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The problem I am working on is,

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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:

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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.

  • 0
    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.2012-11-27

1 Answers 1

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

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

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

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



For your second question: in your work you have:

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

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

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

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:

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

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