`y = ln(cosx) , [0,pi/3]` Find the arc length of the graph of the function over the indicated interval.

Thursday, September 15, 2011

$\displaystyle{y}={\ln{{\left({\cos{{x}}}\right)}}},{\left[{0},\frac{\pi}{{3}}\right]}$ Find the arc length of the graph of the function over the indicated interval.

The arc length of a function of x, f(x), over an interval is determined by the formula below:

$\displaystyle{L}={\int_{{a}}^{{b}}}\sqrt{{{1}+{{\left(\frac{{{\left.{d}{y}\right.}}}{{{\left.{d}{x}\right.}}}\right)}}^{{2}}}}{\left.{d}{x}\right.}$

So using the function given, let us first find $\displaystyle\frac{{{\left.{d}{y}\right.}}}{{{\left.{d}{x}\right.}}}:$

$\displaystyle\frac{{d}}{{{\left.{d}{x}\right.}}}{\left({\ln{{\left({\cos{{\left({x}\right)}}}\right)}}}\right)}={\left(\frac{{1}}{{{\cos{{\left({x}\right)}}}}}\right)}\cdot{\left(-{\sin{{\left({x}\right)}}}\right)}=-\frac{{{\sin{{\left({x}\right)}}}}}{{{\cos{{\left({x}\right)}}}}}=-{\tan{{\left({x}\right)}}}$

We can now substitute this into our formula above:

$\displaystyle{L}={\int_{{a}}^{{b}}}\sqrt{{{1}+{{\left(\frac{{{\left.{d}{y}\right.}}}{{{\left.{d}{x}\right.}}}\right)}}^{{2}}}}{\left.{d}{x}\right.}={L}={\int_{{0}}^{{\frac{\pi}{{3}}}}}\sqrt{{{1}+{{\left(-{\tan{{\left({x}\right)}}}\right)}}^{{2}}}}{\left.{d}{x}\right.}$

Which can then be simplified to:

$\displaystyle{L}={\int_{{0}}^{{\frac{\pi}{{3}}}}}\sqrt{{{1}+{{\tan}}^{{2}}{\left({x}\right)}}}{\left.{d}{x}\right.}={\int_{{0}}^{{\frac{\pi}{{3}}}}}\sqrt{{{{\sec}}^{{2}}{\left({x}\right)}}}{\left.{d}{x}\right.}={\int_{{0}}^{{\frac{\pi}{{3}}}}}{\sec{{\left({x}\right)}}}{\left.{d}{x}\right.}$

Then you find the definite integral as you normally would. (Using the method shown on the link below, you can find the integral of sec(x).)

$\displaystyle{L}={\int_{{0}}^{{\frac{\pi}{{3}}}}}{\sec{{\left({x}\right)}}}{\left.{d}{x}\right.}={\ln}{{\left|{\sec{{\left({x}\right)}}}+{\tan{{\left({x}\right)}}}\right|}_{{0}}^{{\frac{\pi}{{3}}}}}$

$\displaystyle{L}={\ln{{\left({\sec{{\left(\frac{\pi}{{3}}\right)}}}+{\tan{{\left(\frac{\pi}{{3}}\right)}}}\right)}}}-{\ln{{\left({\sec{{\left({0}\right)}}}+{\tan{{\left({0}\right)}}}\right)}}}={\ln{{\left({2}+\sqrt{{{3}}}\right)}}}-{\ln{{\left({1}+{0}\right)}}}$

$\displaystyle{L}={\ln{{\left({2}+\sqrt{{{3}}}\right)}}}-{\ln{{\left({1}\right)}}}={\ln{{\left({2}+\sqrt{{{3}}}\right)}}}\approx{1.32}$

So the exact value of the arc length of the graph of the function over the given interval is $\displaystyle{\ln{{\left({2}+\sqrt{{{3}}}\right)}}}$

which is approximately 1.32.

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Thursday, September 15, 2011

$\displaystyle{y}={\ln{{\left({\cos{{x}}}\right)}}},{\left[{0},\frac{\pi}{{3}}\right]}$ Find the arc length of the graph of the function over the indicated interval.

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Thomas Jefferson's election in 1800 is sometimes called the Revolution of 1800. Why could it be described in this way?

Thursday, September 15, 2011

Find the arc length of the graph of the function over the indicated interval.

No comments:

Post a Comment

Thomas Jefferson&#39;s election in 1800 is sometimes called the Revolution of 1800. Why could it be described in this way?

$\displaystyle{y}={\ln{{\left({\cos{{x}}}\right)}}},{\left[{0},\frac{\pi}{{3}}\right]}$ Find the arc length of the graph of the function over the indicated interval.

Thomas Jefferson's election in 1800 is sometimes called the Revolution of 1800. Why could it be described in this way?