Question

How do you find the length of the curve `y=(2x+1)^(3/2), 0<=x<=2`?

Answer 1

Length is 10.38 units. See details below

Explanation:

The length of a curve between a and b values for x is given by

`L=int_a^bsqrt(1+y´^2)dx`

`y´^2=18x+9`

`L=int_0^2sqrt(18x+10)dx=1/27(18x+10)^(3/2)]_0^2=10.38`

Answer 2

`1/27{46sqrt(46)-10sqrt(10)} ~~ 10.384 \ (3dp)`

Explanation:

The Arc Length of a curve `y=f(x)` from `x=a` to `x=b` is given by:

`L = int_a^b \ sqrt(1+(dy/dx)^2) \ dx`

Sop, for the given curve `y=(2x+1)^(3/2)` for `x in [0,2`, we form the derivative using the power rule for differentiation in conjunction with the chain rule:

`dy/dx = (3/2)(2x+1)^(3/2-1)(2)`

`\ \ \ \ \ \ = 3(2x+1)^(1/2)`

So then, the arc length is:

`L = int_0^2 \ sqrt(1+(3(2x+1)^(1/2))^2 ) \ dx`

`\ \ = int_0^2 \ sqrt(1+9(2x+1) ) \ dx`

`\ \ = int_0^2 \ sqrt(1+18x+9 ) \ dx`

`\ \ = int_0^2 \ sqrt(18x+10 ) \ dx`

And using the power rule for integration, we can integrate to get:

`L = [((18x+10 )^(3/2) )/(3/2) * 1/18]_0^2`

`\ \ = [1/27(18x+10 )^(3/2)]_0^2`

`\ \ = 1/27{(36+10)^(3/2)-10^(3/2)}`

`\ \ = 1/27{46sqrt(46)-10sqrt(10)}`

`\ \ ~~ 10.384 \ (3dp)`