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Math Help - Calc: Problem Solving

  1. #1
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    Calc: Problem Solving

    THe graph of y = f (x) passes through the origin. the arc length of the curve from (0,0) to (x, f(x)) is given by s(x) = (integral from a to x) of the square roote of 1 + e ^ t dt... Identify the function f.
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  2. #2
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    Quote Originally Posted by 09Ashaw View Post
    THe graph of y = f (x) passes through the origin. the arc length of the curve from (0,0) to (x, f(x)) is given by s(x) = (integral from a to x) of the square roote of 1 + e ^ t dt... Identify the function f.
    Formula for arclength: L = \int_{a}^{b} \sqrt{1 + \left( \frac{dy}{dx}\right)^2} \, dx.

    So it looks like you need to solve \left( \frac{dy}{dx} \right)^2 = e^x subject to the boundary condition that y = 0 when x = 0 ....
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  3. #3
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    Hello, 09Ashaw!

    The graph of y \,=\, f(x) passes through the origin.
    The arc length of the curve from (0,0) to (x, f(x)) is given by: . s(x) \:=\:\int^x_0\sqrt{1 + e^t}\,dt
    Identify the function f(x)
    The arc length formula is: . s(x) \;=\;\int^b_a\sqrt{1 + \left(\tfrac{dy}{dx}\right)^2}\,dx

    So we know that: . \left(\frac{dy}{dx}\right)^2 \:=\:e^t \quad\Rightarrow\quad \frac{dy}{dx} \:=\:e^{\frac{x}{2}} \quad\Rightarrow\quad dy \:=\:e^{\frac{x}{2}}\,dx

    Integrate: . \int dy \;=\;\int e^{\frac{x}{2}}\,dx \quad\Rightarrow\quad y \;=\;2e^{\frac{x}{2}} + C


    Since the graph contains the origin (0,0), we have: . 0 \;=\;2e^0 +C \quad\Rightarrow\quad C \:=\:-2


    . . Therefore: . \boxed{f(x) \;=\;2e^{\frac{x}{2}} - 2}

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