arc length/integral

[berni2723]

hi, i am trying to solve the arc length if:

f(x)=(3/2)x^(2/3) on the interval [1,8]
--------------^edit
so first, the formula is s=integral of sqrt(1+[f'(x)]^2)

i solved f'(x) to be x^(-1/3) so [f'(x)]^2 is x^(-2/3)

this gives me the integral for arc length to be

integral of sqrt(1+x^(-2/3))

and now i am stuck...any tips, do i need to do double substitution or is there a way to integrate this as is or with one substitution?

 

[tkhunny]

Better try the 1st derivative again.

 

[berni2723]

Better try the 1st derivative again.

[FONT=Tahoma, Calibri, Verdana, Geneva, sans-serif]correction: the function we are dealing with is

f(x)= (3/2)x^(2/3)

the exponent is not negative, i think that based on that the work is correct, what do you suggest i do next?[/FONT]

 

[tkhunny]

Fair enough.

1) Are you sure the integral exists?
2) Arc Length often leads to intractable integral expressions. This is why we study quantitative methods.
3) "Double Substitution"? Can you find those words, used like that, in your textbook.

 

[berni2723]

Fair enough.

1) Are you sure the integral exists?
2) Arc Length often leads to intractable integral expressions. This is why we study quantitative methods.
3) "Double Substitution"? Can you find those words, used like that, in your textbook.

Thanx for responding again.
Regarding your first 2 points, I wouldn't know how to answer that being that this is the beginning of the semester of calc2, we are simply learning to find arc length, he didn't mention any complications.
Regarding "3)", what I mean is subsituting u=x^(-2/3) du=...then subsituting again for u, being that no matter what I subtitute, u=x^(-2/3) or u=1+x^(-2/3), the result is a mystery to solve. Here is all my work:

let u=1+x^(-2/3),
du=-(2/3)x^(-5/3)dx, -(3/2)x^(5/3)du=dx,

based on ^^, x=(u-1)^(-3/2), and subtituting this value gives:

dx=-(3/2)(u-1)^(-5/2)du

putting all this into my integral, we get:

integral |from x=1 to 8|, u^(1/2)(-3/2)(u-1)^(-5/2)du

rewriting this gives us:

(-3/2) * integral sqrt[u(u-1)^(-5)]du {from x=1 to 8}

does this make sense to you? and back to wether the integral exists, these online integral calculators gave an answer, but i dont want to get the answer from them because i want to be helped to reach it on my own, as that is the best way to learn!!!
thanx!!!

 

[Deleted member 4993]

A more reasonable substitution would be:

x^-1/3 = tan(Θ) ................. don't forget to change the limits of integration accordingly

(-1/3)x^-4/3 dx = sec²(Θ) dΘ

dx = - 3 cot⁴(Θ)*sec²(Θ) dΘ

continue....

 

[soroban]

Hello, berni2723!

I am trying to find the arc length of: .\(\displaystyle f(x)\:=\:\frac{3}{2}x^{\frac{2}{3}}\) on the interval \(\displaystyle [1,8]\)

First, the formula is: .\(\displaystyle \displaystyle S\;=\;\int^b_a \sqrt{1+[f'(x)]^2}\)

I found: \(\displaystyle f'(x) \:=\: x^{-\frac{1}{3}}\), so \(\displaystyle [f'(x)]^2 \:=\: x^{-\frac{2}{3}}\)

This gives me the integral for arc length to be: .\(\displaystyle \displaystyle\int^8_1 \sqrt{1+x^{-\frac{2}{3}}}\,dx \)

and now i am stuck ... . You are doing fine!

