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Math Help - [SOLVED] Differential Equation Geometry/Word Problem Help!

  1. #1
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    [SOLVED] Differential Equation Geometry/Word Problem Help!

    Hi everyone, I'm currently having trouble mentally grasping this question I have:

    Find the equation of the curve that goes through (1,3) for which any tangent and the line from the origin to the point of contact make with the ordinate a triangle having area A.

    I have been trying several approaches to this, but to no avail. I assume it can be set up as a first-order differential equation of some sort. I've defined the curve as a, the tangent line as b, and the line from the point of tangency to the origin as c. I also defined a random point along the curve (alpha, beta). Thus,

    b(x)=beta+a'(alpha)*(x-alpha)
    c(x)=beta/alpha * x

    EDIT: The area of the triangle, A, could also be found with one of the following:
    A_1=int(b(x))-int(c(x))
    A_2=(1/2)(b)(h)=(1/2)(b(0))(alpha)=(1/2)(beta+a'(alpha)*(x-alpha))

    I have verified that, in fact, A_1=A_2.

    Any insight would be greatly appreciated!
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  2. #2
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    [According to Wikipedia, the ordinate is another name for the y-axis. I never came across that terminology before.]

    I find that a, b, c, alpha, beta, ... notation hard to follow, so I'll use other letters. Suppose the curve is given by y=f(x), and let (x_0,y_0) be a point on the curve. The tangent at this point has equation y-y_0=f'(x_0)(x-x_0). It meets the y-axis at the point (0,y_1), where y_1 is given by y_1-y_0 = -f'(x_0)x_0.

    If we take the triangle to have its base along the y-axis then its base is \pm y_1 ( the +/- is needed because we don't know whether y_1 is positive or negative) and its height is x_0. So its area is given by \textstyle A=\pm\frac12x_0y_1 = \pm\frac12x_0(y_0-x_0f'(x_0)).

    At this stage we need to change the notation. Up until now, (x_0,y_0) has been fixed. But in order to get a differential equation for f(x), we need to see what happens as (x_0,y_0) varies. The best thing to do is to drop the zero subscripts and to replace f'(x_0) by dy/dx (or just y'). Then the previous equation becomes x(y-xy') = \pm2A. At last, we have a differential equation!

    You'll need to use an integrating factor to solve the equation, and then to use the initial condition that y=3 when x=1. There are two solutions, corresponding to the +/- sign in the equation.
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  3. #3
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    Hello, bherila!

    Find the equation of the curve that goes through (1,3)
    for which any tangent and the line from the origin to the point of contact
    make with the ordinate a triangle having area A.
    Code:
            |
            *                 * y = f(x)
            |    *
            |        *       *
            |             *  P(p,q)
            |           o /
            |       o   /
            |   o     /
          B o       /
            |     /
            |   /
            | /
        - - * - - - - - - - - - - -
            |O
            |

    We have a function: y \,=\,f(x)

    Let P(p,q) be a point on y \,=\,f(x).

    OP is the segment from the origin to point P.

    BP is the tangent at P.
    . . The slope of this tangent is y' and it contains P(p,q).
    . . Its equation is: . y - q \:=\:y'(x - p)\quad\Rightarrow\quad y \:=\:y'x - y'p + q
    . . Its y-intercept is: . B(0,\,q - py')

    The base of \Delta OBP is: . OB \:=\:q - py'
    The height of the triangle is: . p

    The area of \Delta OBP\!:\;\;\frac{1}{2}(q - py')p \;=\;A

    \. . which simplifies to: . p^2y' - pq \:=\:-2A

    Divide by p^2\!:\;\;y' - \frac{1}{p}\,q \:=\:-\frac{2A}{p^2}

    Hence, we have: . \frac{dy}{dx} - \frac{1}{x}\,y \;=\;-\frac{2A}{x^2}

    Integrating factor: . I \:=\:e^{\int\left(-\frac{1}{x}\right)dx} \:=\:e^{-\ln x} \:=\:e^{\ln\frac{1}{x}} \:=\:\frac{1}{x}

    So we have: . \frac{1}{x}\,\frac{dy}{dx} - \frac{1}{x^2}\,y \;=\;-\frac{2A}{x^3}

    Then: . \frac{d}{dx}\left(\frac{1}{x}\!\cdot\!y\right) \;=\;-2Ax^{-3}

    Integrate: . \frac{1}{x}\!\cdot\!y \;=\;-Ax^{-2} + C

    Multiply by x\!:\;\;y \;=\;-\frac{A}{x} + Cx

    Since (1,3) is on the curve,
    . . 3 \;=\;-\frac{A}{1} + C\quad\Rightarrow\quad C \:=\:A + 3


    Therefore: . \boxed{y \;=\;-\frac{A}{x} + (A + 3)x}

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  4. #4
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    The other solution is \boxed{y =\frac{A}{x} + (3-A)x}.
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  5. #5
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by Soroban View Post
    Hello, bherila!

    Code:
            |
            *                 * y = f(x)
            |    *
            |        *       *
            |             *  P(p,q)
            |           o /
            |       o   /
            |   o     /
          B o       /
            |     /
            |   /
            | /
        - - * - - - - - - - - - - -
            |O
            |

    We have a function: y \,=\,f(x)

    Let P(p,q) be a point on y \,=\,f(x).

    OP is the segment from the origin to point P.

    BP is the tangent at P.
    . . The slope of this tangent is y' and it contains P(p,q).
    . . Its equation is: . y - q \:=\:y'(x - p)\quad\Rightarrow\quad y \:=\:y'x - y'p + q
    . . Its y-intercept is: . B(0,\,q - py')

    The base of \Delta OBP is: . OB \:=\:q - py'
    The height of the triangle is: . p

    The area of \Delta OBP\!:\;\;\frac{1}{2}(q - py')p \;=\;A

    \. . which simplifies to: . p^2y' - pq \:=\:-2A

    Divide by p^2\!:\;\;y' - \frac{1}{p}\,q \:=\:-\frac{2A}{p^2}

    Hence, we have: . \frac{dy}{dx} - \frac{1}{x}\,y \;=\;-\frac{2A}{x^2}

    Integrating factor: . I \:=\:e^{\int\left(-\frac{1}{x}\right)dx} \:=\:e^{-\ln x} \:=\:e^{\ln\frac{1}{x}} \:=\:\frac{1}{x}

    So we have: . \frac{1}{x}\,\frac{dy}{dx} - \frac{1}{x^2}\,y \;=\;-\frac{2A}{x^3}

    Then: . \frac{d}{dx}\left(\frac{1}{x}\!\cdot\!y\right) \;=\;-2Ax^{-3}

    Integrate: . \frac{1}{x}\!\cdot\!y \;=\;-Ax^{-2} + C

    Multiply by x\!:\;\;y \;=\;-\frac{A}{x} + Cx

    Since (1,3) is on the curve,
    . . 3 \;=\;-\frac{A}{1} + C\quad\Rightarrow\quad C \:=\:A + 3


    Therefore: . \boxed{y \;=\;-\frac{A}{x} + (A + 3)x}

    very nice. that was a lot easier than i thought it would be. Good stuff as always, Soroban!!!
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