Question

Find the arc length of x=1/3(y^2+2)^3/2 y=0 y=1

Answer 1

apply formula and done :DDD

Answer 2

I mean yeah I know that I'm just having problems simplifying it

Answer 3

y=((3x)^2/3-2)^0.5

Answer 4

How?

Answer 5

did you find the derivative of your x with respect to y?

Answer 6

first multiply it with 3 for getting the 1/3 from the left side to the right side. Second. take the 2/3 on both side so that the 3/2 power will be gone from the right side. Third, -2 on both side (what you are left with is (3x)^2/3-2=y^2) and Last sqrtroot on both sides.

Answer 7

you don't have to solve for y

Answer 8

y' = (2/3) (y^2+2)^(1/2) (2y)

Answer 9

urg i meant *x' = *

Answer 10

which simplifes to
x' = (y/3) (y^2 + 2)^(1/2)

Answer 11

\[\frac{dx}{dy}=\frac{1}{3} \frac{3}{2} (y^2+2)^\frac{1}{2} (2y)\]

Answer 12

O.O forgot about the 3/2

Answer 13

Lol yeah I noticed it should be 1/2(y^2+2)^1/2 (2y) right?

Answer 14

essentially yes

Answer 15

which can be reduced to y (y^2 + 2)^(1/2)

Answer 16

then use L = ∫ sqrt(1 + (dx/dy)^2]. I believe you'll have a perfect square under the radical after you simplify.

Answer 17

Okay let me try it out to see what I get :)

Answer 18

Okay so I got up to sqrt 1+(y^2(y^2+2) dy, should I distribute the y^2?

Answer 19

yea

Answer 20

Okay and after that what do I do?

Answer 21

did you get y^4 + 2y^2 + 1 ?

Answer 22

Yes

Answer 23

ok, it's a perfect square

Answer 24

So y^2+2y?

Answer 25

no it's (y^2 + 1)^2

Answer 26

since that's under the square root, the exponents cancel leaving you with

∫ (y^2+1)dy, from 0 to 1.

This is just a simple integral.

Answer 27

Is the integral 1/2[1/3u^3]?

Answer 28

no, it's (1/3)y^3 + y

Answer 29

evaluate the limit and you should get (1/3) + 1 = 4/3

Answer 30

Oh lol I thought there was a squared there that's why :p

Answer 31

No wait yeah there is a squared isn't there?

Answer 32

it cancels with root in the formula

Answer 33

Okay got it thank you so so much ^-^