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

• Sep 8th 2012, 01:50 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 $\displaystyle V$ 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$$\displaystyle cm^3 of liquid was lost in the third second. ************************************************** ****************** According to me, one should integrate the gradient function back to the curve function and then integrate again as usual to find area. So, Equation of Curve = \displaystyle \int (-30t)\hspace{1mm}\textup{dt} \displaystyle V=-15t^2+c Arbitrary constant C may be ignored as t=0 when V=0 for physical variables. \displaystyle V=-15t^2 Therefore, Change of Volume between \displaystyle t=2 and \displaystyle t=3 \displaystyle \int_{2}^{3}(-15t^2)= \left[-5t^3\right]_2^3=(-135)-(-40)=-95 The volume lost must be 95\displaystyle cm^3, How far do you agree?(Thinking) • Sep 8th 2012, 02:11 PM MaxJasper Re: Anomaly in Maths Textbook (Integration) or from Me? Lost V(t)=15*t^2 V(3)=15*3^2=135 cm^3 • Sep 13th 2012, 04:57 AM zikcau25 Re: Anomaly in Maths Textbook (Integration) or from Me? Finally I think I have just discovered the right conclusion. Now I admit my previously proposed solution is WRONG and the QUOTED one is the right one I wasn't completely at fault in assuming that: Indeed, Area under a curve is found by integrating the curve equation. But that is a different story with physical variables like Volume, Displacement etc... where one need to be careful in the interpretation of meaningful data. In-fact, Area under a Volume-Time or Displacement-Time graph give area that is equal to the product = \displaystyle Vt (Volume x Time) and \displaystyle Dt (Displacement x Time) respectively, but not just Volume \displaystyle V or Displacement \displaystyle D alone as expected. (Clapping) Fortunately while studying Mechanics topics, I found that: \displaystyle D=\int_{t_1}^{t_2}V$$\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. • Sep 13th 2012, 05:13 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, 05:29 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 \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. • Sep 13th 2012, 05:49 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 \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?
• Sep 13th 2012, 06:27 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,

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

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