WEBVTT
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we went to evaluate the length of the curve,
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as described by the vector function are of tea which
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could be written as the co sign T defector I
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waas sine of t time is the back to the
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J plus the natural log of the two sine of
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t was the vector k and this is on the
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interval, zero to pi over four. So what's
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first? By taking the derivative of R factor function
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. So we have our prime of tea which,
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if we, uh, evaluate the first, the
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derivative of the first component function, we should get
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negative sign of tea. Plus the derivative of sine
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of t is co sign of tea and we have
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to use the chain role to evaluate the third component
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function. So what we'll get is one over co
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sign of tea times negative sign of tea. And
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if you want to simplify about a little bit,
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we can simplify that last term as negative tangent of
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t since sign of her co sign is equal to
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tangent. All right, so we have our derivative
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. Next thing we want to do is find the
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magnitude of that resulting vector function so the magnitude of
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our prime of tea Take the square root of our
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first component function. Negative sign of t squared plus
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co sign of tea Well squared plus negative tangent of
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t quantity squared and simplifying that a little bit We
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end up with sine squared of teeth plus co sine
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squared of tea plus hand agent squared T we can
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recognize are the thuggery and identity is one of our
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trig identities Sine squared plus co sine squared is one
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So we end up with one plus tangent square toe
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t and we can see another one of our factory
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and identities Ah, one plus tangent Square T is
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gonna be seeking squared of tea So we end up
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with a square root of Seacon Square t which is
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equal Teoh Sequent of do you? So now we're
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in a good position to be able Teoh, evaluate
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the length of our curve. So we have l
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equals the integral from zero to pi. Over four
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was the interval. We were given, uh,
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seek it square or sigint of tea. Do you
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t and the integral of second of T is the
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natural log Absolute value of sequent t plus tangent of
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T. I'm gonna use the fundamental theorem of calculus
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to evaluate that from zero to pi over four.
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Sir, what we get is natural long, uh
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, seeking to pi over for plus tangent pi over
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four minus natural log Sequent of zero plus Tangent of
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Zero which we can simplify by evaluating those trick functions
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. So, seeking a pi over four squared of
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to tangent of pi over four is one minus natural
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log. Uh, seeking of zero is one and
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tangent of 00 So this, uh, last term
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ends up being natural log of one which we know
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is zero. So we can cross that out and
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we end up with the natural log, uh,
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squared of two plus one, which represents the length
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of the curve, as described by the vector function
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, are of tea.