Calculate the length of the arc of # y = 2x-x^2 # for #x in [0,2]?

1 Answer
Oct 19, 2017

Arc Length #~~ 2.970 # to 3dp

Explanation:

Arc Length is given by:

# L = int_alpha^beta \ sqrt(1+(dy/dx)^2) \ dx #

We have:

# y = 2x-x^2 #

Differentiating wrt #x# we have:

# dy/dx = 2-2x #

So the Required Arc Length, is given by:

# L = int_0^2 \ sqrt( 1 + (2-2x)^2 ) \ dx #
# \ \ = int_0^2 \ sqrt( 1 + 4-8x+4x^2 ) \ dx #
# \ \ = int_0^2 \ sqrt( 4x^2-8x+5 ) \ dx #

Now, let us estimate this integral numerically by using the Trapezium rules with #10# sub-intervals working to 6dp. The values of the function are tabulated as follows;

Steve M using Excel

# A = 0.2/2 * { 2.236068 + 2.236068 + #
# \ \ \ \ \ \ \ \ \ 2*(1.886796 + 1.56205 + 1.280625 + 1.077033 + #
# \ \ \ \ \ \ \ \ \ 1 + 1.077033 + 1.280625 + 1.56205 + 1.886796) } #
# \ \ \ = 0.1 * { 4.472136 + 2*(12.613008) } #
# \ \ \ = 0.1 * { 4.472136 + 25.226016 } #
# \ \ \ = 0.1 * 29.698152 #
# \ \ \ = 2.969815 #

So we estimate that:

# L = 2.970 # to 3dp

Actual Value

As it happens this integral can be evaluated and we find that

# L = 1/2(sinh^(-1)(2)+2sqrt(5)) = 2.957885715089195 #