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Math Help - Finding the length of the curve (where it gets undefined at t=0)

  1. #1
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    Finding the length of the curve (where it gets undefined at t=0)

    Find the lenght of the curve

    f(x)= x^(1/3)+ x^(2/3), 0≤x≤2

    L= integral of square root of (1+ (dy/dx)^2 (dx) ) or (1+ (dx/dy)^2 (dy) )

    The problem is that i cannot take the inverse function of that and differentiate it.
    For example, if you are given y=x^(1/3) and asked to find the curve between (-8,-2) and (8, 2),
    1. switch x and y (reverse the x y coordinates so (-2,-8), (2,8) and you have x=y^(1/3)
    2. cube both sides x^3=y
    3, differentiate it : 3x^2=dy/dx
    4, plug that into the formula of L and evaluate it from -2 to 2.

    I cannot do the same for the given function f(x)= x^(1/3)+ x^(2/3), 0≤x≤2
    Because, after switching x and y, x=y^(1/3)+y^(2/3) +y^(2/3), even if you differentiate it, you cannot isolate y and define dy/dx only in terms of x.

    What should i do?
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  2. #2
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    I'm not sure I understand what you are doing.

    You just need to calculate \frac{dy}{dx} then use that value to solve

    \int_0^2 \sqrt{1+\left(\frac{dy}{dx}\right)^2} dx

    You do not need to solve for an inverse.
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  3. #3
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    but...

    if you just find the dy/dx of the giv en function, it gets undefined at x=0
    in the graph, the slope becomes vertical at x=0, which is why i attempted to take the inverse of the function
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  4. #4
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    f'(x)(1/3)/(x^(2/3)+(2/3)/(x^1/3))

    See how the denominator takes 0 when x=0 which is why we cannot evaluate the integral like that
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  5. #5
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    You're right, I didn't notice that.

    You can solve for an inverse function like this:

    y=x^{1/3} + x^{2/3}

    Let u=x^{1/3}

    Then: y=u+u^2 \implies u^2+u-y = 0

    Using the quadratic formula yields:

    u = \frac{-1 \pm \sqrt{1+4y}}{2} = x^{1/3}

    x = \left(\frac{-1 \pm \sqrt{1+4y}}{2}\right)^3

    We can see that the relevant portion is where x>0, so we really just want:

    x = \left(\frac{-1 + \sqrt{1+4y}}{2}\right)^3

    I'm not sure if this will lead to an answer though. There's probably some trick I'm not seeing...
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  6. #6
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    Integrals are tougher than that. It doesn't have to exist, it just has to converge.

    \int_{BadPlace}^{b}f(x) \; dx = \lim_{a \to BadPlace} \int_{a}^{b}f(x) \; dx

    IF it converges. This one does. It works either way.

    Try limits [1,2] just for practice.
    Try limits [0.5,1] just for fun.
    Try limits [0.25,0.5] for kicks and jollies.
    Try limits [0.1,0.25] for further enjoyment
    Try limits [0.01,0.1] They should all come out finite.

    Thought question. If you are trying to evaluate an integral on [0,2], and you decide to break it up into two chunks, [0,1],[1,2] aren't you counting x = 1 TWICE?!
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  7. #7
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    diverging and converging

    i haven't learned that concept (improper integrals?)

    I just took a look at it and convergence means the limit is the value of the integral, whereas divergence means the limit is infinitiy..

    it may sound incredibly stupid but how do you find the limit without the help of a calculator.
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  8. #8
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    Quote Originally Posted by TKHunny View Post
    Integrals are tougher than that. It doesn't have to exist, it just has to converge.

    \int_{BadPlace}^{b}f(x) \; dx = \lim_{a \to BadPlace} \int_{a}^{b}f(x) \; dx

    IF it converges. This one does. It works either way.

    Try limits [1,2] just for practice.
    Try limits [0.5,1] just for fun.
    Try limits [0.25,0.5] for kicks and jollies.
    Try limits [0.1,0.25] for further enjoyment
    Try limits [0.01,0.1] They should all come out finite.

    Thought question. If you are trying to evaluate an integral on [0,2], and you decide to break it up into two chunks, [0,1],[1,2] aren't you counting x = 1 TWICE?!
    i think it should be [0,1), [1,2]
    and do the improper integral you taught me
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  9. #9
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    Quote Originally Posted by hangainlover View Post
    it may sound incredibly stupid but how do you find the limit without the help of a calculator.
    Two things:

    1) Calculus is 300+ years old. Calculators only about 40.

    2) That, My Friend, is what calculus is all about.
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  10. #10
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    Quote Originally Posted by hangainlover View Post
    i think it should be [0,1), [1,2]
    You missed the point. It doesn't matter. The definition of the integral includes \Delta x \to 0. It's that "limit" thing again.
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