# Thread: Find the length of the curve

1. ## [SOLVED] Find the length of the curve

Find the exact length of the curve defined by 0= x^3 - 8xy + 3 from (1,(2/6)) to (3, (10/6)) .

How is this problem to be approached. Do I turn the given into a linear equation and use the given x values in a definate integral?
Not sure what to do.

2. ## Re: Find the length of the curve

The length of the curve is found as $\displaystyle L = \int_a^b\sqrt{1+\left(\frac{dy}{dx}\right)^2}$

Notice your equation becomes $\displaystyle y = \frac{1}{8}\left(x^2+\frac{3}{x}\right)$

3. ## Re: Find the length of the curve

Ah. So in order to find this length, you would take the derivative of the rearranged equation and plug it into the formula for "L" yes? That much I understand. My concern lies determining a and b using the givens.

4. ## Re: Find the length of the curve

Originally Posted by andvaka
Ah. So in order to find this length, you would take the derivative of the rearranged equation and plug it into the formula for "L" yes? That much I understand. My concern lies determining a and b using the givens.
$L = \int_a^b \sqrt{1 + \left(\frac{dy}{dx}\right)^2} \, dx$

$a$ and $b$ are x-values of the curve's end points

$L = \int_c^d \sqrt{1 + \left(\frac{dx}{dy}\right)^2} \, dy$

$c$ and $d$ are y-values of the curve's end points

5. ## Re: Find the length of the curve

Originally Posted by andvaka
Find the exact length of the curve defined by 0= x^3 - 8xy + 3 from (1,(2/3)) to (3, (14/3)).
Those points do not lie on that curve, so I really don't understand the question.

6. ## Re: Find the length of the curve

Originally Posted by andvaka
Find the exact length of the curve defined by 0= x^3 - 8xy + 3 from (1,(2/3)) to (3, (14/3)) .

How is this problem to be approached. Do I turn the given into a linear equation and use the given x values in a definate integral?
Not sure what to do.

I recommend you go back and read the problem again. As reckoner said, the points you give, (1,(2/3)) and (3, (14/3)) are NOT on the given curve, $x^3 - 8xy + 3= 0$. If x= 1 and y= 2/3, then $x^3- 8xy+ 3= 1- 16/3+ 3= -4/3$, not 0. And if x= 3, y= 14/3, then $x^3- xy+ 3= 27- 112+ 3= -82$, not 0.
Have you found $\frac{dy}{dx}$ ?