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Math Help - Related Rates Problem

  1. #1
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    Related Rates Problem

    I can't seem to figure this problem out

    A point p is moving along the curve whose equation is y= x^(1/2). Suppose that x is increasing at the rate of 4 units per second when x=3

    a. how fast is the distance between P and the point (2,0) changing at this instant?

    b. How fast is the angle of inclination of the line segment from P to (2,0) changing at this instant?

    Thanks
    Last edited by Linnus; October 11th 2007 at 11:50 AM.
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  2. #2
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by Linnus View Post
    I can't seem to figure this problem out

    A point p is moving along the curve whose equation is y= x^(1/2). Suppose that x is increasing at the rate of 4 units per second when x=3

    a. how fast is the distance between P and the point (2,0) changing at this instant?

    b. How fast is the angle of inclination of the line segment from P to (2,0) changing at this instant?

    Thanks
    Point P is at (x, \sqrt{x}) and we know that \frac{dx}{dt} = 4 when x = 3.

    The distance between point P and (2, 0) is
    s = \sqrt{(x - 2)^2 + (\sqrt{x} - 0)^2}

    s = \sqrt{x^2 - 3x + 4}

    So
    \frac{ds}{dt} = \frac{2x - 3}{2 \sqrt{x^2 - 3x + 4}} \cdot \frac{dx}{dt}

    So when x = 3 the distance is changing at a rate of
    \frac{ds}{dt}(x = 3) = \frac{2 \cdot 3}{2 \cdot \sqrt{3^2 - 3\cdot 3 + 4}} \cdot 4 = 3

    The angle of the line segment formed by this distance is characterized by the slope between the two points:
    m = \frac{\sqrt{x} - 0}{x - 2} = \frac{\sqrt{x}}{x - 2}
    (I know this isn't standard form, but I'm just going to leave it like this for now.)

    Now, geometrically speaking
    m = tan(\theta)
    where \theta is the angle of inclination.

    So
    \theta = tan^{-1}(m) = tan^{-1} \left ( \frac{\sqrt{x}}{x - 2} \right )

    Now,
    \frac{d}{dx}tan^{-1}(x) = \frac{1}{x^2 + 1}.

    Thus
    \frac{d\theta}{dt} = \frac{1}{\left ( \frac{\sqrt{x}}{x - 2} \right )^2 + 1} \cdot \left ( \frac{ \frac{x - 2}{2 \sqrt{x}} - \sqrt{x}}{(x - 2)^2} \right ) \cdot \frac{dx}{dt}

    \frac{d\theta}{dt} = \frac{1}{\frac{x}{(x -2)^2} + 1} \cdot \left ( \frac{(x - 2)  - 2x}{2(x - 2)^2 \sqrt{x}} \right ) \cdot \frac{dx}{dt}

    \frac{d\theta}{dt} = -\frac{(x - 2)^2}{x^2 - 3x + 4} \cdot \left ( \frac{x + 2}{2(x - 2)^2 \sqrt{x}} \right ) \cdot \frac{dx}{dt}

    \frac{d\theta}{dt} = -\frac{x +2}{2(x^2 - 3x + 4)\sqrt{x}} \cdot \frac{dx}{dt}

    \frac{d\theta}{dt} = -\frac{(x +2)\sqrt{x}}{2x(x^2 - 3x + 4)} \cdot \frac{dx}{dt}

    (I'm tired. I'll let you plug in x = 3.)

    -Dan
    Last edited by topsquark; October 12th 2007 at 10:52 AM.
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  3. #3
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    Quote Originally Posted by Linnus View Post
    I can't seem to figure this problem out

    A point p is moving along the curve whose equation is y= x^(1/2). Suppose that x is increasing at the rate of 4 units per second when x=3

    a. how fast is the distance between P and the point (2,0) changing at this instant?

    b. How fast is the angle of inclination of the line segment from P to (2,0) changing at this instant?

    Thanks
    a. how fast is the distance between P and the point (2,0) changing at this instant?

    P is on y = (x)^(1/2)
    So, P is point (x,sqrt(x))

    Distance, D, between P and (2,0).
    D^2 = (x-2)^2 +(sqrt(x) -0)^2
    D^2 = x^2 -4x +4 +x
    D^2 = x^2 -3x +4
    Differentiate both sides with respect to time t,
    2D(dD/dt) = 2x(dx/dt) -3(dx/dt)
    2D(dD/dt) = (2x -3)(dx/dt) -----------------(i)

    When x=3,
    dx/dt = 4 units/sec
    D = sqrt(x^2 -3x +4) = sqrt(9 -9 +4) = 2
    Plug those into (i),
    (2*2)dD/dt = (2*3 -3)(4)
    dD/dt = (12)/4 = 3 unit/sec

    Therefore, at that instant, the distance between P and (2,0) is increasing at the rate of 3 unit/sec. --------answer.

    ------------------------------------------------

    b. How fast is the angle of inclination of the line segment from P to (2,0) changing at this instant?

    Let angle A be the angle in question

    Draw the figure.

    tanA = y / (x -2)
    tanA = sqrt(x) / (x-2)
    Differentiate both sides with respect to time t,
    sec^2(A)*dA/dt = {(x-2)*(1 / 2sqrt(x))*(dx/dt) -(sqrt(x)*1(dx/dt))} / (x-2)^2 -----(ii)

    when x=3,
    dx/dt = 4 units/second
    D = 2 units
    secA = hypotenuse/(adjacent side) = D/(x-2) = 2/(3-1) = 1

    So, substitute those into (ii),
    (1^2)dA/dt = {(1)*(1/ 2sqrt(3))(4) -sqrt(3)*4} /(1^2)
    dA/dt = 2/sqrt(3) -4sqrt(3)
    dA/dt = [2 -4*3] / sqrt(3)
    dA/dt = -10/sqrt(3)
    dA/dt = -(10/3)sqrt(3) = -5.7735 radian/sec

    Therefore, at that instant, the angle of inclination of the line segment from (2,0) to point P is decreasing at the rate of 5.7735 radians/second. --------answer.
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  4. #4
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    confused

    i'm kinda confused...if i do it the inverse trig way...
    tan(A)=m
    A=arctan (m)
    m should be X^(1/2)/ (x-2)
    so
    A= arctan 2^(1/2)/ (x-2)

    when I take the derivative it

    and plug in 3 i got -1.443 radians per second
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  5. #5
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by Linnus View Post
    tan(A)=m
    A=arctan (m)
    m should be X^(1/2)/ (x-2)
    so
    A= arctan 2^(1/2)/ (x-2)
    How did you get from line 2 to line 3?

    -Dan
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  6. #6
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    Quote Originally Posted by topsquark View Post
    How did you get from line 2 to line 3?

    -Dan
    my bad, I mistyped

    A=arctan (x^(1/2)/x-2)
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  7. #7
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by Linnus View Post
    my bad, I mistyped

    A=arctan (x^(1/2)/x-2)
    Okay. Looking at this I noticed that I had my slope flipped over. I fixed it in my original post, so you can refer to it there. Oddly enough all it did was throw a negative sign onto my original answer. Weird.

    Anyway, I plug in x = 3 and get \frac{d\theta}{dt} = -\frac{5\sqrt{3}}{6}~rad/s \approx -1.44338~rad/s so apparently I agree with you.

    -Dan
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