Mathematics Corner :)

^Implicit differentiation like differentiating x but shoving dy/dx in its face.
 
i love calculaters.... who ever invented them was a genious!!!!! <3

<---- has to use fingers and toes to count. :-(


he he he he he he

~Grofica
 
Okay, so alhamdullilah I understand implicit differentiation now .. And I kind of get integration, I just want to read up some more about, check out more examples.

In the link bro Saifur-Rahman posted, I found this:
link

That's the kind I've been doing at school.

Does this kind of integration have any specific name? Because I've tried googling for more practice, and I get all sorts of results, such as "integration by parts", or "integration by substitution", etc. which are completely different. So what is this way called, and does anyone have any more links about it?
 
^ the link basically outlines the fundamental principles of what exactly Integration entails. I don't think it has a specific name though. Its just the basic concept, and from that you'll progress to 'Integration by substitution/parts' etc. Maybe someone else can explain it better inshaAllaah. Maths is my worst subject!
 
^^ Yup, exactly...essentially, this is the basic method you follow for integrating polynomials (well, for integrating anything of the form x^n; n can be almost any number...)

As the expression you're trying to integrate gets more and more complicated, you have to resort to things like substitution, and integrating by parts (even then, you're likely to have to integrate a polynomial at one stage or another, lol...)
 
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:sl:
I apologize if I'm disturbing a flow but I still need help on area and volume for the following: Prisms,pyramids,rectangle,triangle,and square.... Any help will be appreciated.
 
For square, it's length x base. But since length and base of a square is the same, it's lengthxlength, which = length squared. Have you done indices and powers? If not, then don't worry about l^2 or length squared. Just accept this:

Area of square = length x length.
 
:salamext:

Can someone please help me out on the following problem:

DifferentialEqn-1.png


I've checked the markscheme and that, but to be honest, I'm literally clueless on all this Differential eqn stuff;.....anyway I'd REALLY appreciate it if someone could also explain it inshaAllaah.

JazakAllaahu khayr!!
 
(i) dr/dt = k/(t+1) --- This is the differential eqn
(ii) using the info that dr/dt = 0.5 at t=3 sub in (i)

k/(3+1) = 0.5 , k = 4*0.5 = 2.

(iii) solving the diff eqn in (i)
integrating the right hand side

r= 0.5ln(t+1) + C. (Where are the initial conditions in the problem statement?)

(iv)k/(t+1)(t+2) ---> k/(3)(4) = 0.5 , k=6.

(v) dr/dt = 6/(t+1)(t+2) ---> [6/(t+1)]-[6/(t+2)], integrating you get
6*ln(t+1)-6ln(t+2) + C, 6(ln(t+1)/(t+2)) + C.
 
:sl:+o(

I always disliked the calculus word problems that dealt with rates that were proportionally to some ooglay expression : /
 
GreyKode said:
r= 0.5ln(t+1) + C. (Where are the initial conditions in the problem statement?)
Click to expand...

The radius of the oil drop, r, would be 0 when t = 0, I think...you can then substitute these into the above equation to find C (which also turns out to be 0, lol) and finally, use that to find r when t = 10. And part (vi) is exactly the same :).
 
Asalamu alaikum again.
So I'm back in the math thread, lol.
I've been doing practice exercises on implicit differentiation, and for some reason, I keep getting them wrong. Obviously I have some major mistake which I'm doing throughout all the problems, but I can't figure out what it is.
I'd appreciate it if anyone could look at these and tell me what I'm doing wrong..


a- 3xy^2 + cosy^2 = 2x^3 + 5
2y(dy/dx) - sin2y(dy/dx) = 6x^2
dy/dx (2y - sin2y) = 6x^2
dy/dx = 6x^2 / (2y-sin2y)

b- 5x^2 - x^3siny +5xy = 10
10x - 3x^2cosy(dy/dx) = 0
dy/dx (3x^2cosy) = -10x
dy/dx = -10x / (3x^2cosy)

c- x - cosx^2 + y^2/x + 3x^5 = 4x^3
1 + sin2x + [2y(dy/dx)]/x^2 + 15x^4 = 12x^2
dy/dx (2y/x^2) = 12x^2 - 15x^4 - sin2x -1
dy/dx = [x^2 (12x^2 - 15x^4 - sin2x -1)] / 2y

edit- WAIT. Duh. Okay, I found my mistake, it was a silly one, as they always are. I'd completely forgotten about actually having to use the product rule. Hmph. I typed all that up...What a waste! Oh well, lol, jazakum Allah khair anyway. :)
 
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The Ruler said:
I'm having difficulties understanding the domain and range with any function of x. Explain?
Click to expand...

you can think of a function as a machine,that takes some input,performs some operation on that input, and then gives u an output.the operation that the machine performs defines the function,the set of input values that a function can take is called the domain of the function,and the set of corresponding output values is called the range.
for example,
let,f(x)=x^2
it will take every real no.as input,(means its domain consists of all real no.s)
operates on it by squaring it,and then gives u output(ie,range),since square of a no. is not -ve,the range consists of +ve no.s only....
hope that clarifies:)
 
sis crayon,
check out ocw.mit.edu ,it has lecture notes of about 1800 MIT courses,including math as well and they are very helpful.u can also watch video lectures @ MIT channel on youtube.they explain calculus very well.and i recommend u a calculus book,that is just amazing.i hated math more than everything cuz i had not taken math at school in A-levels and now in college,after a two year gap,i was just helpless in calculus.but thanks to that book,after studying it for three months,i have literally fallen in love with calculus.it seems fun to evaluate complicated triple integrals lol:D
by the way the book is
Calculus:one and several variables,by Salas,Hille and Etgen.
publishers:john wiley and sons.
hopefully u could find that book in ur college library:)
 
~Raynn~ said:
The radius of the oil drop, r, would be 0 when t = 0, I think...you can then substitute these into the above equation to find C (which also turns out to be 0, lol) and finally, use that to find r when t = 10. And part (vi) is exactly the same :).
Click to expand...

But this way the model defies physics.
How can a something of zero radius, expand i.e. something out of noting
This is more like the big bang theory except, you can get around this problem by defining a singularity at the origin, but so far in reality that never happens when you perform such an experiment.
 
^ Lol, I see what you mean, but I think it's a simplification this mathematical model would allow...the question's saying that the time t = 0 is the point just before the oil drop hits the water - at that point, obviously, there's no circle of oil formed on the water yet...the radius of this 'circle' is 0. But it's not that the oil appears on the water out of nothing...

Such a model probably wouldn't have taken into account any actual physics, lol...it'd just be about finding an equation that fits what's observed, or any measurements taken...
 
Wow, no, not until today...I understand it's a function which is infinite at x = 0, and is zero for all other values of x...? Could that be applied here?? It does remind me of a black hole or big bang singularity (as a point of infinite density, and all), as you mentioned...
 

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