So I'm having trouble with this problem:
Let $$x=\frac{y^2+2y}{8}-\ln(y+1)$$ Find the arclength for $0\leq y\leq 2$.
My work. I know the Arc length formula is $(1+ (x'^2))^{1/2}$ in this case but when I take $x'$ I get $$x' = (y/4 + 1/4 - 1/(y+1))$$ and when I square that I get something way too hectic and adding $1$ to all of that makes it even more of a mess. Am I missing something here?
1 Answer
You are going in the right direction. However note that after squaring and adding $1$ a perfect square turns out: $$1+(x'(y))^2=1+\left(\frac{y+1}{4}-\frac{1}{y+1}\right)^2 =\frac{(y+1)^2}{16}+\frac{1}{(y+1)^2}+\frac{1}{2}=\left(\frac{t}{4}+\frac{1}{t}\right)^2.$$ where $t=y+1$. Now integrate the square root of the right-hand side for $t$ in $(1,3)$ and you will get the result.
