Lost V(t)=15*t^2
V(3)=15*3^2=135 cm^3
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 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.
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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 =
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?
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.
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:
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.
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.
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.
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.