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# 2011 Calculus BC free response #1d

Arc length for a curve. Created by Sal Khan.

## Want to join the conversation?

• It would be awesome if there was a video for just arc length in Calculus.
• Can we get an example of how to find the arc length without using a calculator, (except maybe for basic number crunching?)
• Wouldn't you just evaluate the integral by hand?
• Just to clarify, if they had asked displacement, would one find the distance between points (0, -4) and (21, -3.226)? When I first saw the question, this is what I thought.
• You have to be careful to distinguish between the distance (just the length of the straight line between two points) and the length of a traced curve between two points (arc length)... It also was my first thought!
• I thought arc length was the square root of 1 + (dy/dx)^2 at . Why is it the pythag and not that?
• If you could convert the parameterized equations he is using into an equation for y in terms of x then that formula would be helpful ...

What we are being shown by Sal is how to derive the 'arc length formula' ... if you watch from to , we are shown that:
`ds = √( (dx)² + (dy)² )`

divide both sides by dx:
`ds/dx = √( (dx)² + (dy)² ) / dx`

move the dx inside the square root, and then distribute it:
`ds/dx = √( (dx)² + (dy)² / (dx)² )`
`ds/dx = √( (dx/dx)² + (dy/dx)² )`

simplify & rearrange:
`ds/dx = √( (1)² + (dy/dx)² )`
`ds/dx = √( 1 + (dy/dx)² )`
`ds = √( 1 + (dy/dx)² ) dx`

... when integrated this gives you the arc length ...

You may also find this helpful:
http://en.wikipedia.org/wiki/Arc_length

HTH
• why is Sal finding the arc length (the small piece) when he doesn't know how the actual graph looks like? The actual might not even have the arc shape, rather it can be a straight line connecting from (0,=4) to (21,-3.226)
• Why is the distance the arc length? I haven't worked too much with parametric equations and usually the total distance is the absolute value of the integral of the velocity function with respects to x.
(1 vote)
• But isn't the distance almost the same as the displacement from 0 to 21
that is the same value ?
(1 vote)
• I thought the length formula was the square root of 1+dx/dt^2 + dydt^2
• Maybe it makes more sense if you think of the nonparametric equation as a special case of the parametric one.

This is the parametric version.
sqrt(dx/dt^2 + dy/dt^2)dt

If you parametrise in terms of x instead of t, as the nonparametric, regular version does, you get
sqrt(dx/dx^2 + dy/dx^2)dx
which of course simplifies to
sqrt(1 + dy/dx^2)dx