- #1
- 8
- 0
1. The problem statement, all variables and given/known
Find the length of the curve $$y=ln(x),\frac{1}{2}<=x<=2$$
Using hyperbolic trig isn't necessary, but it's how my text (Serge Lang's A First Course in Calculus) approaches most square roots, and as a result, it's what I've grown accustomed to. I am aware of other solutions (thanks to Wolfram Alpha). I want to know where the heck I'm going wrong.
Arc length can be calculated by $$ \int_a^b \sqrt{1+(\frac{dx}{dy})^{2}}dx$$. Since y=ln(x), then dx/dy=1/x dx, which gives our first integral of $$ \int_\frac{1}{2}^2 \sqrt{1+\frac{1}{x}^{2}} dx$$. Combining the two terms under the radical and simplifying, we have $$\int_\frac{1}{2}^2 \frac{\sqrt{x^{2}+1}}{x} dx$$. Next, let x=sinh(t), then dx=cosh(t) dt. Substitution gives us the indefinite integral (with limits of integration to be used after finding an indefinite integral) $$\int \frac {\sqrt{sinh^{2}t+1}}{sinht}dt$$. Renaming sinh^2(t)+1 into cosh^2(t), and then taking the square root of that, then multiplying it by cosht, gives us $$\int \frac {cosh^{2}t}{sinht}dt$$. Renaming cosh^2(t) into sin^2(t)+1, and splitting the fraction into its two equivalent pieces, then simplifying, we now have $$ \int sinh(t)dt+\int \frac{dt}{sinh(t)}$$. The left integral simplifies into sinht dt, whose integral is cosht, and this allows us to substitute back for an explicit expression (in terms of x), namely, $$\sqrt{x^{2}+1}$$.
The right side is my issue. I did a number of things, but what worked the best was to rename sinh(t) in terms of e. I simplified the fraction and found that I had $$\int \frac{2 dt}{e^{t}-e^{-t}}$$. I multiplied this integral by the clever one (e^t)/(e^t), which gave me $$2 \int \frac{e^{t}dt}{e^{2t}-1}$$. Because this looked like a very difficult integrand, I substituted back for x, since t=arcsinh(x) (which we know). Then we have $$2 \int \frac {x+\sqrt{x^2+1}dx}{(x+\sqrt{x^2+1})^{2}-1}$$. Simplifying this I found $$2 \int \frac {x+\sqrt{x^2+1}}{2x^{2}+2x\sqrt{x^{2}+1}}$$. Factoring out a 1/2, which cancels the constant on the outside, we have what appears to simplify into a very simple integrand, that is, $$\int \frac{dx}{x}$$.
Evaluating this from 1/2 to 2 would give me the answer $$\frac{\sqrt{5}}{2}-ln(\frac{1}{4})$$. This is approximately equal to 2.504. This is not the right answer. The book's answer is given as $$\frac{\sqrt{5}}{2}+ln(\frac{4+2\sqrt{5}}{1+\sqrt{5}})$$, which is approximately equal to 2.080.
Phew! That was a long write up from me - any help would be appreciated. Also, some advice: Are the hyperbolic trig functions a good or bad idea here? I know we can also use tan^2(theta)+1 to our advantage here as well.
Find the length of the curve $$y=ln(x),\frac{1}{2}<=x<=2$$
Homework Equations
Using hyperbolic trig isn't necessary, but it's how my text (Serge Lang's A First Course in Calculus) approaches most square roots, and as a result, it's what I've grown accustomed to. I am aware of other solutions (thanks to Wolfram Alpha). I want to know where the heck I'm going wrong.
The Attempt at a Solution
Arc length can be calculated by $$ \int_a^b \sqrt{1+(\frac{dx}{dy})^{2}}dx$$. Since y=ln(x), then dx/dy=1/x dx, which gives our first integral of $$ \int_\frac{1}{2}^2 \sqrt{1+\frac{1}{x}^{2}} dx$$. Combining the two terms under the radical and simplifying, we have $$\int_\frac{1}{2}^2 \frac{\sqrt{x^{2}+1}}{x} dx$$. Next, let x=sinh(t), then dx=cosh(t) dt. Substitution gives us the indefinite integral (with limits of integration to be used after finding an indefinite integral) $$\int \frac {\sqrt{sinh^{2}t+1}}{sinht}dt$$. Renaming sinh^2(t)+1 into cosh^2(t), and then taking the square root of that, then multiplying it by cosht, gives us $$\int \frac {cosh^{2}t}{sinht}dt$$. Renaming cosh^2(t) into sin^2(t)+1, and splitting the fraction into its two equivalent pieces, then simplifying, we now have $$ \int sinh(t)dt+\int \frac{dt}{sinh(t)}$$. The left integral simplifies into sinht dt, whose integral is cosht, and this allows us to substitute back for an explicit expression (in terms of x), namely, $$\sqrt{x^{2}+1}$$.
The right side is my issue. I did a number of things, but what worked the best was to rename sinh(t) in terms of e. I simplified the fraction and found that I had $$\int \frac{2 dt}{e^{t}-e^{-t}}$$. I multiplied this integral by the clever one (e^t)/(e^t), which gave me $$2 \int \frac{e^{t}dt}{e^{2t}-1}$$. Because this looked like a very difficult integrand, I substituted back for x, since t=arcsinh(x) (which we know). Then we have $$2 \int \frac {x+\sqrt{x^2+1}dx}{(x+\sqrt{x^2+1})^{2}-1}$$. Simplifying this I found $$2 \int \frac {x+\sqrt{x^2+1}}{2x^{2}+2x\sqrt{x^{2}+1}}$$. Factoring out a 1/2, which cancels the constant on the outside, we have what appears to simplify into a very simple integrand, that is, $$\int \frac{dx}{x}$$.
Evaluating this from 1/2 to 2 would give me the answer $$\frac{\sqrt{5}}{2}-ln(\frac{1}{4})$$. This is approximately equal to 2.504. This is not the right answer. The book's answer is given as $$\frac{\sqrt{5}}{2}+ln(\frac{4+2\sqrt{5}}{1+\sqrt{5}})$$, which is approximately equal to 2.080.
Phew! That was a long write up from me - any help would be appreciated. Also, some advice: Are the hyperbolic trig functions a good or bad idea here? I know we can also use tan^2(theta)+1 to our advantage here as well.