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

2 Answers
F. Javier B.
Apr 29, 2018

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#

Steve M
Apr 29, 2018

# 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)#

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