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

• chewy
In summary: for clearing that up, my confusion was purely academic anyways :)ahh, i see, i was taking it too literaly, thks very much for clearing that up, my confusion was purely academic anyways :)
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

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

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!

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?

>>i then stuck this in the formula

Which formula ?

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,

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).

arildno said:
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?

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.

arildno said:
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??

chewy said:
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.

Dick said:
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

## 1. What is the equation for calculating the length of a line from y=1 to y=3?

The equation for calculating the length of a line from y=1 to y=3 is √(1 + (y2-y1)^2), where y1=1 and y2=3.

## 2. How do you determine the coordinates of the endpoints when calculating the length of a line from y=1 to y=3?

The coordinates of the endpoints are the y-values given in the equation, y1=1 and y2=3.

## 3. Can you use the Pythagorean Theorem to calculate the length of a line from y=1 to y=3?

Yes, the equation used to calculate the length of a line from y=1 to y=3 is derived from the Pythagorean Theorem, a^2 + b^2 = c^2.

## 4. What is the significance of calculating the length of a line from y=1 to y=3?

Calculating the length of a line from y=1 to y=3 can help determine the distance between two points on a graph or the slope of a line.

## 5. Is it necessary to use the given formula to calculate the length of a line from y=1 to y=3?

No, there are other methods for calculating the length of a line, such as using the distance formula or using the slope and coordinates of the endpoints.

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