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Math Help - Apps of differentiation

  1. #1
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    Apps of differentiation

    I don't have a problem solving these problems but i cant seem to find the equation(s) of this problem. A right triangle is formed in the first quadrant by the x and y axes and a line through the point (1,2). a0Write the length L of the hypotenuse as a function of x.

    the second part asks for me to find the minimum length but i can do that on my own. I can't find the equations for it. I believe the primary is L^2= x^2+y^2 but i cant determine the secondary one to sub in for y. I appreciate any help.
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  2. #2
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    Hello, jarny!

    A right triangle is formed in the first quadrant by the x and y axes
    and a line through the point (1,2).
    a) Write the length L of the hypotenuse as a function of x.
    Code:
            |
       (0,b)*
            | *
            |   * (1,2)
            |     o
            |       *
            |         *
        - - + - - - - - * - -
            |         (a,0)

    The line with intercepts 0,b)" alt="(a,0),\0,b)" /> has the equation: . \frac{x}{a} + \frac{y}{b} \:=\:1

    Since (1,\,2) is on this line: . \frac{1}{a} + \frac{2}{b} \:=\:1\quad\Rightarrow\quad b \:=\:\frac{2a}{a-1}\;\;{\color{blue}[1]}

    The length of the hypotenuse is given by: . L^2 \:=\:a^2+b^2\;\;{\color{blue}[2]}

    Substitute [1] into [2]: . L^2 \;=\;a^2 + \left(\frac{2a}{a-1}\right)^2 \;=\;\frac{a^4 - 2a^3 + 5a^2}{(a-1)^2}

    Therefore: . L \;=\;\frac{\sqrt{x^4-2x^2 + 5x^2}}{x-1}

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