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find the arc length of f(x)= ln(sec(x)) on the interval [0, (pi/4)] around the x-axis.
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Hey Sam :) What part we stuck on?
We take a curve and break it into a bunch of small pieces of "length" \(\Large\rm ds\). To find arc length, we simply add up all of those pieces.\[\Large\rm S=\int\limits ds\]\[\Large\rm S=\int\limits \sqrt{1+\left(\frac{dy}{dx}\right)^2}~dx\]
We can go over that ds = (stuff) if you need. Otherwise, looks like we just need a derivative to get things rolling.
\[\Large\rm f(x)=\ln(\sec x)\]\[\Large\rm f'(x)=?\]
Oh `around the x-axis`. So this is a surface area problem?
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\[(1\div \sec x)*secxtanx\]
\[\Large\rm f'(x)=\frac{1}{\sec x}~\sec x \tan x\]Ok good :o simplify
tanx
dont I have to use the formula |dw:1426736520655:dw|
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