Question

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.

Answer 1

\[ y=\frac{3}{8} (x^{4/3}-2x^{2/3})\]

\[1 \lt x \lt 8\]

????????

Answer 2

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
\[ds = \sqrt{dx^2 + dy^2} \ = \sqrt{ \frac{9}{4}p + \frac{9}{16}(p-1)^2 \ } \ \ dp\]

Answer 3

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.

Answer 4

Did you get 63/8, because I got 37.

Answer 5

\[y=\frac{ 3 }{ 8 }(x ^{4/3}-2x ^{2/3})\]
\[y=\frac{ 3 }{ 8 }x ^{4/3}-\frac{ 3 }{ 4 }x ^{2/3}\]

Answer 6

\[\frac{ dy }{ dx }=\frac{ 1 }{2 }x ^{1/3}-\frac{ 1 }{ 2 }x ^{-1/3}\]

Answer 7

\[\frac{ dy }{ dx }=\frac{ 1 }{ 2 }x ^{1/3}-\frac{ 1 }{2 }x ^{-1/3}\]

Answer 8

Can you screenshot me the problem?

Answer 9

@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?

Answer 10

\[\sqrt{ \frac{9}{4}p + \frac{9}{16}(p-1)^2 \ } \]
\[=\sqrt{ \frac{9}{4}p + \frac{9}{16}(p^2- 2p +1) \ } \]
\[=\sqrt{ \frac{9}{16}(p^2- 2p +4p +1) \ } \]
\[=\sqrt{ \frac{9}{16}(p +1)^2 \ } \]
\[\implies \int\limits_{p=1}^{4}\frac{3}{4}(p+1) \ dp\]

Answer 11

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

\(= \frac{1}{2} \sqrt{(x^\frac{1}{3} + x^\frac{-1}{3})^2} \times dx \)

Answer 12

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.