Length of curve proble please verify my answer the curve is y=(9-x^(2/3))^(3/2) x=1 to x=27?

Question

My result is length = 36 is this correct?

I worked on this more and now believe length = 24 vice 36 am I correct?

Answer 1

Answer length= 36 or 24 seem be doubtful. For asteroid
x2/3 + y2/3 = a2/3 Whole length of 4branches = 6*a
Your curve: x(2/3) + y(2/3)= 9=27^(2/3). In your case, a=27 so whole length = 6*27= 162

HOW???? For each value of x=1, there are 2-values of y. and only in case of x=27 there is 1=value of y=0. So curve is symetrical about x-axis and y-axis. See the slope of curve, at x=27, is zero, and at x=0 slope=infinite. (extend curve, slope becomes x-axis and y-axis.

Length= ∫ sqrt[1+(dy/dx)^2] dx = ∫ 3 [x^(-1/3] dx -----(1)

Whole length = 2*length (from x= -27 to x=27)
……..x= 27………………………..27
= 2 * ∫ 3 [x^(-1/3)] dx = 9[x^(2/3)] ……………(2)
……x= - 27……………………..- 27

Property of definite integral
……..x= 27
= 9[x^2/3] =
……x= - 27
.............-1............0
= 9[x^2/3]+9[x^2/3] +
...........-27...........-1
..........1...........27
9[x^2/3]+9[x^2/3]
..........0............1
=9[1 - 9]+(9)[0 - 1]+9[1 - 0]+.9[9 - 1]
= [72]+[9]+[9][72] = 162

Whenever you are asked to find length of curve between x=1 to x=27, you have to see whether for given x, the value of y ONE or many (say 2). If y has 2 values then it will have symmetry about x-axis, and length or area will be automatically doubled. i.e. 36*2 = 72. I bet 36 answer will be treated as WRONG, and zero marks will be given because the mistake had taken place right at the ONSET of question. So
………..x= 27
Length= Limit = 9[9-1]=72
…………x= 1

In changing answer from L=36 earlier to L=24 (though both wrong for given limits), you must have mad mistake of omitting Factor “3” (1) or factor (2/3) in denominator of integral. Explanation of equ (2): when curve is symmetrical about both axes, then for calculation of arc length we use

X = 2x, Y=2y and then ds^2= (dX)^2 + d(Y)^2 = 4[dx^2+dy^2]
length ds = 2 sqrt (dx^2+dy^2) that is why length is DOUBLED in equation (2)

Settle my both questions: enough of ASTROID on my time.

Answer 2

Yes, you are correct. Infinitesimal length of the curve ds = sqrt(dx^2 + dy^2). In your case:

dy = (3/2)*(9 - x^(2/3))^(1/2)*(-2/3)*x^(-1/3) dx
= - (9 - x^(2/3))^(1/2)*x^(-1/3)*dx.

dy^2 = (9 - x^(2/3)*x^(-2/3)*dx^2
= [9*x^(-2/3) - 1]*dx^2.

Therefore: ds^2 = dy^2 + dx^2 = 9*x^(-2/3)*dx^2, and
ds = 3*x^(-1/3)*dx. The length of your curve is:

s = int[from 1 to 27](3*x(^-1/3)*dx)
= 3*(3/2)*x^(2/3) [from 1 to 27]
= (9/2)*[9 - 1] = 36.

Answer 3

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