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:

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The volume $\displaystyle V$ of the liquid in a container leaks out at the rate of $\displaystyle 30t$ $\displaystyle cm^3s^_-1$ where $\displaystyle t$ is the time in seconds.

Find the amount of liquid lost in the third second.

*Solution ...*

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

The third second is between $\displaystyle t=2$ and $\displaystyle t=3$.

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

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

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

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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 = $\displaystyle \int (-30t)\hspace{1mm}\textup{dt}$

$\displaystyle V=-15t^2+c$

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

$\displaystyle V=-15t^2$

Therefore,

Change of Volume between $\displaystyle t=2$ and $\displaystyle t=3$

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

The volume lost must be 95$\displaystyle cm^3$, How far do you agree?(Thinking)

Re: Anomaly in Maths Textbook (Integration) or from Me?

Lost V(t)=15*t^2

V(3)=15*3^2=135 cm^3

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 = $\displaystyle Vt$ (Volume x Time) and $\displaystyle Dt$ (Displacement x Time) respectively, but not just Volume $\displaystyle V$ or Displacement $\displaystyle D$ alone as expected.

(Clapping) Fortunately while studying Mechanics topics, I found that:

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

Where $\displaystyle D$ is Displacement

$\displaystyle V$ is Velocity (Rate of Change of $\displaystyle D$).

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

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

$\displaystyle Area= v\times t=D$

(Bow)Therefore,

Same for our Volume example,

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

Where $\displaystyle f'(t)$ = gradient function of $\displaystyle V$

Area representing change in Volume between any time intervals $\displaystyle t_1$ and $\displaystyle t_2$ is calculated from it's gradient function graph([volume/time]/time) graph.

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.

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 $\displaystyle t_1$ and $\displaystyle t_2$.

Calculus is the simplest and general solution to all equations as it get complex, ex. $\displaystyle v=\frac{t^7}{7}-\frac{3t^8}{8}$ 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.

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 $\displaystyle V$ of the liquid in a container leaks out at the rate of $\displaystyle 30t$ $\displaystyle cm^3s^_-1$ where $\displaystyle t$ is the time in seconds.

Find the amount of liquid lost in the third second.

*Solution ...*

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

The third second is between $\displaystyle t=2$ and $\displaystyle t=3$.

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

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

Hence $\displaystyle 75$$\displaystyle 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 = $\displaystyle \int (-30t)\hspace{1mm}\textup{dt}$

$\displaystyle V=-15t^2+c$

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

$\displaystyle V=-15t^2$

Therefore,

Change of Volume between $\displaystyle t=2$ and $\displaystyle t=3$

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

The volume lost must be 95$\displaystyle cm^3$, How far do you agree?(Thinking)

Are you sure it's dV/dx = -30t and not dV/dt = -30t?

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,

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

Hope if you can edit that mistake please in the first post.