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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


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    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:
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