Question

FInd length of curve, without calculator?

y= (1/3)x^3 + 1/4x {1, 2}

I know what the arc length formula is but im having a tough time integrating this one without a calculator

Answer 1

Looks like you will have to use trigonometric substitution once you get\[\int\limits_{1}^{2}\sqrt{1 + (x ^{2} + \frac{ 1 }{ 4 })} dx\]

Answer 2

except that there is an x in the denominator of the 1/4

Answer 3

Is your original equation term (1/4)x or 1/(4x) ?

Answer 4

1 / (4x)

Answer 5

ok, that changes everything.

Answer 6

yea :S

Answer 7

ok, got it, hold on.

Answer 8

ok

Answer 9

You should come down to :\[\int\limits_{1}^{2}\frac{ 4x ^{4}+ 1 }{ 4x ^{2} }dx\] which can be computed pretty easily after further simplifying. Did you get to this step?

Answer 10

hold up

Answer 11

i got to

Answer 12

sqrt ( 1 + [( 4x^4 -1 )/4x^2] )

Answer 13

You have to square [( 4x^4 -1 )/4x^2]. Once you do that, you can add the one and the -8x^4 will become +8x^4 and then both the numerator and denominator will be squares and the square root will be easy to get. You will come down to my integral listed most recently. Piece of cake at that point.

Answer 14

how do you root that?

Answer 15

i got to the point to where you add the one.... now how do i root that?

Answer 16

the numerator specifically... sorry

Answer 17

[( 4x^4 -1 )/4x^2]^2 = (16x^8 - 8x^4 + 1)/(16x^4) When you add 1 you get (16x^8 + 8x^4 + 1)/(16x^4) and the sqrt is (4x^4 + 1)/(4x^2) Now you have my integral above.

Answer 18

ok thanks

Answer 19

Is it ok now? I flew through it so I'm counting on you to check me for accuracy.

Answer 20

nope its correct. i just worked through the problem. thanks a lot

Answer 21

You're quite welcome!