Exact length of the curve analytically

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Homework Statement



a curve is given by y=(9-x^(2/3))^(3/2) for 1 ≤ x ≤ 8. Find the exact length of the curve analytically by antidifferentiation


Homework Equations





The Attempt at a Solution


\int_a^b \sqrt{1+\frac{dy}{dx}} dx
I use this formula right?

i took the derivative of the equation and i got \sqrt{9-x^(2/3)} times X^(-1/3)

How do I integrate it with the negative exponent... I know its probably simple but I'm just not getting it
 
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volleygirl292 said:

Homework Statement



a curve is given by y=(9-x^(2/3))^(3/2) for 1 ≤ x ≤ 8. Find the exact length of the curve analytically by antidifferentiation


Homework Equations





The Attempt at a Solution


\int_a^b \sqrt{1+\frac{dy}{dx}} dx
I use this formula right?

i took the derivative of the equation and i got \sqrt{9-x^(2/3)} times X^(-1/3)

How do I integrate it with the negative exponent... I know its probably simple but I'm just not getting it


Your equation above is wrong. It should be

\int_a^b \sqrt{1 + (\frac{dy}{dx}})^2 dx

When you use this form of the equation for the arc length, you will find a couple of things simplify.
 
ok so now i have it down to integral of sqrt of 1-(9-x^(2/3)/(x^(2/3))

but i still have a exponent in the denominator
 
Let me first fix your expression by putting in the "tex" commands:

volleygirl292 said:
ok so now i have it down to

\int_a^b \sqrt{1 + (\frac{9-x^(2/3)}{x^2/3}}) dx

That doesn't look right.

Now,

y = (9 - x^{\frac{2}{3}})^{\frac{3}{2}}

has a derivative that looks like

\frac{dy}{dx} = -(9 - x^{\frac{2}{3}})^{\frac{1}{2}} x^{-\frac{1}{3}}

Now when you square that mess so you can put it into your expression for the arc length there will be simplifications that will allow you to do the integration.
 
I got it down to the integral of sqrt of 1+ (27-x)/x

but i still have an x in the denominator
 
volleygirl292 said:
ok so now i have it down to integral of sqrt of 1-(9-x^(2/3)/(x^(2/3))

but i still have a exponent in the denominator

It's 1+(9-x^(2/3)/(x^(2/3)), isn't it? Simplify the algebraically and then integrate.
 
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