We have:. . \(\displaystyle \sqrt{1 + x^{\text{-}\frac{2}{3}}} \;=\;\sqrt{1 + \frac{1}{x^{\frac{2}{3}}}} \;=\;\sqrt{\frac{x^{\frac{2}{3}} + 1}{x^{\frac{2}{3}}}} \;=\;\frac{\sqrt{x^{\frac{2}{3}} + 1}}{\sqrt{x^{\frac{2}{3}}}} \;=\;\frac{\sqrt{x^{\frac{2}{3}} + 1}}{x^{\frac{1}{3}}} \)

Then: .\(\displaystyle \displaystyle S \;=\;\int^8_1\left(x^{\frac{2}{3}} + 1\right)^{\frac{1}{2}}\left(x^{\text{-}\frac{1}{3}}\,dx\right) \)

Let \(\displaystyle u \:=\: x^{\frac{2}{3}} + 1 \quad\Rightarrow\quad du \:=\:\frac{2}{3}x^{\text{-}\frac{1}{3}}dx \quad\Rightarrow\quad x^{\text{-}\frac{1}{3}}dx \:=\:\frac{3}{2}du\)

Substitute: .\(\displaystyle \displaystyle S \;=\;\int\left(u^{\frac{1}{2}}\right)\left(\tfrac{3}{2}du\right) \;=\;\tfrac{3}{2}\int u^{\frac{1}{2}}du \;=\;u^{\frac{3}{2}} + C\)

Back-substitute: .\(\displaystyle S \;=\;\left(x^{\frac{2}{3}}+1\right)^{\frac{3}{2}} + C \)


Evaluate: .\(\displaystyle S \;=\;\left(x^{\frac{2}{3}}+1\right)^{\frac{3}{2}} \,\bigg]^8_1\)

. . . . . . . .\(\displaystyle S \;=\;\left(8^{\frac{2}{3}}+1\right)^{\frac{3}{2}} - \left(1^{\frac{2}{3}}+1\right)^{\frac{3}{2}} \)

. . . . . . . .\(\displaystyle S \;=\;5^{\frac{3}{2}} - 2^{\frac{3}{2}} \)

. . . . . . . .\(\displaystyle S \;=\;5\sqrt{5} - 2\sqrt{2}\)

 

[berni2723]

Hello, berni2723!


We have:. . \(\displaystyle \sqrt{1 + x^{\text{-}\frac{2}{3}}} \;=\;\sqrt{1 + \frac{1}{x^{\frac{2}{3}}}} \;=\;\sqrt{\frac{x^{\frac{2}{3}} + 1}{x^{\frac{2}{3}}}} \;=\;\frac{\sqrt{x^{\frac{2}{3}} + 1}}{\sqrt{x^{\frac{2}{3}}}} \;=\;\frac{\sqrt{x^{\frac{2}{3}} + 1}}{x^{\frac{1}{3}}} \)

Then: .\(\displaystyle \displaystyle S \;=\;\int^8_1\left(x^{\frac{2}{3}} + 1\right)^{\frac{1}{2}}\left(x^{\text{-}\frac{1}{3}}\,dx\right) \)

Let \(\displaystyle u \:=\: x^{\frac{2}{3}} + 1 \quad\Rightarrow\quad du \:=\:\frac{2}{3}x^{\text{-}\frac{1}{3}}dx \quad\Rightarrow\quad x^{\text{-}\frac{1}{3}}dx \:=\:\frac{3}{2}du\)

Substitute: .\(\displaystyle \displaystyle S \;=\;\int\left(u^{\frac{1}{2}}\right)\left(\tfrac{3}{2}du\right) \;=\;\tfrac{3}{2}\int u^{\frac{1}{2}}du \;=\;u^{\frac{3}{2}} + C\)

Back-substitute: .\(\displaystyle S \;=\;\left(x^{\frac{2}{3}}+1\right)^{\frac{3}{2}} + C \)


Evaluate: .\(\displaystyle S \;=\;\left(x^{\frac{2}{3}}+1\right)^{\frac{3}{2}} \,\bigg]^8_1\)

. . . . . . . .\(\displaystyle S \;=\;\left(8^{\frac{2}{3}}+1\right)^{\frac{3}{2}} - \left(1^{\frac{2}{3}}+1\right)^{\frac{3}{2}} \)

. . . . . . . .\(\displaystyle S \;=\;5^{\frac{3}{2}} - 2^{\frac{3}{2}} \)

. . . . . . . .\(\displaystyle S \;=\;5\sqrt{5} - 2\sqrt{2}\)

thanx a million, that is the answer that i needed to reach but didnt know how. thank you!!!!!