curve
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because we know that this length is going to be
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five. We move five units. So we're going
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from from our starting point to our ending point,
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we'll figure out what that interval is. In a
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moment. We still need to find the magnitude off
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the derivative of that vector and then integrate that with
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respect to t. So let's go ahead and find
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the magnitude of the derivative first, um, So
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first we need to take the derivative with respect to
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T, which is going to give us three co
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sign of tea four and negative three sine of t
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and then we want to take the magnitude of that
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. So we're going to get magnitude of our prime
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of tea is equal to the square root of our
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first component function squared. So we get nine co
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sine squared of tea plus 16 plus nine sine squared
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a t going Teoh Rearrange that a little bit.
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So we're gonna end up with 16 plus nine times
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and factoring the nine out of the first and last
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term. So nine times co sine squared of tea
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plus sine squared of tea which we know from our
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trig identities is equal Teoh one. So we end
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up with the square root of 16 plus nine,
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which is the square root of 25 or five.
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We also want to think about, um, the
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the starting point for T s 01 of the ways
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that I would do this is look for the simplest
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of our three component functions. So I would say
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that our simplest is 40 and that is our our
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why component. So if I look in the starting
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point, why is equal to zero? So I
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could think about the fact that when why is equal
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to zero for key is equal to zero. So
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that means that are starting value for T is going
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to be zero If we saw that for team.
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So we want to find out. Um what the
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That's our starting point for tea. We want to
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find out what the ending point is for tea,
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so we can ultimately calculate, um, are triple
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coordinate triple. So now that we know what our
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starting point is, we want to figure out we'll
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set up our expression for the length of the curve
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. So we would be going from zero to t
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because we don't know where we're ending at five units
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. The derivative are the magnitude of the derivative of
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vector Are we found was five. So this is
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going to be five or enter it great with respect
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to you, which is just our dummy variable,
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And then we can, right, this is five
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you and use the fundamental theorem of calculus to evaluate
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this. So we have five times t times there
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minus five times zero, which is just zero.
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So we know this much and we know that we
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have moved five units along the curve. So in
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this case, the length of the curve that we're
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interested in is five. And then if we take
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that information, this right here tells us that we
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end. Where are parameter? T is equal to
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one. So if we want to find out the
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coordinate triple that we end up with, we can
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plug that into your three component functions. So are
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three component functions are three sine of T. It's
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four of tea and three co sign of tea.
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So then we can look at our ending point,
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which is t equals one and value each of those
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. So we end up with three times the sign
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of one. Why we just end up with four
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times one or four frizzy. We end up with
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three co sign of one. So if you want
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to write that as a coordinate triple, we can
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say that we end up at the point. Three
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sign of one for three co sign off one And
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again, that is, after traveling five units along
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the curve represented by our of team