The Asteroid (implicit differentiation)

[TicTacToe]

Homework Statement

Calculate the length of the graph/equation: x^(2/3) + y^(2/3) = a^(2/3)

The graph is formed as an s.c. asteroid, almost like a diamond. It seems to be some sort of modified unit circle.

Homework Equations

The length of the graph between x₁ and x₂ can be described as L=∫ √(1+(y´(x))²) dx

Might as well throw in the Pythagorian theorem: a² + b² = c²

The Attempt at a Solution

The problem can reformed this way.

x^(2/3) + y^(2/3) - a^(2/3) = 0

Implicit differentiate the LHS with respect to x (dy/dx)=

(2x^(-1/3))/3 + (2y^(-1/3))/3 * y´ = 0 (I thought of a as a constant)

y´ = - ((2x^(-1/3))/3) / ((2y^(-1/3))/3) which leads to

y´= - y^(1/3) / x^(1/3)

Now, I'm not quite sure from here and onwards.

y = (-x^(2/3) + a^(2/3))^(3/2) <--- is this correct? if so then the length of the graph could be described as

L = ∫ √(1+(-((-x^(2/3)+a^(2/3))^(3/2))^(1/3) / x^(1/3)))²

but I can't really figure this out. I mean, I still have 2 variables and it's already messy as it is. The uncertainty lies in what y should be written as. So my direct question is, could anybody solve out y from this equation and make it possible to integrate the L-equation? Thanks in advance.

 

[LCKurtz]

Homework Statement

Calculate the length of the graph/equation: x^(2/3) + y^(2/3) = a^(2/3)

The graph is formed as an s.c. asteroid, almost like a diamond. It seems to be some sort of modified unit circle.

Homework Equations

The length of the graph between x₁ and x₂ can be described as L=∫ √(1+(y´(x))²) dx

Might as well throw in the Pythagorian theorem: a² + b² = c²

The Attempt at a Solution

The problem can reformed this way.

x^(2/3) + y^(2/3) - a^(2/3) = 0

Implicit differentiate the LHS with respect to x (dy/dx)=

(2x^(-1/3))/3 + (2y^(-1/3))/3 * y´ = 0 (I thought of a as a constant)

y´ = - ((2x^(-1/3))/3) / ((2y^(-1/3))/3) which leads to

y´= - y^(1/3) / x^(1/3)

Now, I'm not quite sure from here and onwards.

y = (-x^(2/3) + a^(2/3))^(3/2) <--- is this correct? if so then the length of the graph could be described as

L = ∫ √(1+(-((-x^(2/3)+a^(2/3))^(3/2))^(1/3) / x^(1/3)))²

but I can't really figure this out. I mean, I still have 2 variables and it's already messy as it is. The uncertainty lies in what y should be written as. So my direct question is, could anybody solve out y from this equation and make it possible to integrate the L-equation? Thanks in advance.

It would be much more readable if you use the X² button or simple Tex. You are on the right track you have

[tex]y' = -\frac{y^{\frac 1 3}}{x^{\frac 1 3}}[/tex]

so

[tex]y'^2 = \frac{y^{\frac 2 3}}{x^{\frac 2 3}} =
\frac{a^{\frac 2 3}-x^{\frac 2 3}}
{x^{\frac 2 3}}[/tex]

Now simplify 1 + y'² by combining the two terms and you might be surprised.

 

[TicTacToe]

Thanks LCKurtz!

Now I get the formula:

√ (a^(2/3)/x^(2/3)) dx = The length of the curve between two x-values

I guess I should integrate this next. I'll do it tomorrow and if I encounter any problems I'll ask for further guidance.

Can't really get the hang of those TEX stuff

 

[TicTacToe]

Alright, I might have the answer to this now but I'm not quite sure. I have integrated the expression:

L = ∫ √ (a^(2/3)/x^(2/3)) dx =

= ∫ a^(1/3)/x^(1/3) dx

to

[ ∛a * (3/2) * x^(2/3) ]

Known by the graph on my analog paper a is represented as the max value of x and y, both positive and negative. In the quadrangular diamond (the asteroid) a is represented as the max value of the four directions to further explain. So therefore I set x₁ to -a and x₂ as 0 and calculated the equation given above. I received the answer

L = -(3/2)*a

So I assumed length of the whole diamond is 4*(3/2)*a since length in this case can't be described as negative (?) and the diamond is a quadrangular where each curve is separated by the x and y angles and the curves are identical.

Does this seem to make sense? That the length of each curve is the 3/2 times longer than the max x and y value (which are the same value as a). I must admit, I am a bit skeptical to the "length can't be described as negative"-part. All help and input is greatly appreciated and everyone that helps other people in here are true heroes.

 

[Dickfore]

It's called ASTROID, not ASTEROID.

 

[TicTacToe]

It's called ASTROID, not ASTEROID.

Aight, thanks but do you have any else that might help me with the problem?

 

[Dickfore]

Yes, Google astroid. There is a formula for the perimeter of this curve.

 

[TicTacToe]

Could someone give me a hint how the formula

s(θ) = (8*(a-b)*b*(sin (aθ/4b))^2)/a

is derived. s is the arc length of the astroid's curve.

where θ is the angle of the circle with (a-b) radius.

Thanks in advance!

the formula can be find at http://mathworld.wolfram.com/Hypocycloid.html

 

[TicTacToe]

Well, I'll try with a more simple question then, I guess. All hypocycloid can be described as a smaller circle rotating inside a larger, with a point P on the rim of the circle creating the curve. Why is then the smaller circle radius (b) * the angle in the small circle (ϑ) the same as the big circle's radius minus the small circle's radius (a-b) * the angle of the big circle (Φ)?

Simply, why is (a-b)Φ=bϑ ?

I can't seem to find any logical explanations. Do they develop the same arc length?

I really wish someone could help me with this one.

Thanks in advance.