# Find the arc length of the graph of the function over the indicated interval: y =2/3(x − 1)^3/2, 1 ≤ x ≤ 7

Hello!

The formula for the arc length for the graph of a differentiable function `f ( x ) ` is

`L = int_a^b sqrt ( 1 + ( f ' ( x ) )^2 ) dx .`

This formula is derived using chord length as an approximation of a small...

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Hello!

The formula for the arc length for the graph of a differentiable function `f ( x ) ` is

`L = int_a^b sqrt ( 1 + ( f ' ( x ) )^2 ) dx .`

This formula is derived using chord length as an approximation of a small curve part length. When the maximum length of the parts tends to zero (and the number of intervals tends to infinity), this integral arises.

It is interesting that a similar approach does not work in general for a surface area. There exists an example of a good enough surface such that sums of areas of triangles under the surface tend to infinity. To get proper approximation, pieces of tangent planes are used instead.

Here `a = 1 , ` `b = 7 , ` `f ( x ) = 2 / 3 ( x - 1 )^( 3 / 2 ), ` `f ' ( x ) = 2 / 3 * 3 / 2 ( x - 1 )^( 1 / 2 ) = ( x - 1 )^( 1 / 2 ) .`

This way, the integral is

`int_1^7 sqrt ( 1 + ( x - 1 ) ) dx = int_1^7 sqrt ( x ) dx = ( 2 / 3 x^( 3 / 2 ) )_(x=1)^7 = 2 / 3 ( 7 sqrt ( 7 ) - 1 ) . `

It is approximately 11.68 linear units.

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