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Curve and tangent line problem, finding the area of enclosed region

  1. May 5, 2013 #1
    1. The problem statement, all variables and given/known data

    Given that the curve y = x^3 has a tangent line that passes through point (0, 2), find the area of the region enclosed by the curve and the line by the following steps.

    2. Relevant equations

    3. The attempt at a solution

    Let f(x) = x^3 and let the coordinates of the point of tangency be (t, t^3)
    f'(x) = 3x^2, f'(t) = 3t^2
    The equation of the tangent line is:
    y = 3t^2(x-t)+t^3
    = 3t^2x - 2t^3

    I am having trouble understand this part of the process. Particularly this step, y = 3t^2(x-t) + t^3. I have no idea what is happening at this step. Everything else I am completely aware of what is happening. If someone could help that would be great. Thanks!
  2. jcsd
  3. May 5, 2013 #2
    1) try to express x with respect to t, thus: y = 3t^2(x-t)+t^3 = x^3, further x = -2t

    2) put x = -2t into the tangent line and use (0, 2), then you get; y - 2 = f ' (-2t)*(x - 0) = 3*(-2t)^2*x

    then; y = 12t^2*x + 2

    3) then: y = 12t^2*x + 2 = x^3
    for a = 0,5 and x = -1
    for a = 0,5 and x = 2

    finally you can find the area...
  4. May 5, 2013 #3


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    The method works like this:
    Suppose the tangent line touches the curve at the point (t, t3). You wish to find t.
    At that point, the gradient of the curve will be 3t2, so that is the gradient of the tangent line. Hence you can write the equation of the tangent line as y = (3t2)x + c.
    This line must also pass through the point (t, t3). Substituting that for (x, y) gives you an expression for c:
    Finally, you need that to be the equation of a line that passes through the point (0, 2). What value of t achieves that?
  5. May 5, 2013 #4
    Oh I see it makes so much more sense now! Thank you
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