Determining the Length of a Curve

Key Questions

  • How do you find the length of the curve #y=x^5/6+1/(10x^3)# between #1<=x<=2# ?

    We can find the arc length to be #1261/240# by the integral
    #L=int_1^2sqrt{1+({dy}/{dx})^2}dx#

    Let us look at some details.

    By taking the derivative,
    #{dy}/{dx}={5x^4)/6-3/{10x^4}#

    So, the integrand looks like:
    #sqrt{1+({dy}/{dx})^2}=sqrt{({5x^4)/6)^2+1/2+(3/{10x^4})^2#
    by completing the square
    #=sqrt{({5x^4)/6+3/{10x^4})^2}={5x^4)/6+3/{10x^4}#

    Now, we can evaluate the integral.
    #L=int_1^2({5x^4)/6+3/{10x^4})dx=[x^5/6-1/{10x^3}]_1^2=1261/240#

    Wataru · 11 · Sep 11 2014
  • How do you find the arc length of #x=2/3(y-1)^(3/2)# between #1<=y<=4#?

    It can be found by #L=int_0^4sqrt{1+(frac{dx}{dy})^2}dy#.

    Let us evaluate the above definite integral.

    By differentiating with respect to y,
    #frac{dx}{dy}=(y-1)^{1/2}#

    So, the integrand can be simplified as
    #sqrt{1+(frac{dx}{dy})^2}=sqrt{1+[(y-1)^{1/2}]^2}=sqrt{y}=y^{1/2}#

    Finally, we have
    #L=\int_0^4y^{1/2}dy=[frac{2}{3}y^{3/2}]_0^4=frac{2}{3}(4)^{3/2}-2/3(0)^{3/2}=16/3#

    Hence, the arc length is #16/3#.

    I hope that this helps.

    Wataru · 5 · Aug 26 2014
  • How do you find the length of a curve using integration?

    If you want to find the arc length of the graph of #y=f(x)# from #x=a# to #x=b#, then it can be found by
    #L=int_a^b sqrt{1+[f'(x)]^2}dx#

    Wataru · · Sep 11 2014

Questions

Applications of Definite Integrals