Arc Lengths

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October 22nd 2009, 12:14 AM
superman69

1 Attachment(s)

Arc Lengths

Find the length of the curve defined by http://hosted.webwork.rochester.edu/...6934f23841.png from http://hosted.webwork.rochester.edu/...4e4e3fa271.png to http://hosted.webwork.rochester.edu/...655a5ac7d1.png.

Ok so first I know that we have to take the derivative which is Attachment 13472 And What would I do from then?
October 22nd 2009, 12:17 AM
mr fantastic

Quote:

Originally Posted by superman69

Find the length of the curve defined by http://hosted.webwork.rochester.edu/...6934f23841.png from http://hosted.webwork.rochester.edu/...4e4e3fa271.png to http://hosted.webwork.rochester.edu/...655a5ac7d1.png.

Ok so first I know that we have to take the derivative which is Attachment 13472 And What would I do from then?

There are numerical errors in your derivative. After you have fixed them you should simplify the result, substitute it into the arclength formula (which I assume you have been taught) and then do the resulting integration.
October 22nd 2009, 12:20 AM
superman69

1 Attachment(s)

Quote:

Originally Posted by mr fantastic

There are numerical errors in your derivative. After you have fixed them you should simplify the result, substitute it into the arclength formula (which I assume you have been taught) and then do the resulting integration.

Yes. But is this how you would set the equation Attachment 13473
October 22nd 2009, 12:30 AM
superman69

Oh I see the mistake, 9 were suppose to be a 4
October 22nd 2009, 12:36 AM
The Second Solution

Quote:

Originally Posted by superman69

Yes. But is this how you would set the equation Attachment 13473

Once you have made the necessary correction, the thing you are square rooting will simplify to $\left(\frac{x^2 + 16}{x^2 - 16}\right)^2$ .
October 22nd 2009, 12:38 AM
superman69

1 Attachment(s)

Quote:

Originally Posted by The Second Solution

Once you have made the necessary correction, the thing you are square rooting will simplify to $\left(\frac{x^2 + 16}{x^2 - 16}\right)^2$ .

So this was what I got so far but I don't seem sure if this is correct
Attachment 13474
October 22nd 2009, 01:59 AM
mr fantastic

Quote:

Originally Posted by superman69

So this was what I got so far but I don't seem sure if this is correct
Attachment 13474

Look, you need to get the derivative correct first. Please post your simplified answer for it and show all your working for how you got it.

Then ..... post #5 tells you what the $1 + \left( \frac{dy}{dx}\right)^2$ will simplify to.