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Applications of Derivatives. Needs Checking

  1. May 15, 2012 #1
    1. A 1000 L tank is draining such that the volume V of water remaining in the tank after t minutes is v=1000(1-t/60)2. Find the rate at which the water is flowing out of the tank after 10 min.

    Calculate dV/dt using chain rule
    u = 1 - t/60:
    u = 1 - t/60 ==> du/dt = -1/60
    V = 1000u^2 ==> dV/du = 2000u = 2000(1 - t/60)

    So we get:
    dV/dt = dV/du * du/dt = 2000(1 - t/60) * -1/60 = -100(1 - t/60)/3

    Plugging in t = 10, we get:
    dV/dt = -100(1 - 10/60)/3 = -250/9 ? -27.778

    So water is draining out at about 27.778 L/min


    2. When a certain object is placed in an oven at 540°C, its temperature T(t) rises according to the equation T(t) = 540(1 – e–0.1t), where tis the elapsed time (in minutes). What is the temperature after 10 minutes and how quickly is it rising at this time?

    Temp after t minutes is given. Just plug in t = 10 mins in the equation ie T = 540(1-1/e)
    rate of change is dT/dt (at t = 10) = 54/e


    3. The amount of daylight a particular location on Earth receives on a given day of the year can be modelled by a sinusoidal function. The amount of daylight that Windsor, Ontario will experience in 2007 can be modelled by the function D(t) = 12.18 + 3.1 sin(0.017t – 1.376), where t is the number of days since the start of the year.

    a. On January 1, how many hours of daylight does Windsor receive?

    D(T) = 12.18 + 3.1sin(-1.376)

    b. What would the slope of this curve represent?

    Slope represents the rate of change of daylight received.

    c. The summer solstice is the day on which the maximum amount of daylight will occur. On what day of the year would this occur?

    It happens when sin(0.017t – 1.376) = 1 since other constants won't change with change in t, only this sin function will change with change in t and the maximum value t hat a sine function can get is at pi/2 ie sin pi/2 = 1
    i.e 0.01t-1.376 = pi/2
    Calculate your t from this equation

    d. Verify this fact using the derivative.

    Verify it by taking derivative,
    d[D(t)]/dt = 3.1*0.01(cos(0.01t-1.376) = 0
    Which implies that cos(0.01t-1.376) = 0
    ie 0.01t-1.376 = pi/2 ..same condition as we got in part c

    e. What is the maximum amount of daylight Windsor receives?

    Maximum amount of daylight can be found out by putting t obtained from parts (c) or (d) in the equation.

    f. What is the least amount of daylight Windsor receives?

    Can do it in two ways, i will do it by trig method.
    least amount is 12.18 - 3.1 = 9.08......you don't even have to take derivative to do this, just do it by the fact that minimum value of a sine is -1
     
  2. jcsd
  3. May 16, 2012 #2

    Bacle2

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    Seems fine; just a few things:

    1)In 3 c,d, Why are you rounding off 0.017 to 0.01 ?

    in 3f , Did you take into account that the values, pi, pi/2 will never be assumed by

    0.017t-1.376 ?

    Kind of strange that "summer solstice" in 3c, happens at the end of October, tho.
     
  4. May 16, 2012 #3
    Thnx for checking, I didn't get what are you trying to say for 3f.
     
  5. May 16, 2012 #4

    SammyS

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    The context of the problem #3 suggests that t is an integer.

    Therefore, 0.017t – 1.376 will not be exactly equal to 3π/2 , so that sin(0.017t – 1.376) will not be exactly equal to -1 for any integer value of t .
     
  6. May 17, 2012 #5
    So what should be the answer for f.)
     
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