anonymous
  • anonymous
Find the length of the curve y=x^5/6+ 1/10x^3 from 1< or equal to x < or equal to 2 Arc length equals?
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
amistre64
  • amistre64
integrate the length of a hypotenuse
anonymous
  • anonymous
huh? should i be using the arc length formula? What should my final answer be?
amistre64
  • amistre64
|dw:1333496165901:dw|

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anonymous
  • anonymous
?
amistre64
  • amistre64
as all these terms go infinitesimal we get:\[ds^2=dx^2+dy^2\] \[ds = \sqrt{dx^2+dy^2}\] when y is a function of x, this turns out to be:\[ds = \sqrt{1+(y')^2}\]
amistre64
  • amistre64
the length of the curve is then the sum of all the itty bitty ds parts:\[\int_C\ ds=\int_{a}^{b}\sqrt{1+(y')^2}\ dx\]
amistre64
  • amistre64
so what we need to find is y' in order to plug it in what have you got for y' ?
anonymous
  • anonymous
how would I compute the derivative of y? 5/6+1 bring the 5/6 up front?
amistre64
  • amistre64
if 5/6 is the exponent then your almost got it :) 5/6 up front and 5/6-1 left
anonymous
  • anonymous
and then for +1/10x^3 i bring the 3 up front and subtract 1 from exponent, so 3/10x^2?
anonymous
  • anonymous
srry, I'm so confuse. I appreciate you sitting down and working with me, step by step would help.
amistre64
  • amistre64
\[y'=\frac{5}{6}x^{1/6}+\frac{3}{10}x^2\] is fine
amistre64
  • amistre64
y'^2 looks like a nightmare to do, doable, but a pain
anonymous
  • anonymous
srry.
amistre64
  • amistre64
\[y'^2=\frac{25}{36}x^{2/6}+\frac{9}{100}x^4+\frac{30}{60}x^{2/6}\] \[y'^2=\frac{25}{36}x^{2/6}+\frac{9}{100}x^4+\frac{3}{6}x^{2/6}\] \[y'^2=\frac{25}{36}x^{2/6}+\frac{9}{100}x^4+\frac{18}{36}x^{2/6}\] \[y'^2=\frac{43}{36}x^{1/3}+\frac{9}{100}x^4\] whew!! i hope i kept that straight
amistre64
  • amistre64
i see one slight error tho; 5/6 - 6/6 = -1/6 so i have to adjust for that mistake
anonymous
  • anonymous
Thank you so much!! ill use the procedure you put down for my other questions. I just needed someone to help me get started.
anonymous
  • anonymous
ah, no worries, I can just look where you went wrong. Thanks again.
amistre64
  • amistre64
good luck :) those fractions scare me at the moment since I got no idea how thats gonna integrate; unless you can do it by computer
anonymous
  • anonymous
doing the derivative gets tedious at times especially with the whole computation with 1+ and square root- Yuck. Thanks for walking me through the problem!
anonymous
  • anonymous
yeah, I totally understand. say, how would I do it on a comp?
amistre64
  • amistre64
id type it up onto wolframalpha.com and see if it can plow through it; im woking that angle at the moment to see what a mess this things gonna be
anonymous
  • anonymous
oh.
amistre64
  • amistre64
http://www.wolframalpha.com/input/?i=sqrt%281%2B%28derivative+x%5E%285%2F6%29%2Bx%5E3%2F10%29%5E2%29 and thats just before you have to integrate it ... ewwww
amistre64
  • amistre64
to be sure: its spose to be:\[y=x^{5/6}+\frac{1}{10}x^3\]?????
anonymous
  • anonymous
uh its 1/10^2/3 but i guess no difference. Other than that, yes.
anonymous
  • anonymous
1< than or equal to x < than or equal to 2
amistre64
  • amistre64
\[y=x^{5/6}+(\frac{1}{10})^{2/3}\] theres no x in the end there right? if so than thats just a constant that disappears in the derivative
amistre64
  • amistre64
im thinking your best bet might be to say that sin(t) = y' and use the trig substitution for it
amistre64
  • amistre64
tan(t) that is ; its a 1+d^2 which comforms better to tan
anonymous
  • anonymous
no limit
anonymous
  • anonymous
wait, so what is your final answer? Ill try to input it and see if it's correct.
amistre64
  • amistre64
http://www.wolframalpha.com/input/?i=integrate+sqrt%2825%2F%2836x%5E%281%2F3%29%29%2B1%29+from+1+to+2 if i did it right; 1.2695

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