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Math Help - Having trouble. Involves rate of change.

  1. #1
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    Having trouble. Involves rate of change.

    I am not exactly sure how to start. If someone can please help me with one of these word problems I would be extremely grateful.

    An airplane is flying at an altitude of 3 miles at a rate of 400 miles per hour. Find the rate of change of angle a at the moment that the observer is standing at a horizontal distance of 4 miles from the plane,




    A ladder 15 ft long is leaning against the wall of a house. The base of the ladder is pulled away from the house at a rate of 2 ft per second.
    Consider the triangle created by the ground the wall and the ladder. How fast is the area changing when the ladder’s top is 12 ft up and falling at a rate of 1/2 ft/sec?
    Last edited by rachelkh; April 12th 2012 at 04:55 AM.
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  2. #2
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    Re: Having trouble. Involves rate of change.

    Draw a right angled triangle with height 3, base x and hypotenuse (that is line from observer to plane) making angle A with the base. If the plane is approaching observer dx/dt=-400 We are trying to find dA/dt
    Now dA/dt=dx/dt times dA/dx
    From the diagram tanA=3/x So x=3cotA dx/dA=-3cosecAcotA So dA/dx=1/(-3cosecAcotA)=-1/3sinAtanA
    So dA/dt=-400times -1/3sinAtanA
    When x=4 tanA=3/4 and using Pythagoras sinA=3/5 Put in these values to work out dA/dt
    If plane is flying away from observer use dx/dt=+400 instead.
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  3. #3
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    Re: Having trouble. Involves rate of change.

    Hello, rachelkh!

    A ladder 15 ft long is leaning against the wall of a house.
    The base of the ladder is pulled away from the house at a rate of 2 ft per second.
    Consider the triangle created by the ground the wall and the ladder.
    How fast is the area changing when the ladder’s top is 12 ft up and falling at a rate of 1/2 ft/sec?

    Code:
          |
        P *
          |\
          | \
          |  \
        y |   \ 15
          |    \
          |     \
          |      \
          * - - - * - -
          R   x   Q
    The ladder is PQ = 15.
    Let x = RQ,\;y = PR.

    The area of \Delta PQR is: . A \:=\:\tfrac{1}{2}xy .[1]


    From the right triangle: . x^2 + y^2 \:=\:15^2 \quad\Rightarrow\quad x \:=\:\sqrt{225 - y^2}

    Substitute into [1]: . A \;=\;\tfrac{1}{2}(225-y^2)^{\frac{1}{2}}y


    Differetiate with respect to time:

    . . \frac{dA}{dt} \;=\;\tfrac{1}{2}\left[\tfrac{1}{2}(225-y^2)^{-\frac{1}{2}}(-2y)\cdot y + (225-y^2)^{\frac{1}{2}}\right]\,\frac{dy}{dt}

    . . \frac{dA}{dt}\;=\;\tfrac{1}{2}\left[\frac{-y^2}{\sqrt{225-y^2}} + \sqrt{225-y^2}\right]\,\frac{dy}{dt} \;=\;\tfrac{1}{2}\left[\frac{-y^2 + 225 - y^2}{\sqrt{225-y^2}}\right]\,\frac{dy}{dt}

    . . \frac{dA}{dt} \;=\;\frac{225-2y^2}{2\sqrt{225-y^2}}\,\frac{dy}{dt}


    When y = 12,\;\frac{dy}{dt} = -\tfrac{1}{2}

    . . \frac{dA}{dt} \;=\;\frac{225 - 2(12^2)}{2\sqrt{225-12^2}}\left(-\frac{1}{2}\right) \;=\;\left(\frac{-63}{2\cdot9}\right)\left(-\frac{1}{2}\right) \;=\;+\frac{7}{4}


    \text{The area is increasing at }1\tfrac{3}{4}\text{ ft}^2\!/\text{sec.}
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  4. #4
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    Re: Having trouble. Involves rate of change.

    Alternative solution for ladder problem (with same notation)

    A=1/2xy A^2=1/4x^2y^2

    4A^2=x^2y^2
    4A^2=(225-y^2)y^2=225y^2-y^4

    Differentiate both sides with respect to t

    8AdA/dt=450ydy/dt-4y^3dy/dt

    When y=12 x=9 A=54 dy/dt=-1/2

    Feed in these values and get dA/dt=1.75
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