Find the length of the curve

**andvaka** · June 24th 2012, 01:37 PM

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.

Your guidance is appreciated.

**pickslides** · June 24th 2012, 01:48 PM

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

**andvaka** · June 24th 2012, 01:58 PM

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.

**skeeter** · June 24th 2012, 02:18 PM

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

**Reckoner** · June 24th 2012, 02:23 PM

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.

**HallsofIvy** · June 24th 2012, 06:34 PM

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.

Your guidance is appreciated.

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.

**andvaka** · June 24th 2012, 09:11 PM

I can read the problem again, but that wouldn't bring me any close to a solution would it? The problem is exactly as I stated it. I'm just trying the most constructive way to approach this. I suppose I will have to ask my calculus professor to elaborate so that I know what it is she wants me to do here.

**pickslides** · June 24th 2012, 09:21 PM

Have you found $\frac{dy}{dx}$ ?

I think you should ask your professor if these points are supposed to be on the curve? You can illustrate they are not.

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