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Math Help - Two Related Rates problems I can't figure out

  1. #1
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    Two problems

    ** oops, messed up with the edits.. check the quote for the question **
    Last edited by eekoz; April 30th 2006 at 06:25 PM.
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  2. #2
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    Quote Originally Posted by eekoz
    So I've been practicing solving related rates problems.. I have two questions which I can't figure out, could someone help me out please?

    1. The ends of a water trough are equilateral triangles whose sides are 2 m long. The length of the trough is 15m. Water is being poured into the trouth at the rate of 3m^3/h. Find the rate at which the water level is rising when the depth of the water is 1m.


    2. A softball diamond has the shape of a square with sides 20 m long. Suppose a player is running from the first base to the second base at 7 m/s. Find the rate at which the distance between the player and home plate is changing when the player is 5m from the first base.

    Thanks in advance
    I'm pretty sure I got an answer for #2 (just need a confirmation), but #1 is giving me a lot of trouble
    The trick in doing word problems is to know what is to find. Then relate that to the given data to establish relations or equations. Then continue from there.

    1.) Find the rate at which water level is rising.

    Rate is per time. Water level is height. So if height is y, and time is t, then
    rate = y/t, or dy/dt for instantenous.

    Given data:
    Trough, triangular prism; 15m long; equilateral triangle, 2m each side, cross section.
    Rate of volume of water in the trough, or dV/dt = 3 cu.m./hr.

    There, we can relate the dy/dt with the dV/dt.

    Draw on paper, or imagine the figure.
    It is a cross sectional area of the trough.
    It is an equilateral triangle whose one vertex is sitting on the ground.

    Volume of water in the trough at any time:
    V = (1/2)(y)(x)(15) = 7.5xy
    where y = height of water; x = width of the water level.

    We express x in terms of y, for simplicity
    If we drop a vertical line from midpoint of x to the vertex at the bottom, the 60-degree angle there is bisected, so,
    tan(30deg) = (x/2)/y = x /2y
    Then, x = 2ytan(30deg) = 2y(1/sqrt(3) = (2/sqrt(3))*y
    Hence, V = 7.5[(2/sqrt(3))*y]*y
    V = [15/sqrt(3)](y^2)
    V = [5sqrt(3)](y^2) --------(i)
    Differentiate both sides of (i) with respect to time t,
    dV/dt = [10sqrt(3)]*y *dy/dt -----------(ii)

    (ii) is at any time t.
    At t when y=1m,
    3 = [10sqrt(3)](1)(dy/dt)
    So,
    dy/dt = 3/[10sqrt(3)] = (3/30)sqrt(3) = 0.1732 m/hr ---------***

    Therefore, when the depth of the water is 1 meter, the water level is rising at the rate of 0.1732 meter per hour. ----------answer.

    --------------------------------
    2.) Find the rate at which the distance between the player and home plate is changing when the player is 5m from the first base.

    So if x = the said distance, then we are looking for dx/dt.

    Let y = distance travelled by the runner.
    By Pythagorean theorem,
    x^2 = y^2 +20^2
    x^2 = y^2 +400 ------(iii)
    Differentiate both sides with respect to time t,
    2x(dx/dt) = 2y(dy/dt)
    x(dx/dt) = y(dy/dt) ----------(iv)

    Given:
    dy/dt = 7 m/sec; y=5m
    So, using (iii), x^2 = 5^2 +400 = 425
    x = sqrt(425) when y=5 ------------------***

    Substitute those into (iv),
    sqrt(425) *(dx/dt) = 5*7
    dx/dt = 35 /sqrt(425) = 1.7 m/sec ------***

    Therefore, when the runner is at 5m away from first base, the distance of the runner from the home plate is increasing at the rate of 1.7 meter per second. -----------answer.
    Last edited by ticbol; April 29th 2006 at 07:15 PM. Reason: +, not =.
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  3. #3
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    Thanks a lot, ticbol!
    I got 3*sqrt(3)/10 m/h for the first question, I guess I did something wrong.
    Going over your answer now, makes sense so far
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