Length of C

**Undefdisfigure** · March 24th 2008, 10:28 PM

I'm not too bad on this section of the book but this question has got me stumped.

Let C be the curve of intersection of the parabolic cylinder x^2 = 2y and the surface 3z = xy. Find the exact length of C from the origin to the point (6, 18, 36).

I can't remember where in my notes we tackled a problem like this.

**mr fantastic** · March 25th 2008, 03:24 AM

Originally Posted by Undefdisfigure

I'm not too bad on this section of the book but this question has got me stumped.

Let C be the curve of intersection of the parabolic cylinder x^2 = 2y and the surface 3z = xy. Find the exact length of C from the origin to the point (6, 18, 36).

I can't remember where in my notes we tackled a problem like this.

The curve can be defined parametrically as:

$x = t$

$y = \frac{t^2}{2}$

$z = \frac{xy}{3} = \frac{t^3}{6}$

The point (0, 0, 0) corresponds to t = 0. The point (6, 18, 36) corresponds to t = 6.

And you know how to get the arc length of a curve defined parametrically using the formula

$L = \int_{t = a}^{t = b} \sqrt{\left( \frac{dx}{dt} \right)^2 + \left( \frac{dy}{dt} \right)^2 + \left( \frac{dz}{dt} \right)^2} \, dt$ ,

right?

ps: You might find the following identity useful: $t^4 + 4t^2 + 4 = (t^2 + 2)^2$ .....

**snaes** · September 18th 2009, 02:14 PM

Hey could you show me/get me started on how you transformed the two equations to one parametric equation? It would be greatly appreciated

Thanks!

**mr fantastic** · September 18th 2009, 02:53 PM

Originally Posted by snaes

Hey could you show me/get me started on how you transformed the two equations to one parametric equation? It would be greatly appreciated

Thanks!

Since z is defined in etrms of x and y, you only need to think about how to define the curve x^2 = 2y parametrically. There ar an infinite number of choices. I made the simplest one: let x = t. If x = t, what does y have to equal?

**snaes** · September 18th 2009, 04:22 PM

oh ok that how...thanks!

**Peter1765** · October 1st 2009, 12:30 AM

Mr. Fantastic i followed the steps you provided and my end result was 1/3 +2 which i converted to 7/3 however thats still wrong why's that?

and as to how i got to that answer
______________________
L = integral from 0-1 ( t^2 +2 t)^2
which when integrated gives you 1/3t^3 + 2t (limit form 0-1)

... in the end my answer is 7/3 ... how's that wrong?

**mr fantastic** · October 1st 2009, 01:59 AM

Originally Posted by Peter1765

Mr. Fantastic i followed the steps you provided and my end result was 1/3 +2 which i converted to 7/3 however thats still wrong why's that?

and as to how i got to that answer
______________________
L = integral from 0-1 ( t^2 +2 t)^2
which when integrated gives you 1/3t^3 + 2t (limit form 0-1)

... in the end my answer is 7/3 ... how's that wrong?

The correct integral (after simplifying) is ${\color{red}\frac{1}{2}} \int_0^{\color{red}6} t^2 + 2 \, dt$ .

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