Another Arc Length Problem

[Jason76]

POST EDITED - 5/5/13

y = L = length of arc

evaluated at lower bound 2 and upper bound 3.

Formula: \(\displaystyle \int dy = \int\limits_a^b [1 + ( (\dfrac{dy}{dx}(x)))^{2}]^{1/2} dx\)

\(\displaystyle y = \dfrac{x^{3}}{24} + 2x^{-1}\)

\(\displaystyle \dfrac{dy}{dx} = \dfrac{x^{2}}{8} - 2x^{-2}\)

\(\displaystyle \int dy = \int\limits_2^3[1 + (\dfrac{x^{2}}{8} - 2x^{-2})^{2})]^{1/2} dx\) Note that the \(\displaystyle \dfrac{dy}{dx}\) was a perfect square.

\(\displaystyle \int dy = \int[1 + (\dfrac{x^{4}}{64} - \dfrac{1}{2} + 4x^{-4})]^{1/2} dx\) This comes from FOIL.

\(\displaystyle \int dy = \int[\dfrac{x^{4}}{64} + \dfrac{1}{2} + 4x^{-4})]^{1/2} dx\) After the 1 was added.

Now here is the main question.

\(\displaystyle \dfrac{x^{4}}{64} + \dfrac{1}{2} = 4x^{-4} = x^{-4}(\dfrac{x^{8}}{64} + \dfrac{x^{4}}{2} + 4) = x^{-4}(\dfrac{x^{4}}{ 8} + 2)^{2}\)

= \(\displaystyle (x^{-2})^{2}(\dfrac{x^{4}}{8} + 2)^{2} = (\dfrac{x^{2}}{8} + 2x^{-2})^{2}\) What is going on this line and the one above it?

\(\displaystyle \int dy = \int[\dfrac{x^{2}}{8} + 2x^{-2})^{2} ]^{1/2} dx\)

\(\displaystyle \int dy = \int(\dfrac{x^{2}}{8} + 2x^{-2}) dx\)

\(\displaystyle y + C_y = \dfrac{x^{3}}{24} - 2x^{-1} + C_x\)

\(\displaystyle y + C_y - C_y = \dfrac{x^{3}}{24} - 2x^{-1} + C_x - C_y \)

\(\displaystyle y = \dfrac{x^{3}}{24} - 2x^{-1} + C \)

Now evaluating the definite integral:

At x = 2 (lower bound)

\(\displaystyle \dfrac{(2)^{3}}{24} - 2(2)^{-1} = -\dfrac{2}{3}\)

At x = 3 (upper bound)

\(\displaystyle \dfrac{(3)^{3}}{24} - 2(3)^{-1} = -\dfrac{11}{24}\)

y= \(\displaystyle \dfrac{11}{24} - \dfrac{2}{3} = \dfrac{9}{8}\) (The length of the arc or L)

 

[MarkFL]

You should be able to show that \(\displaystyle 1+\left(\dfrac{dy}{dx} \right)^2\) is a perfect square.

 

[Jason76]

You should be able to show that \(\displaystyle 1+\left(\dfrac{dy}{dx} \right)^2\) is a perfect square.

Right, I went back and edited it and wrote "FOIL" next to the line. But my main problem is the two lines where there was some simplification (that I pointed out).

 

[MarkFL]

This is how I see it:

\(\displaystyle \dfrac{x^4}{64}+\dfrac{1}{2}+\dfrac{4}{x^4}=\dfrac{x^8+32x^4+256}{64x^4}=\left(\dfrac{x^4+16}{8x^2} \right)^2\)

and the rest follows.

 

[Deleted member 4993]

L = length of arc

evaluated at lower bound 2 and upper bound 3.

Formula: \(\displaystyle L = \int dy = \int\limits_a^b [1 + ( (\dfrac{dy}{dx}(x)))^{2}]^{1/2} dx\)

\(\displaystyle y = \dfrac{x^{3}}{24} + 2x^{-1}\)

\(\displaystyle \dfrac{dy}{dx} = \dfrac{x^{2}}{8} - 2x^{-2}\)

\(\displaystyle \int dy = \int\limits_2^3[1 + (\dfrac{x^{2}}{8} - 2x^{-2})^{2})]^{1/2} dx\) Note that the \(\displaystyle \dfrac{dy}{dx}\) was a perfect square.

\(\displaystyle \int dy = \int[1 + (\dfrac{x^{4}}{64} - \dfrac{1}{2} + 4x^{-4})]^{1/2} dx\) This comes from FOIL.

\(\displaystyle \int dy = \int[\dfrac{x^{4}}{64} + \dfrac{1}{2} + 4x^{-4})]^{1/2} dx\) After the 1 was added.

Now here is the main question.

\(\displaystyle \dfrac{x^{4}}{64} + \dfrac{1}{2} = 4x^{-4} = x^{-4}(\dfrac{x^{8}}{64} + \dfrac{x^{4}}{2} + 4) = x^{-4}(\dfrac{x^{4}}{ 8} + 2)^{2}\)

= \(\displaystyle (x^{-2})^{2}(\dfrac{x^{4}}{8} + 2)^{2} = (\dfrac{x^{2}}{8} + 2x^{-2})^{2}\) What is going on this line and the one above it?

Are you saying that you can see:

x⁴/64 - 1/2 + 4x^-4 = [x²/8 - 2x^-2]²

And you DON'T see:

x⁴/64 + 1/2 + 4x^-4 = [x²/8 + 2x^-2]²


\(\displaystyle \int dy = \int[\dfrac{x^{2}}{8} + 2x^{-2})^{2} ]^{1/2} dx\)

\(\displaystyle \int dy = \int(\dfrac{x^{2}}{8} + 2x^{-2}) dx\)

\(\displaystyle y + C_y = \dfrac{x^{3}}{24} - 2x^{-1} + C_x\)

\(\displaystyle y + C_y - C_y = \dfrac{x^{3}}{24} - 2x^{-1} + C_x - C_y \)

\(\displaystyle y = \dfrac{x^{3}}{24} - 2x^{-1} + C \)

Now evaluating the definite integral:

At x = 2 (lower bound)

\(\displaystyle \dfrac{(2)^{3}}{24} - 2(2)^{-1} = -\dfrac{2}{3}\)

At x = 3 (upper bound)

\(\displaystyle \dfrac{(3)^{3}}{24} - 2(3)^{-1} = -\dfrac{11}{24}\)

L = \(\displaystyle \dfrac{11}{24} - \dfrac{2}{3} = \dfrac{9}{8}\)

.

 

[Jason76]

.

I assume that the result from the section you have highlighted are due to fractional addition/subtraction (which I know how to do).

 

[Deleted member 4993]

I assume that the result from the section you have highlighted are due to fractional addition/subtraction (which I know how to do).

No..

I am saying that

the first equality (x⁴/64 - 1/2 + 4x^-4 = [x²/8 - 2x^-2]² )uses:

(a-b)² = a_² - 2ab + b²

and

the second equality (x⁴/64 + 1/2 + 4x^-4 = [x²/8 + 2x^-2]²) uses:

(a+b)² = a_² + 2ab + b²


Nothing to do with fraction - only change of signs.