Question

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?

Answer 1

integrate the length of a hypotenuse

Answer 2

huh? should i be using the arc length formula? What should my final answer be?

Answer 3

?

Answer 4

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:\[ds = \sqrt{1+(y')^2}\]

Answer 5

the length of the curve is then the sum of all the itty bitty ds parts:\[\int_C\ ds=\int_{a}^{b}\sqrt{1+(y')^2}\ dx\]

Answer 6

so what we need to find is y' in order to plug it in

what have you got for y' ?

Answer 7

how would I compute the derivative of y? 5/6+1 bring the 5/6 up front?

Answer 8

if 5/6 is the exponent then your almost got it :) 5/6 up front and 5/6-1 left

Answer 9

and then for +1/10x^3 i bring the 3 up front and subtract 1 from exponent, so 3/10x^2?

Answer 10

srry, I'm so confuse. I appreciate you sitting down and working with me, step by step would help.

Answer 11

\[y'=\frac{5}{6}x^{1/6}+\frac{3}{10}x^2\] is fine

Answer 12

y'^2 looks like a nightmare to do, doable, but a pain

Answer 13

srry.

Answer 14

\[y'^2=\frac{25}{36}x^{2/6}+\frac{9}{100}x^4+\frac{30}{60}x^{2/6}\]
\[y'^2=\frac{25}{36}x^{2/6}+\frac{9}{100}x^4+\frac{3}{6}x^{2/6}\]
\[y'^2=\frac{25}{36}x^{2/6}+\frac{9}{100}x^4+\frac{18}{36}x^{2/6}\]
\[y'^2=\frac{43}{36}x^{1/3}+\frac{9}{100}x^4\]
whew!! i hope i kept that straight

Answer 15

i see one slight error tho; 5/6 - 6/6 = -1/6 so i have to adjust for that mistake

Answer 16

Thank you so much!! ill use the procedure you put down for my other questions. I just needed someone to help me get started.

Answer 17

ah, no worries, I can just look where you went wrong. Thanks again.

Answer 18

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

Answer 19

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!

Answer 20

yeah, I totally understand. say, how would I do it on a comp?

Answer 21

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

Answer 22

oh.

Answer 23

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

Answer 24

to be sure: its spose to be:\[y=x^{5/6}+\frac{1}{10}x^3\]?????

Answer 25

uh its 1/10^2/3 but i guess no difference. Other than that, yes.

Answer 26

1< than or equal to x < than or equal to 2

Answer 27

\[y=x^{5/6}+(\frac{1}{10})^{2/3}\]

theres no x in the end there right? if so than thats just a constant that disappears in the derivative

Answer 28

im thinking your best bet might be to say that sin(t) = y' and use the trig substitution for it

Answer 29

tan(t) that is ; its a 1+d^2 which comforms better to tan

Answer 30

no limit

Answer 31

wait, so what is your final answer? Ill try to input it and see if it's correct.

Answer 32

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