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$y = 1/3 \sqrt{x} (x-3)$ from 1-9

In my book it is actually $x = 1/3 \sqrt{y} (y-3)$ but I prefer to work with y so I just swap the two variables and I think everything should be the same.

The first thing I need to do is get an integral.

$y = 1/3 \sqrt{x} (x-3)$ = $\frac{x^\frac{3}{2} - 3x^\frac{1}{2}}{3}$

$\sqrt{x} + \frac {\sqrt{x} - 3}{2\sqrt{x}}$

Plug that into the arc length formula

$\int_1^9 \sqrt{1 + (\sqrt{x} + \frac {\sqrt{x} - 3}{2\sqrt{x}})^2}$

$\int_1^9 \sqrt{1 + x - 2+ x^\frac{1}{4} + \frac{x}{4} -\frac{3}{2} - \frac{9}{4x}}$

From here I am very lost.

4 Answers 4

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It is very similar your previous problem. Magically, when we add $1$ to $(f'(x))^2$, we get something whose square root is pleasant. But we must differentiate and do the algebra flawlessly!

Let $f(x)$ be our function. Then $f(x)=\frac{1}{3}\left(x^{3/2}-3x^{1/2}\right)$, and therefore $f'(x)=\frac{x^{1/2}}{2}-\frac{x^{-1/2}}{2}$.

Square. We get $\frac{x}{4}-\frac{1}{2}+\frac{x^{-1}}{4}$

Add $1$. We get $\frac{x}{4}+\frac{1}{2}+\frac{x^{-1}}{4}$. Take the square root. This turn out to be $\frac{x^{1/2}}{2}+\frac{x^{-1/2}}{2}$. Finally, integrate.

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    Look at what we got when we squared $\frac{x^{1/2}}{2}-\frac{x^{-1/2}}{2}$. There is a middle term of $-\frac{1}{2}$. When we add $1$, we get the same thing, except for a middle term of $+\frac{1}{2}$. So our expression must be the result of squaring $\frac{x^{1/2}}{2}+\frac{x^{-1/2}}{2}$. You can check by squaring the just mentioned expression.2012-06-06
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I try to use logarithmic differentiating wherever appropriate when dealing with functions involving rational functions and/or radicals as factors. The advantage is that one has to deal with additive terms instead of factors which are often easy to manipulate and reduce:

$\log y = \log\frac{1}{3} +\frac{1}{2}\log x +\log (x-3)$ $\frac{y'}{y}=\frac{1}{2x}+\frac{1}{x-3}=\frac{3}{2}\frac{x-1}{x(x-3)}$ $y'=\frac{1}{2}\frac{x-1}{\sqrt{x}}$

$dl = \sqrt{1+y'^2}dx=\sqrt{1+\frac{1}{4}\frac{(x-1)^2}{x}}dx=\frac{1}{2\sqrt{x}}\sqrt{4x+x^2-2x+1}dx=\frac{1}{2\sqrt{x}}\left(x+1\right)dx=\frac{1}{2}\left(\frac{1}{\sqrt{x}}+\sqrt{x}\right)dx$

One method that I myself find very handy when you get used to it is "pushing" expressions under $d$ using the rule: $d\left(f(x)\right)=f'(x)dx$ All it takes is remembering some standard derivatives from the table. All it gives is breaking down complex expressions into manageable blocks without having to introduce new variables. Changing variables then is understood much like giving names to chunks of programming code which you have to reuse consistently and no "magic" at all. If you look up similar questions I have answered here, you will see that I am using the same "method" all around. Below, just by noticing that $\left(\sqrt{x}\right)'=\frac{1}{2\sqrt{x}}$ one can avoid dealing with fractional exponents, reduce the limits and achieve greater transparency in calculations:

$l=\frac{1}{2}\int_1^9\left(\frac{1}{\sqrt{x}}+\sqrt{x}\right)dx=\int_1^9\left(1+\left(\sqrt{x}\right)^2\right)d\left(\sqrt{x}\right)=\int_1^3\left(1+t^2\right)dt=\left.\left(t+\frac{t^3}{3}\right)\right|_1^3=3+9-1-\frac{1}{3}=\frac{32}{3}$

