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.
Wednesday, October 30, 2019
Calculus of a Single Variable, Chapter 7, 7.4, Section 7.4, Problem 12
Subscribe to:
Post Comments (Atom)
Summarize the major research findings of "Toward an experimental ecology of human development."
Based on findings of prior research, the author, Bronfenbrenner proposes that methods for natural observation research have been applied in ...
-
Find the indefinite integral $\displaystyle \int \sec^4 \left( \frac{x}{2} \right) dx$. Illustrate by graphing both the integrand and its an...
-
Determine $\displaystyle \frac{dy}{dx}$ of $y^5 + x^2y^3 = 1 + x^4 y$ by Implicit Differentiation. $\displaystyle \frac{d}{dx}(y^5) + ...
-
Determine the area of the region bounded by the hyperbola $9x^2 - 4y^2 = 36$ and the line $ x= 3$ By using vertical strips, Si...
-
Find the integral $\displaystyle \int^1_0 \frac{1}{\sqrt{16 t^2 + 1}} dt$ If we let $u = 4t$, then $du = 4dt$, so $\displaystyle dt = \frac{...
-
Determine the integral $\displaystyle \int \frac{\sin^3 (\sqrt{x})}{\sqrt{x}} dx$ Let $u = \sqrt{x}$, then $\displaystyle du = \frac{1}{2 \s...
-
Gertrude's comment "The lady protests too much, methinks" in act 3, scene 2, of Shakespeare's Hamlet exposes her own guilt...
-
Given y=cos(2x), y=0 x=0,x=pi/4 so the solid of revolution about x-axis is given as V = pi * int _a ^b [R(x)^2 -r(x)^2] dx here R(x) =cos(2x...
No comments:
Post a Comment