happykiddo
  • happykiddo
Explain how to get the arclength of y=3/8(x^(4/3)-2x^(2/3)) from x = 1 to x = 8. I know the answer is 63/8, but don't understand how to get there.
Mathematics
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happykiddo
  • happykiddo
Explain how to get the arclength of y=3/8(x^(4/3)-2x^(2/3)) from x = 1 to x = 8. I know the answer is 63/8, but don't understand how to get there.
Mathematics
jamiebookeater
  • jamiebookeater
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IrishBoy123
  • IrishBoy123
\[ y=\frac{3}{8} (x^{4/3}-2x^{2/3})\] \[1 \lt x \lt 8\] ????????
IrishBoy123
  • IrishBoy123
if this is correct, i'd go for a sub, \(p = x^{\frac{2}{3}}\) so \(y = \frac{3}{8} (p^2 -2 p) \rightarrow dy = \frac{3}{8} (2p .dp -2 dp) = \frac{3}{4}.dp.(p-1)\) \(x = p^{\frac{3}{2}}, dx = \frac{3}{2}p^\frac{1}{2} .dp\) and so \[ds = \sqrt{dx^2 + dy^2} \ = \sqrt{ \frac{9}{4}p + \frac{9}{16}(p-1)^2 \ } \ \ dp\]
happykiddo
  • happykiddo
Could you explain the u-substitution that you use here? Why you chose the "p" you did?Why use u-sub twice?....ETC......Or is there a rule, this falls under? Like always your help is much appreciated.

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happykiddo
  • happykiddo
Did you get 63/8, because I got 37.
anonymous
  • anonymous
\[y=\frac{ 3 }{ 8 }(x ^{4/3}-2x ^{2/3})\] \[y=\frac{ 3 }{ 8 }x ^{4/3}-\frac{ 3 }{ 4 }x ^{2/3}\]
anonymous
  • anonymous
\[\frac{ dy }{ dx }=\frac{ 1 }{2 }x ^{1/3}-\frac{ 1 }{ 2 }x ^{-1/3}\]
anonymous
  • anonymous
\[\frac{ dy }{ dx }=\frac{ 1 }{ 2 }x ^{1/3}-\frac{ 1 }{2 }x ^{-1/3}\]
anonymous
  • anonymous
Can you screenshot me the problem?
IrishBoy123
  • IrishBoy123
@happykiddo the sub comes from messing around with it for a bit. those rational exponents just look really annoying, especially as you have the \(\sqrt{}\) to deal with. sorry i can't be more precise. i guess if you do enough of these, you will just get a feel for it and the answer is 63/8 as advertised. the limits become \(1 \le p \le 4\), right?
IrishBoy123
  • IrishBoy123
\[\sqrt{ \frac{9}{4}p + \frac{9}{16}(p-1)^2 \ } \] \[=\sqrt{ \frac{9}{4}p + \frac{9}{16}(p^2- 2p +1) \ } \] \[=\sqrt{ \frac{9}{16}(p^2- 2p +4p +1) \ } \] \[=\sqrt{ \frac{9}{16}(p +1)^2 \ } \] \[\implies \int\limits_{p=1}^{4}\frac{3}{4}(p+1) \ dp\]
IrishBoy123
  • IrishBoy123
but you don't actually need the sub because \(dy = \frac{1}{2}(x^{\frac{1}{3}} - x^{\frac{-1}{3}})dx\) \(\sqrt{dx^2 + dy^2} = \sqrt{1 + \frac{1}{4}(x^\frac{2}{3}+x^\frac{-2}{3}-2)} \times \ dx\) which eventually becomes \(= \frac{1}{2} \sqrt{(x^\frac{1}{3} + x^\frac{-1}{3})^2} \times dx \)
happykiddo
  • happykiddo
Thanks for the help guys! Chrome on my computer crashed, and man was it a journey to get it working again. Anyway I understand, and appreciate the help.

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