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    yep, corrected this and more2012-06-06
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Your original function is

$y(x) =\frac{1}{3}\sqrt x (x-3)$

It's derivative is

$y'(x) =\frac{1}{3}\frac{1}{2\sqrt x} (x-3)+\frac{1}{3}\sqrt x$

$\eqalign{ & y'(x) = \frac{1}{3}\frac{{x - 3}}{{2\sqrt x }} + \frac{1}{3}\sqrt x \cr & y'(x) = \frac{1}{3}\frac{{x - 3 + 2x}}{{2\sqrt x }} \cr & y'(x) = \frac{1}{3}\frac{{3x - 3}}{{2\sqrt x }} \cr & y'(x) = \frac{{x - 1}}{{2\sqrt x }} \cr & y'(x) = \frac 1 2\left( \sqrt x -\frac{{ 1}}{{\sqrt x }} \right)\cr} $ Then

$1 + y'{\left( x \right)^2} = \frac{{{{\left( {x - 1} \right)}^2}}}{{4x}} + 1$ $\eqalign{ & y'{(x)^2} = \frac{1}{2}\left( {\sqrt x - \frac{1}{{\sqrt x }}} \right) = \frac{x}{4} - \frac{1}{2} + \frac{1}{{4x}} \cr & 1 + y'{\left( x \right)^2} = \frac{x}{4} + \frac{1}{2} + \frac{1}{{4x}} = \frac{x}{4} + 2\frac{{\sqrt x }}{2}\frac{1}{{2\sqrt x }} + \frac{1}{{4x}} \cr} $

Try to emulate the answers to your other questions on this topic.

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The length for a curve $y = f(x)$ from $x=a$ to $x=b$ is given by $L = \int_a^b \sqrt{1 + \left( \dfrac{dy}{dx} \right)^2} dx$

In your case, $y = \dfrac13 \left( x^{3/2} - 3x^{1/2}\right)$ with $a=1$ and $b=9$. Hence, $\dfrac{dy}{dx} = \dfrac13 \left( \dfrac32 x^{3/2-1} - 3 \times \dfrac12 x^{1/2-1}\right)= \dfrac13 \left( \dfrac32 x^{1/2} - \dfrac32 x^{-1/2}\right) = \dfrac12 \left( x^{1/2} - x^{-1/2}\right)$ Hence, $\dfrac{dy}{dx} = \dfrac12 \left( \sqrt{x} - \dfrac1{\sqrt{x}}\right)$ Hence, $L = \int_1^9 \sqrt{1 + \left( \dfrac{dy}{dx} \right)^2} dx = \int_1^9 \sqrt{1 + \left( \dfrac12 \left( \sqrt{x} - \dfrac1{\sqrt{x}}\right) \right)^2} dx$ Hence, $L = \int_1^9 \sqrt{1 + \dfrac14 \left( (\sqrt{x})^2 + \left(\dfrac1{\sqrt{x}} \right)^2 -2 \right)} dx = \int_1^9 \sqrt{1 + \dfrac14 \left( x + \dfrac1x -2 \right)} dx$ $L = \int_1^9 \sqrt{\dfrac14 \left(4 + x + \dfrac1x -2 \right)} dx = \int_1^9 \sqrt{\dfrac14 \left(2 + x + \dfrac1x\right)} dx = \int_1^9 \sqrt{\dfrac14 \left(\sqrt{x} + \dfrac1{\sqrt{x}}\right)^2} dx$ $L = \int_1^9 \dfrac12 \left(\sqrt{x} + \dfrac1{\sqrt{x}}\right)dx = \dfrac12 \left(\dfrac{x^{3/2}}{3/2} + \dfrac{x^{1/2}}{1/2} \right)_1^9 = \left(\dfrac{x^{3/2}}{3} + x^{1/2} \right)_1^9$ $L = \left(\dfrac{9^{3/2}}{3} + 9^{1/2} \right) - \left(\dfrac{1^{3/2}}{3} + 1^{1/2} \right) = (9+3) - (1/3 + 1) = \dfrac{32}{3}$

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    @Jorda$n$: Use of profanity is not tolerated here. I have removed it from your above comment, and any further issues may be cause for suspension.2012-06-07