Question

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

Answer 1

where are you stuck?

Answer 2

After I input it into the formula, I'm having trouble simplifying it

Answer 3

can you show what you have so far?

Answer 4

integrate (1/2) sqrt(6 + y^2) from 0 to1

Answer 5

@TuringTest sqrt 1+ (y(y^2+2)^1/2)^2

Answer 6

@pi3 where did the 6 come from?

Answer 7

simplify \((y(y^2+2)^{1/2})^2\) by using the rule \((x^a)^b=x^{ab}\)

Answer 8

alongside the rule \((xy)^a=x^ay^a\)

Answer 9

okay so i use that rule with the (y^2+2)^1/2 first?

Answer 10

the order doesn't matter

Answer 11

\[(y(y^2+2)^{1/2})^2=(y)^2[(y^2+1)^{1/2}]^2=?\]

Answer 12

where did the 1 come from?

Answer 13

should have been a 2, sorry

Answer 14

I just distributed the exponent

Answer 15

\[(y(y^2+2)^{1/2})^2=(y)^2[(y^2+2)^{1/2}]^2=?\]

Answer 16

oh okay lol

Answer 17

so i would get (y)^2 [(y^2+2)]^5/2?

Answer 18

not quite\[[(y+2)^a]^b=(y+2)^{ab}\]what are a and b in your case?

Answer 19

they are 1/2 and 2, are they suppose to multiply?

Answer 20

yes