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25 November, 22:35

Find the exact length of the curve. y2 = 4 (x + 1) 3, 0 ≤ x ≤ 1, y > 0

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  1. 25 November, 23:46
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    We can find this using the formula: L = ∫√1 + (y') ² dx

    First we want to solve for y by taking the 1/2 power of both sides:

    y = (4 (x+1) ³) ^1/2

    y=2 (x+1) ^3/2

    Now, we can take the derivative using the chain rule:

    y'=3 (x+1) ^1/2

    We can then square this, so it can be plugged directly into the formula:

    (y') ² = (3√x+1) ²

    (y') ²=9 (x+1)

    (y') ²=9x+9

    We can then plug this into the formula:

    L = ∫√1+9x+9 dx * I can't type in the bounds directly on the integral, but the upper bound is 1 and the lower bound is 0

    L = ∫ (9x+10) ^1/2 dx * use u-substitution to solve

    L = ∫u^1/2 (du/9)

    L = 1/9 ∫u^1/2 du

    L = 1/9[ (2/3) u^3/2]

    L = 2/27 [ (9x+10) ^3/2] * upper bound is 1 and lower bound is 0

    L = 2/27 [19^3/2-10^3/2]

    L = 2/27 [√6859 - √1000]

    L=3.792318765

    The length of the curve is 2/27 [√6859 - √1000] or 3.792318765 units.
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