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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.

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up vote 2 down vote accepted

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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I can't figure ut the complete the square formula for this since none of the my powers are 2. – user138246 Jun 6 '12 at 22:45
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. – André Nicolas Jun 6 '12 at 23:40

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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Based on our previous experience with Jordan, I think it is best to guide him. It is better for him to understand how to solve the problems, than to have the solution served. – Pedro Tamaroff Jun 6 '12 at 22:03
@PeterTamaroff I have tried with him in the past. You can see here (…) and here (…). Read his comments to my answers there. To be honest, Jordan is only looking for the solution and is not interested in taking time to understand the idea or the procedure. – user17762 Jun 6 '12 at 22:08
Even though he wants full solutions, I think it is nocive for him to get them if he doesn't understand what we're doing. He'll stay dependant on others when solving math problems when he should be improving his skills and independance. – Pedro Tamaroff Jun 6 '12 at 22:16
Seeing some of your answers, I think the problem is that you introduce certain notation Jordan is not familiar with, such as $d(\exp(z))$. You can explain what you're doing in some sentences and be a little more "pedagogical". – Pedro Tamaroff Jun 6 '12 at 22:19
I am interested in learning as much as I can within reason but I do have a lot of work to get done with not a lot of time to do it, hence sometimes all I want to learn is the subject at hand and not a dozen different nifty techniques to solve a problem. Marvis your answer in those linked questions are a good example of a bad answer. It is just basically you showing off in a way that not a single person taking calc 2 will understand. – user138246 Jun 6 '12 at 22:33

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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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:


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I think you mean $$y'(x) = \frac{{x - 1}}{{2\sqrt x }}$$ – Pedro Tamaroff Jun 6 '12 at 21:56
yep, corrected this and more – Valentin Jun 6 '12 at 22:08

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