arc length x=9-sqrt(-y^2+2y+8) from y=1 to y=4

[stars2099]

Hey guys, I'm stuck on this problem:

Find the arc length x=9-sqrt(-y^2+2y+8) from y=1 to y=4

I'm stuck on how to approach this. I found the derivative but then I didn't know where to go from there. I tried PFD but I didn't know how to solve it.
Any help would be appreciated. Thanks.

 

[royhaas]

You should be evaluating the integral of \(\displaystyle \frac{3}{\sqrt{-y^2+2y+8}}\).

 

[stars2099]

how'd you get to that integral? after finding X', did you find a common denominator and add?

 

[soroban]

Hello, stars2099!

Looks like you need some baby-step algebra . . .

\(\displaystyle \text{Find the arc length: }\,x\:=\:9-\sqrt{8+2y-y^2}\:\text{ from }y=1\text{ to }y=4\)

\(\displaystyle \text{Formula: }\;L \;=\;\int^b_a\sqrt{1 + \left(\frac{dx}{dy}\right)^2}\,dy\)

\(\displaystyle \text{We have: }\:x \;=\;9 - \left(8+2y-y^2\right)^{\frac{1}{2}}\)

. . \(\displaystyle \text{Then: }\:\frac{dx}{dy}\;=\;\tfrac{1}{2}\left(8+2y-y^2\right)^{\text{-}\frac{1}{2}}(2-2y) \;=\;\frac{1-y}{\sqrt{8+2y-y^2}}\)

. . \(\displaystyle \left(\frac{dx}{dy}\right)^2 \;=\;\left(\frac{1-y}{\sqrt{8+2y-y^2}}\right)^2 \;=\;\frac{(1-y)^2}{8+2y-y^2}\)

. . \(\displaystyle 1 + \left(\frac{dx}{dy}\right)^2 \;=\;1 + \frac{(1-y)^2}{8+2y-y^2} \;=\; \frac{8+2y-y^2}{8+2y-y^2} + \frac{1-2y + y^2}{8+2y-y^2} \;=\;\frac{9}{8+2y-y^2}\)


\(\displaystyle \text{Therefore: }\;\sqrt{1 + \left(\frac{dx}{dy}\right)^2} \;=\;\sqrt{\frac{9}{8+2y-y^2}} \;=\;\frac{\sqrt{9}}{\sqrt{8+2y-y^2}} \;=\;\frac{3}{\sqrt{8+2y-y^2}}\)

Got it?

 

[stars2099]

thanks so much!

 

[stars2099]

my final answer is 3pi/2. Is that correct?

 

[galactus]

Yep, that's it.