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

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• Sep 8th 2012, 02:50 PM
zikcau25
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?(Thinking)
• Sep 8th 2012, 03:11 PM
MaxJasper
Re: Anomaly in Maths Textbook (Integration) or from Me?
Lost V(t)=15*t^2

V(3)=15*3^2=135 cm^3
• Sep 13th 2012, 05:57 AM
zikcau25
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.

(Clapping) 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:

$Area= v\times t=D$

(Bow)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.
• Sep 13th 2012, 06:13 AM
a tutor
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.
• Sep 13th 2012, 06:29 AM
zikcau25
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.
• Sep 13th 2012, 06:49 AM
Prove It
Re: Anomaly in Maths Textbook (Integration) or from Me?
Quote:

Originally Posted by zikcau25
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?(Thinking)

Are you sure it's dV/dx = -30t and not dV/dt = -30t?
• Sep 13th 2012, 07:27 AM
zikcau25
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.