StudyOriented
New member
- Joined
- Jan 4, 2014
- Messages
- 8
Hello,
First of all, sorry if my English is bad, that because i am not from an English-speaking country.
But, i think you could understand what i say, because mathematics is universal.
Well, my brother give me an example of his calculus exam test. He's already in his first year the university, while i am still High School.
He said that it will use the natural logarithm ( ln )
dx/dt= k x
dx/x = k dt
integral both side get;
ln x = ∫ k dt
= k (t) + C
x = (e^(kt)) (e^C) and e^C is constant
and that applied to the question below.
As far as i calculate the formula for the salt is
dx/dt = 4 - (6x/120-2t)
the trouble is there is still an x
but how to apply the formula that stated before? ( x = (e^(kt))(e^C) )
and how to find the pure formula of x , x(t)=?
First of all, sorry if my English is bad, that because i am not from an English-speaking country.
But, i think you could understand what i say, because mathematics is universal.
Well, my brother give me an example of his calculus exam test. He's already in his first year the university, while i am still High School.
He said that it will use the natural logarithm ( ln )
dx/dt= k x
dx/x = k dt
integral both side get;
ln x = ∫ k dt
= k (t) + C
x = (e^(kt)) (e^C) and e^C is constant
and that applied to the question below.
There is a tank full of water 120 Litres.
Saltwater from a pipe goes to the tank with the flowing 4 Litres per second and the composition of the saltwater is 1 gram per Litre.
The tank always stir the water inside it automatically.
And the stirred water inside it then flowed out from another pipe flowing 6 Litres of stirred water out per second.
The question is, how many grams of salt left at the time of "t"?
As far as i calculate the formula for the salt is
dx/dt = 4 - (6x/120-2t)
the trouble is there is still an x
but how to apply the formula that stated before? ( x = (e^(kt))(e^C) )
and how to find the pure formula of x , x(t)=?