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 of the liquid in a container leaks out at the rate of where is the time in seconds.

Find the amount of liquid lost in the third second.

*Solution ...*

(decreasing)

The third second is between and .

So we find (value of when ) (value of when using a definite integral.

Change of volume =

Hence 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 =

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

Therefore,

Change of Volume between and

The volume lost must be 95 , 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 = (Volume x Time) and (Displacement x Time) respectively, but not just Volume or Displacement alone as expected.

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

dt

Where is Displacement

is Velocity (Rate of Change of ).

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

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

(Bow)Therefore,

Same for our Volume example,

dt

Where = gradient function of

Area representing change in Volume between any time intervals and 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 and .

Calculus is the simplest and general solution to all equations as it get complex, ex. 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

of the liquid in a container leaks out at the rate of

where

is the time in seconds.

Find the amount of liquid lost in the third second.

*Solution ...* (decreasing)

The third second is between

and

.

So we find (value of

when

)

(value of

when

using a definite integral.

Change of volume =

Hence

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 =

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

Therefore,

Change of Volume between

and

The volume lost must be 95 , 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,

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