Length of the Curve...CALCULUS HELP!
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Length of the Curve...CALCULUS HELP!

[From: ] [author: ] [Date: 12-04-20] [Hit: ]
y^3 = (9/4)x^2 ==> y = (9/4)^(1/3)x^(2/3).The former equation is the easiest to work with,x = (2/3)y^(3/2).With x = (2/3)y^(3/2),dx/dy = (2/3)(3/2)y^(1/2) = √y ==> (dx/dy)^2 = y.y varies from 0 to 3,......
What is the length of the curve of 9x^2 = 4y^3 from (0, 0) to (2√3, 3)

I can't figure this one out. I know the answer is 14/3 but I do not know how to get that.

Thanks!

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We can solve 9x^2 = 4y^3 for either x or y. If we solve this equation for x, we have:
x^2 = (4/9)y^3 ==> x = (2/3)y^(3/2).
(Note that the positive square root is taken since x > 0 throughout the entire path.)

Alternatively, solving for y yields:
y^3 = (9/4)x^2 ==> y = (9/4)^(1/3)x^(2/3).

The former equation is the easiest to work with, so we will use:
x = (2/3)y^(3/2).

With x = (2/3)y^(3/2), we see that:
dx/dy = (2/3)(3/2)y^(1/2) = √y ==> (dx/dy)^2 = y.

y varies from 0 to 3, so the required arc length is:
L = ∫ √(1 + y) dy (from y=0 to 3)
= (2/3)(1 + y)^(3/2) (evaluated from y=0 to 3)
= (2/3)(8 - 1)
= 14/3, as required.

I hope this helps!

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Note that the equation of the curve can be re-arranged as:

x = 2/3*y^(3/2)

dx/dy = y^(1/2)

So therefore the length of the curve is:

∫√(1 + y)) dy from 0 to 3

2/3*(1 + y)^(3/2) eval. from 0 to 3

= 16/3 - 2/3 = 14/3
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