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Math Help - Anomaly in Maths Textbook (Integration) or from Me?

  1. #1
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    Question Anomaly in Maths Textbook (Integration) or from Me?

    So far as I have learned,
    one has to integrate the curve f(x) function only in order to find the area under the curve.


    But I Quote as follows:
    ************************************************** ******************
    The volume V of the liquid in a container leaks out at the rate of  30t cm^3s^_-1 where t is the time in seconds.
    Find the amount of liquid lost in the third second.


    Solution ...

    \frac{\mathrm{d}V}{\mathrm{d} x}=-30t (decreasing)

    The third second is between t=2 and t=3.

    So we find (value of V when  t=3) - (value of V when  t=2 using a definite integral.

    Change of volume = \int_{2}^{3}\frac{\mathrm{d}V}{\mathrm{d} t} dt=\int_{2}^{3}(-30t) dt= \left[-15t^2\right]_2^3=-135-(-60)=-75

    Hence  75  cm^3 of liquid was lost in the third second.

    ************************************************** ******************

    According to me, one should integrate the gradient function back to the curve function and then integrate again as usual to find area.

    So,
    Equation of Curve = \int (-30t)\hspace{1mm}\textup{dt}

    V=-15t^2+c

    Arbitrary constant C may be ignored as t=0 when V=0 for physical variables.

    V=-15t^2

    Therefore,
    Change of Volume between t=2 and t=3

    \int_{2}^{3}(-15t^2)= \left[-5t^3\right]_2^3=(-135)-(-40)=-95

    The volume lost must be 95 cm^3, How far do you agree?
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  2. #2
    Senior Member MaxJasper's Avatar
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    Re: Anomaly in Maths Textbook (Integration) or from Me?

    Lost V(t)=15*t^2

    V(3)=15*3^2=135 cm^3
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    Lightbulb Re: Anomaly in Maths Textbook (Integration) or from Me?

    Finally I think I have just discovered the right conclusion.
    Now I admit my previously proposed solution is WRONG and the QUOTED one is the right one

    I wasn't completely at fault in assuming that:

    Indeed,
    Area under a curve is found by integrating the curve equation.
    But that is a different story with physical variables like Volume, Displacement etc... where one need to be careful in the interpretation of meaningful data.

    In-fact,
    Area under a Volume-Time or Displacement-Time graph give area that is equal to the product = Vt (Volume x Time) and Dt (Displacement x Time) respectively, but not just Volume V or Displacement D alone as expected.


    Fortunately while studying Mechanics topics, I found that:

    D=\int_{t_1}^{t_2}V \hspace{2mm}dt

    Where D is Displacement
    V is Velocity (Rate of Change of D).

    It clearly shows that the value of Displacement isn't calculated from it's Displacement-time graph between any intervals of time  t_1 and t_2 but from the velocity-time graph which can be considered as the gradient function graph of D ([s/t]/t ).

    It makes sense that the Area under the Velocity-Time graph gives Displacement as:

    Area= v\times t=D


    Therefore,
    Same for our Volume example,

    Volume=\int_{t_1}^{t_2}f'(t) \hspace{2mm}dt

    Where f'(t) = gradient function of V


    Area representing change in Volume between any time intervals  t_1 and t_2 is calculated from it's gradient function graph([volume/time]/time) graph.
    Last edited by zikcau25; September 13th 2012 at 06:57 AM.
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    Re: Anomaly in Maths Textbook (Integration) or from Me?

    You don't really need calculus for this.

    At t=2 it's leaking at a rate of 60 cm^3/s.
    At t=3 it's leaking at a rate of 90 cm^3/s.

    The average rate of leakage during the third second is (60+90)/2 cm^3/s = 75 cm^3/s so the volume of liquid leaked in the third second is 75 cm^3.
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    Re: Anomaly in Maths Textbook (Integration) or from Me?

    I agree with you sir .

    But the problem above is simple equation and short interval between t_1 and t_2.


    Calculus is the simplest and general solution to all equations as it get complex, ex. v=\frac{t^7}{7}-\frac{3t^8}{8} etc..
    Would be tedious without Calculus.

    I think people should use Calculus more often to get familiar with it's magical range of applications in most cases.
    Last edited by zikcau25; September 13th 2012 at 05:42 AM.
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    Re: Anomaly in Maths Textbook (Integration) or from Me?

    Quote Originally Posted by zikcau25 View Post
    So far as I have learned,
    one has to integrate the curve f(x) function only in order to find the area under the curve.


    But I Quote as follows:
    ************************************************** ******************
    The volume V of the liquid in a container leaks out at the rate of  30t cm^3s^_-1 where t is the time in seconds.
    Find the amount of liquid lost in the third second.


    Solution ...

    \frac{\mathrm{d}V}{\mathrm{d} x}=-30t (decreasing)

    The third second is between t=2 and t=3.

    So we find (value of V when  t=3) - (value of V when  t=2 using a definite integral.

    Change of volume = \int_{2}^{3}\frac{\mathrm{d}V}{\mathrm{d} t} dt=\int_{2}^{3}(-30t) dt= \left[-15t^2\right]_2^3=-135-(-60)=-75

    Hence  75  cm^3 of liquid was lost in the third second.

    ************************************************** ******************

    According to me, one should integrate the gradient function back to the curve function and then integrate again as usual to find area.

    So,
    Equation of Curve = \int (-30t)\hspace{1mm}\textup{dt}

    V=-15t^2+c

    Arbitrary constant C may be ignored as t=0 when V=0 for physical variables.

    V=-15t^2

    Therefore,
    Change of Volume between t=2 and t=3

    \int_{2}^{3}(-15t^2)= \left[-5t^3\right]_2^3=(-135)-(-40)=-95

    The volume lost must be 95 cm^3, How far do you agree?
    Are you sure it's dV/dx = -30t and not dV/dt = -30t?
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    Re: Anomaly in Maths Textbook (Integration) or from Me?

    Thank you for pointing that copying error.
    Due to troublesome Manual Latex editing without a straightforward latex software

    Evidentl it is,

     \frac{\mathrm{d} v}{\mathrm{d} t}=-30t


    Hope if you can edit that mistake please in the first post.
    Last edited by zikcau25; September 13th 2012 at 06:33 AM.
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