Question

Find the length of the curve: y = 1+2x^(3/2)

Answer 1

\[0 \le x \le 1\]

Answer 2

L = \[\int\limits_{0}^{1} \sqrt{1 + \frac{ dy }{ dx }}^{2} dx\]

Answer 3

^yep

Answer 4

2.268 would be the length when you integrate from 0 to 1

Answer 5

Well, I to L = \[\int\limits_{0}^{1}\sqrt{1+(3x ^{1/2})}\] And I'm stuck there now

Answer 6

it'll be sqrt(1+3x)

Answer 7

So you are integrating from 0 to 1 of sqrt(1+9x). you can get this if you simplify the equation.

Answer 8

it'll be sqrt(1+9x)

Answer 9

substitute (1+9x)= t^2

Answer 10

does it help u?

Answer 11

I'm not sure how you get from sqrt(1+3x^(1/2)) to 1+9x

Answer 12

Its 3x^(3/2)

Answer 13

actually it's integral of sqrt( 1+ (dy/dx)^2)

Answer 14

OH I forgot about the square

Answer 15

and dy/dx= 3 sqrt(x)

Answer 16

When you square both sides you get the equation that I wrote above

Answer 17

Okay, I have L = \[\int\limits_{0}^{1} \sqrt{1+9x}dx\]

Answer 18

yes...now substitute 1+9x= t^2

Answer 19

or u can write directly the integral of (1+9x)^1/2 = {(1+9x)^(1/2+1)}/9

Answer 20

if u do the substitution then u'll have to change the limit of integration also.

Answer 21

so now u can solve by ur own?