Homework Section

[Patton]

Drop a soluble product on a glass of water. In instants in (t), in minutes, the amount of the product not dissolved its, in grams, given by the function:



Question - How much is the amount of the product droped initially on the water?

Since your function is time dependent you need to determine which amount to solve it for. If you take that the moment when you drop the solute in the water as t = 0min, that is the time that you can use since the reaction most likely won't start immediately, and the amount of the solute in water will still be at its maximum. Therefore if you replace the time variable you will get something like this:

Also, for larger amounts of time you will get remaining amount of solute. After 30 minutes, you'll have 5,05g left and after 60 minutes there will be 0,27g of the solute left.

Protip: if you want to have your solute dissolved in the solvent faster add heat or put your glass into the ultrasound cleaner while stirring.

[Daco]

There's really no point as most of math solutions are available on the net. But sure if you want some for your own pleasure I can give you some integration exercises. ∫(2dx)/[x(x²+2x+2)] ∫dx/(x²+4x+8). There you go, have fun.

too easy bro

[Jubin]

Interesting way of finding the constant a there. Also how hard of a integrals you want?

[Daco]

Interesting way of finding the constant a there.

There's nothing into it. Find a quotient where the denominator can be factorised such as this function:


Easy to notice that it can be factorized:


Do this then:


If you want A you multiply by its respective denominator and replace the x by the value that makes that denominator 0 (for A it's -3, for B it's 1)

Easy to find that A= -1/4 and B= 1/4

So:


It's called partial fraction decomposition in Algebra, but I guess you already know that...

Also how hard of a integrals you want?

Impossible tier... :cool:
Post Merge: April 22, 2012, 09:33:40 am

Drop a soluble product on a glass of water. In instants in (t), in minutes, the amount of the product not dissolved its, in grams, given by the function:



Question - How much is the amount of the product droped initially on the water?

INB4 THAT'S 3^t

[JDC]

Daco = :cool:

In our country, math for 12 year olds is still where you are introduced to Algebra. Our entire education system is basically shit (which is why I preferred immersing in books for hours compared to the classroom lessons, I'd learn even more) even if they are adding two years.

[Jubin]

Daco yeah, I know how to do it but you for some reason wrote there a= lim(x->0) 2/(x^2+2x+2)

[duffman]

Post Merge: April 22, 2012, 09:33:40 amINB4 THAT'S 3^t

No, its something called Deriv or smth.

[Daco]

Daco yeah, I know how to do it but you for some reason wrote there a= lim(x->0) 2/(x^2+2x+2)

too obvious since 2/(x^2+2x+2) is continuous at 0 so f(0) = lim x->0 f(x)

No, its something called Deriv or smth.

derivate 3??? lol

[duffman]

derivate 3??? lol

ye

[Daco]

ye

3' = 0

.....................

[duffman]

3' = 0

.....................

OBVIOUSLY, SO PATTON IS WRONG?

[Patton]

ye

Then it's even easier. Use the same value for time (t = 0min); you'll get something like this as the amount of solute at the beginning:

[duffman]

Then it's even easier. Use the same value for time (t = 0min); you'll get something like this as the amount of solute at the beginning:

What you did to the 0.09t ?

[Daco]

OBVIOUSLY, SO PATTON IS WRONG?

then it's 60/5= 12g

[Patton]

What you did to the 0.09t ?

I took the time as 0 since we can assume that at the very beginning of the reaction (time when the solute was added to the water) there is still the maximum (i.e. all) of the solute present in the non-dissolved state. So, 0,09 * 0 is obviously 0 and e⁰ is 1.