Wednesday, October 30, 2019

Calculus of a Single Variable, Chapter 7, 7.4, Section 7.4, Problem 12

The arc length of a function of x, f(x), over an interval is determined by the formula below:
L=int_a^bsqrt(1+((dy)/(dx))^2)dx
So using the function given, let us first find (dy)/(dx):
d/(dx)(ln(cos(x)))=(1/(cos(x)))*(-sin(x))=-(sin(x))/(cos(x))=-tan(x)
We can now substitute this into our formula above:
L=int_a^bsqrt(1+((dy)/(dx))^2)dx=L=int_0^(pi/3)sqrt(1+(-tan(x))^2)dx
Which can then be simplified to:
L=int_0^(pi/3)sqrt(1+tan^2(x))dx=int_0^(pi/3)sqrt(sec^2(x))dx=int_0^(pi/3)sec(x)dx
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).)
L=int_0^(pi/3)sec(x)dx=ln|sec(x)+tan(x)|_0^(pi/3)
L=ln(sec(pi/3)+tan(pi/3))-ln(sec(0)+tan(0))=ln(2+sqrt(3))-ln(1+0)
L=ln(2+sqrt(3))-ln(1)=ln(2+sqrt(3))~~1.32
So the exact value of the arc length of the graph of the function over the given interval is ln(2+sqrt(3))
which is approximately 1.32.

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