# Arc Length Problem: Assimilation of A Constant

The problem I am working on is,

I understand the motivation of the step, but I am not quite sure how the author the solution manual absorbs the constant into the other term. What is the process?

EDIT:

Here is another one:

Here is my attempt to try and figure out what they are doing:

$1+[y']^2=[\frac{1}{2}(e^x-e^-x)]+1$

$1+[y']^2=[\frac{1}{2}(e^x-e^-x)+1]^2$

$1+[y']^2=[\frac{1}{2}(e^x-e^-x+2)]^2$ This is where I get stuck, because I can't truly see any sort of manipulation that would get rid of the 2 and put a plus sign between the exponential functions.

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If you try multiplying out the brackets at the point where you have drawn the arrows, you will see that they are really equal. The crucial thing is that the -2 inside the bracket seems to change to +2, but if you notice, there is a factor of 1/4 outside, so ... – Old John Nov 27 '12 at 13:08
Yeah, I did notice that, so I figured 4 must have been added to it. But I just wanted to make sure I was processing it correctly. – Mack Nov 27 '12 at 13:12
The identity $(x+y)^2=x^2+2xy+y^2$ is familiar. Set $x=t$ and $y=\frac{1}{t}$. Then $2xy=2$. It follows that $\left(t+\frac{1}{t}\right)^2=t^2+2+\frac{1}{t^2}$. There is a similar expression for $\left(t-\frac{1}{t}\right)^2$. These come up moderately often in textbook (and exam!) arclength problems, because there are very few arclength problems for which the integration is oable. Many of the doable ones involve this trick. – André Nicolas Nov 27 '12 at 17:24

$\quad\quad\quad$First question: $$1 + \frac14\left(x^8 - 2 +\frac{1}{x^8}\right)\tag{1.1}$$ $$=\frac14\cdot 4 + \frac14\left(x^8 - 2 + +\frac{1}{x^8}\right)\quad=\quad\frac14\left(4 + x^8 -2 +\frac{1}{x^8}\right)\tag{1.2}$$

$$=\frac14\left(x^8 + 2 + \frac{1}{x^8}\right)\tag{1.3}$$

$$=\frac14\left(x^4 + \frac{1}{x^4}\right)^2\tag{1.4}$$

$$(1.3) \to (1.4):\quad\quad\text{Note that}$$ $$\frac14 \left(x^4 + \frac{1}{x^4}\right)^2\tag{1.4}$$ $$= \frac14\left(x^{2\cdot 4} + 2\frac{x^4}{x^4} + \frac{1}{x^{2\cdot 4}}\right) = \frac14\left(x^8 + 2 + \frac{1}{x^8}\right)\tag{1.3}$$

$$1+[y']^2=[\frac{1}{2}(e^x-e^{-x})]+1\tag{a}$$

$$1+[y']^2=[\frac{1}{2}(e^x-e^{-x})+1]^2\tag{b}$$

$$1+[y']^2=[\frac{1}{2}(e^x-e^{-x}+2)]^2\tag{c}$$

How did you get from $(a)\to(b)$? Did you forget the exponent in $(a)$? Shouldn't $(a)$ be $$1+[y']^2=[\frac{1}{2}(e^x-e^{-x})]^2+1\quad?$$And if so, you cannot bring the constant term $1$ into an expression that is exponentiated without first making appropriate computations on the exponentiation expression. Failing that, your move from $(b)\to (c)$ is affected.

In this case, first expand $\left[\dfrac{1}{2}(e^x-e^{-x})\right]^2$, then worry about adding the $1$ term:

$$1+[y']^2=\left[\frac{1}{2}(e^x-e^{-x})\right]^2+1\tag{2.1}$$ $$=\frac14\left(e^{2x} -2\frac{e^{2x}}{e^{2x}} + \frac{1}{e^{2x}}\right) + 1\tag{2.2}$$

$$=\frac14\left(e^{2x} -2 + \frac{1}{e^{2x}}\right) + \frac14\cdot4 \quad=\quad\frac14\left(e^{2x} -2 +\frac{1}{e^{2x}}+4\right)\tag{2.3}$$ $$= \left(\frac12\right)^2\left(e^{2x} +2 +\frac{1}{e^{2x}}\right)\quad=\quad\left[\frac12(e^x + e^{-x})\right]^2\tag{2.4}$$

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