A community for students.
Here's the question you clicked on:
 0 viewing
happykiddo
 one year ago
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.
happykiddo
 one year ago
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.

This Question is Closed

IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.1\[ y=\frac{3}{8} (x^{4/3}2x^{2/3})\] \[1 \lt x \lt 8\] ????????

IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.1if this is correct, i'd go for a sub, \(p = x^{\frac{2}{3}}\) so \(y = \frac{3}{8} (p^2 2 p) \rightarrow dy = \frac{3}{8} (2p .dp 2 dp) = \frac{3}{4}.dp.(p1)\) \(x = p^{\frac{3}{2}}, dx = \frac{3}{2}p^\frac{1}{2} .dp\) and so \[ds = \sqrt{dx^2 + dy^2} \ = \sqrt{ \frac{9}{4}p + \frac{9}{16}(p1)^2 \ } \ \ dp\]

happykiddo
 one year ago
Best ResponseYou've already chosen the best response.0Could you explain the usubstitution that you use here? Why you chose the "p" you did?Why use usub twice?....ETC......Or is there a rule, this falls under? Like always your help is much appreciated.

happykiddo
 one year ago
Best ResponseYou've already chosen the best response.0Did you get 63/8, because I got 37.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0\[y=\frac{ 3 }{ 8 }(x ^{4/3}2x ^{2/3})\] \[y=\frac{ 3 }{ 8 }x ^{4/3}\frac{ 3 }{ 4 }x ^{2/3}\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0\[\frac{ dy }{ dx }=\frac{ 1 }{2 }x ^{1/3}\frac{ 1 }{ 2 }x ^{1/3}\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0\[\frac{ dy }{ dx }=\frac{ 1 }{ 2 }x ^{1/3}\frac{ 1 }{2 }x ^{1/3}\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Can you screenshot me the problem?

IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.1@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
 one year ago
Best ResponseYou've already chosen the best response.1\[\sqrt{ \frac{9}{4}p + \frac{9}{16}(p1)^2 \ } \] \[=\sqrt{ \frac{9}{4}p + \frac{9}{16}(p^2 2p +1) \ } \] \[=\sqrt{ \frac{9}{16}(p^2 2p +4p +1) \ } \] \[=\sqrt{ \frac{9}{16}(p +1)^2 \ } \] \[\implies \int\limits_{p=1}^{4}\frac{3}{4}(p+1) \ dp\]

IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.1but 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
 one year ago
Best ResponseYou've already chosen the best response.0Thanks 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.
Ask your own question
Sign UpFind more explanations on OpenStudy
Your question is ready. Sign up for free to start getting answers.
spraguer
(Moderator)
5
→ View Detailed Profile
is replying to Can someone tell me what button the professor is hitting...
23
 Teamwork 19 Teammate
 Problem Solving 19 Hero
 Engagement 19 Mad Hatter
 You have blocked this person.
 ✔ You're a fan Checking fan status...
Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.