1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: Calculus problem (integrals)

  1. Jan 4, 2005 #1
    y(x) is defined by this equation:

    [tex]\int_0^y{\sqrt[3]{1+t^2} dt} + \int_x^0{(2t+1)^2 dt} = 0[/tex]

    How do I find the tangent line to y(x) at x=0?

    Here's what I tried to do:

    [tex]y(x) = \int_0^y{\sqrt[3]{1+t^2} dt} = \int_0^x{(2t+1)^2 dt}[/tex]

    Taking the derivative of both sides of the equation:

    [tex]y'(x) = ... = (2x+1)^2 = 4x^2 + 4x + 1[/tex]
    [tex]y'(x=0) = 1[/tex]

    First, is this method correct?

    Second, how do I find y(x=0)? I'm guessing that it's 0, but I'm not sure.

  2. jcsd
  3. Jan 4, 2005 #2
    seems simple enough...

    just substitue the first and second integrals using the y and x as if they're values and equate them to eachother like you have done. find y'. may have to use implicit differentiation. then equate to 0
  4. Jan 4, 2005 #3


    User Avatar
    Science Advisor

    You say that y is defined by
    [tex]\int_0^y{\sqrt[3]{1+t^2} dt} + \int_x^0{(2t+1)^2 dt} = 0[/tex]

    It does NOT follow that
    [tex]y(x) = \int_0^y{\sqrt[3]{1+t^2} dt} = \int_0^x{(2t+1)^2 dt}[/tex]
    There is no reason for the first "y= ". Saying that y is "defined by" the formula does not mean you can just set y equal to both parts- it means that for any given value of x, y must have the value that makes the two integrals equal. In particular, when x= 0, the right hand side is 0 so y must be such that [itex]\int_0^y{\sqrt[3]{1+t^2}dt}= 0[/itex] and that means that y(0)= 0, not 1.

    You can use the "fundamental theorem of calculus" to find the derivative of y:

    The derivative of the left side is [itex]\sqrt[3]{1+ y^2}y'[/itex] while the the derivative of the right side is -(2x+1)2. In particular, at x= 0, y= 0,
    [itex]\sqrt[3]{1+0^2}y'(0)= -(2(0)+1)^2[/itex] or y'(0)= -1. The tangent line goes through (0,0) and has slope -1: y= -x.
  5. Jan 4, 2005 #4
    Thanks Ivy. :smile:
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook