Finding the length of a curve

  • Thread starter Thread starter rappa
  • Start date Start date
  • Tags Tags
    Curve Length
Click For Summary
The discussion focuses on finding the length of the curve defined by the equation 24xy = x^4 + 48 between the points (2, 4/3) and (3, 43/24), with the expected answer being 9/8. Participants analyze the derivative dy/dx and the integration process required to calculate the curve length using the formula involving the square root of 1 plus the square of the derivative. There is confusion regarding the correct application of the integration steps and the proper handling of square roots in the calculations. The conversation emphasizes the importance of accurately squaring the derivative and integrating directly from the square root expression to achieve the correct result. The thread concludes with a call for clarity in mathematical operations to avoid errors in the integration process.
rappa
Messages
8
Reaction score
0
Question: Find the length of the curve 24xy = x^4 + 48 between (2, 4/3) and (3, 43/24). Answer is supposed to be 9/8.

This is the formula (I think): square root(1 + (dy/dx)^2 dx)

I tried to integrate and I did it like this:

dy/dx = 3/24 x^2 - 2x^-2 or 3/24 x^2 - 2/x^2

then: square root(1 + (3/24 x^2 -2x^-2)^2)

= square root(1 + (1/64 x^4 - 4x^-4))

now integration:

(1 + (1/64 x^4 - 4x^-4))^(1/2)

= (2/3) (1 + (1/64 x^4 - 4x^-4))^(3/2) (x + (1/320 x^5 + 4/3 x^-3)

That's how far I got, I think it's wrong. If anyone could help me out, I would appreciate it. Thanks.
 
Last edited:
Physics news on Phys.org
How did you come up with your y'
 
24xy = x^4 + 48

y = (x^4 + 48) / 24x

y = 1/24 x^3 + 2/x or 2x^-1

dy/dx = 3/24 x^2 - 2x^-2
 
Now show me how you do this:

then: square root(1 + (3/24 x^2 -2x^2)^2)

= square root(1 + (1/64 x^4 - 4x^-4))

[You probably mean 1 + (3/24 x^2 - 2/x^2)^2)]
 
the 1+ part isn't squared, just the inside bracket. I just left the square root part, but I only squared the inside.
 
Last edited:
My post does not indicate that it is, does it?

My first point was that you wrote 2x^2 whereas you meant to write 2/x^2 or 2x^(-2)

My second point is show me how (3/24 x^2 - 2/x^2)^2 = 1/64 x^4 - 4x^(-4)
 
is it: 1/64 x^4 - 1/2 + 4/x^4 ?
 
That looks better
 
ok so now I have:

(2/3) (1 + (1/64 x^4 - 1/2 + 4/x^4))^(3/2) (x + (1/320 x^5 - 1/2x - 4/3 x^-3)
 
  • #10
I have no idea how you got that, you have to show more steps, for example next step would be to show what happens when you add 1 to what you had and take sqrt
 
  • #11
We can't add the one, and we can't just take the square root either. Have to just integrate it straight from the square root and plug in the numbers to get the length.

This is what I did for the integration

(1/64 x^4 - 1/2 + 4/x^4)^(1/2)

= (2/3) (1 + (1/64 x^4 - 1/2 + 4/x^4))^(3/2) (x + (1/320 x^5 - 1/2x - 4/3 x^-3)
 
  • #12
...why not?

The steps are:

1) Take the derivative
2) Square it
3) add 1 to it
4) take square root of that
5) integrate
 
  • #13
Ok here goes, I hope that works...

(1 + (1/64 x^4 - 1/2 + 4/x^4)

65/64 x^4 + 1/2 + 5/x^4

so now square root of that:

root(65/65) x^2 + root (1/2) + root (5) / x^2


that looks bad, that's why I thought to do it the other way:

(1/64 x^4 - 1/2 + 4/x^4)^(1/2)

integrates to:

(2/3) (1 + (1/64 x^4 - 1/2 + 4/x^4))^(3/2) (x + (1/320 x^5 - 1/2x - 4/3 x^-3)
 
  • #14
You should know from algeba that sqrt(a+b) is NOT sqrt(a) + sqrt(b), stop thinking that.

To work with what you have, find the common denominator and then use the property that sqrt(a/b) DOES equal sqrt(a)/sqrt(b)
 

Similar threads

  • · Replies 10 ·
Replies
10
Views
2K
Replies
2
Views
1K
Replies
4
Views
2K
Replies
12
Views
2K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 6 ·
Replies
6
Views
2K
  • · Replies 9 ·
Replies
9
Views
2K
Replies
15
Views
2K
  • · Replies 105 ·
4
Replies
105
Views
6K
  • · Replies 25 ·
Replies
25
Views
2K