Calculating Length of Line from y=1 to y=3

[chewy]

Homework Statement

finding the lengths of a line

Homework Equations

x = (y^3/3) + 1/(4y) from y =1 to y=3
hint:: 1 + (dx/dy)^2 is a perfect square

The Attempt at a Solution

I found the solution, i did this by just finding the derivative and then putting it into the equation for finding a line and then messing around with it until i could get rid of the sqaure root in the equation, as usual.
what i don't understand is the hint, what is the perfect square part all about?? what does it mean?? how does it help??

thks
jason

 

[arildno]

DO post what you did, because if you did it CORRECTLY, then you'd immediately see the relevance of perfect squareness!

 

[Google_Spider]

finding the lengths of a line

Find the length of which line ? The equation you have posted will graph out to be a complex curve!

 

[chewy]

ok, i found the derivative first which was y^2 - 1/(4y^2), i then squared this which gave me y^4 + 1/(16y^4) - 1/2, i then stuck this in the formula and added the one, which gave y^4 +1/(16y^4) + 1/2, i then put the terms together (16y^8 + 8y^4 + 1)/16y^4, i then seen the numerator is the a perfect square (4y^4 + 1)^2/16y^4, which enabled me to get rid of the square root and integrate (4y^4 + !)/4y^2 with no probs. what i don't understand is how 1 + (dx/dy)^2 is a perfect square and where it helps?

 

[Google_Spider]

>>i then stuck this in the formula

Which formula ?

 

[chewy]

The formula for length of a line y = f(x)

Length = int [sqrt(1+ (dx/dy)^2)] dx

in this case the variable is y,

 

[chewy]

Find the length of which line ? The equation you have posted will graph out to be a complex curve!

the line will be continuous from y =1 to y= 3, not quite sure what u mean, there is only one line x = (y^3/3) + 1/(4y).

 

[chewy]

DO post what you did, because if you did it CORRECTLY, then you'd immediately see the relevance of perfect squareness!

ok, i found the derivative first which was y^2 - 1/(4y^2), i then squared this which gave me y^4 + 1/(16y^4) - 1/2, i then stuck this in the formula and added the one, which gave y^4 +1/(16y^4) + 1/2, i then put the terms together (16y^8 + 8y^4 + 1)/16y^4, i then seen the numerator is the a perfect square (4y^4 + 1)^2/16y^4, which enabled me to get rid of the square root and integrate (4y^4 + !)/4y^2 with no probs. what i don't understand is how 1 + (dx/dy)^2 is a perfect square and where it helps?

 

[arildno]

i then seen the numerator is the a perfect square

Well, not all students would have seen this by themselves.
You did, so you didn't need the hint. Other students would have needed the hint to see what you saw on your own.

 

[chewy]

Well, not all students would have seen this by themselves.
You did, so you didn't need the hint. Other students would have needed the hint to see what you saw on your own.

ok i see, thks, but I am still confused on how 1 + (dy/dx)^2 is a perfect square, doesn't there need to be a third term??

 

[Dick]

ok i see, thks, but I am still confused on how 1 + (dy/dx)^2 is a perfect square, doesn't there need to be a third term??

You just showed that 1+(dy/dx)^2 is a perfect square in this very special case. They aren't claiming that it is ALWAYS an algebraic perfect square. It definitely isn't.

 

[chewy]

You just showed that 1+(dy/dx)^2 is a perfect square in this very special case. They aren't claiming that it is ALWAYS an algebraic perfect square. It definitely isn't.

ahh, i see, i was taking it too literaly, thks very much