Math Questions

Well, I'm expecting to have a great deal of math questions in the future because of a course I'm taking, so I guess I'm creating this thread for asking about some questions (mostly differential equations stuff) and for anyone else who might need to

Okay, first:

Given

R(dQ/dt) + Q/C = V

R: resistance
C: capacitance:
V: voltage
Q: charge

find Q(t) [treat R/C/V as constants]

Attempt/thoughts: (k is constant for after integration)

R(dQ/dt) = V-Q/C
RC(dQ/dt) = CV-Q
(dQ/dt)/(CV-Q) = 1/(RC)
ln(CV-Q) = t/(RC) + k
CV - Q = ke^(t/RC)
Q(t) = CV-ke^(t/RC)

Q(0)=0 [this is provided in prob] so
k = CV so

Q(t) = CV-CVe^(t/RC)

correct answer provided:

Q(t) = CV-CVe^(-t/RC)

So....kinda similar to what I got. So can anyone see what I did wrong? What I did seems pretty elementary and I can't find any problems with it

I won't spell it out for you, but my hint is to try differentiating line 4 using the chain rule to see if you get the original integrand. Should be clear where the minus sign has crept in then.

It would help if you could let us know what the question is for (homework, exam practice, coursework etc), the amount of help I can give you depends on whether it's being marked or not.

Yeah, I realized it myself. Blasted chain rule....

This is for homework so its not like for a test or anything (I don't think most classes do them anymore or do they?)

Okay....another one. I really have no idea how to get this:

Assuming you have a 1st order ordinary differential equation of the form

y'+p(t)y = g(t) and A is a function of t, (exp means e^(whatever) ; Int means integral[whatever])

the general solution is

y=A(t)exp[-Int(p(t)dt)] (denoted by *)

By substituting for y, show that A(t) satisfies this:

A'(t)=g(t)exp[Int(p(t)dt)]

Thoughts:

Okay, I suppose, you can solve for y in the very first thing to get:

y = (1/p(t))[g(t)-y']

and then, take the derivative of * so you can substitute it in for y', but that seems like its going to be very nasty and potentially not very promising

Alternative route:

Solve for A(t) in * and then take the derivative. Substitute in for y and y' as necessary. Again, seems to be very nasty and probably down to a dead end.

So....any ideas. This is just for homework, btw

Differentiate y with respect to t using the product rule and chain rule, then substitute your expressions for y and y' back into the original ODE.

You should find everything dependent on A(t) cancels, leaving you with only g(t) and a term multiplied by A'(t). Rearrange to solve for A'(t).

Yeah, the answer hit me this morning. And that is why you don't do homework when you haven't slept for like 20 hours......

what?!?!?

Yeah, I've got some more basic homework I need help with. I'll provide a sample question only, as I want to do the actual questions from the assignment myself:

Each limit represents the derivative of some function f at some number a. State such an f and a in each case.

The limit as h approaches zero(h->0) is:
(1 + h)^10 - 1
h

The given answer is:
f(x)=x^10, a=1
or
f(x)=(1 + x)^10, a=0

Through my calculations, I was able to find the limit of 1/2. However, I have no idea how those answers came up. Any help(hopefully today) would be much appreciated.

Well, I believe the limit that you have reached is incorrect. It should be 10. (You can just graph the limit, and go to tables and look at what 0.000001 is, which is roughly 10)

Slope of any function is (y-y0)/(x-x0)

In order to find the instantaneous slope at that point, you want the difference between y/yo and x/xo to approach 0. Thus, you have the limit of h-> 0 of [f(x+h)-f(x)]/h as a basic way to find the derivative. Given the limit you had above, you would have [(1+h)^10 + 1^10]/h

Take out the h from (1+h)^10 and you get 1^10, which would be f(x)=x^10 and the limit is looking for f'(1).

Alternatively, you can look at it as if that the 1 is not a substitution for a value of x, but merely a constant within the function and that the real value of x is actually 0. Since (1+0+h)^10 is redundant, you can simply just write (1+h)^10
Thus, you would have f(x)=(1+x)^10 and the limit is looking for f'(0)

Umm....I hope this is kinda clear. If it isn't, would take another look at the definition of a derivative

Great, thx. I think I understood it. I handed in my homework, anyways. :yes:
Just to verify my understanding of this topic... So, if:

the limit as h approaches 0 is:
4/16 + h + 2
h

Note that the first part in the numerator is the 4th root of (16 + h). Yeah, it's kind of illegible.
Anyways, I could say that f(x)= the 4th root of (16 + x), and a=0. Right?

EDIT: I believe I can also say that f(x)=the 4th root of x, and a=2. Interesting.

ok... what math are you taking

@Meth

yeah, thats right. It all depends on your initial parameters

@Veho Nex

Methuselah is taking essentially Calculus I which covers basic differentiation and integration

I am taking differential equations which basically is something like given f(t,y,y',y'',....)=0, what is y(t)?

Edit:

Actually, got another homework problem

Given the logistic equation
dy/dt=r[(1-y)/K]y-Ey where K=(r/a) and E is a positive constant, show that if E<r, the two equilibrium points are at y=0 and y=K[(1-E)/r]>0

Thoughts:

y=0 is pretty obvious there, but the other one not so much. My work so far:

r[(1-y)/K]y-Ey=0
r[(1-y)/K]y=Ey
r[(1-y)/K]=E
(1-y)/K=E/r
1-y=(KE)/r
y=1-(KE)/r

which is still off from y=K[(1-E)/r]. My question here is that I don't see how if E<r, the equilibrium point will be changed since the constant values should be irrelevant if you're just symbolically solving for when dy/dt=0. Furthermore, what would the ">0" mean in the equilibrium point context?

Edit 2:

Well....scratch everything I just posted b/c I misunderstood the original problem. It was 1-y/k actually so the prob is really easy actually

Quick question here, I seem to forgot how to derive the solution to the Verhulst/logistic equation of

dy/dt=ry(1-y/K)

Basically, I get

y/(1-y/K)=Ce^(rt) where C=y0/[1-(y0/K)]

and I just keep on forgetting how to isolate y in this instance. I know the general form of the solution,

y= (y0K)/[y0+(K-y0)e^(-rt)], but I just can't seem to simplify the above equation into this one

ok I know f(x) = Y

Is function of set parameters but what does y(t) mean?

(sorry to butt in just kinda wanna learn a bit of calc before i head off into next semester)

Well, it means that y is defined in terms of t, which would be time most commonly

Here's an example:

Y= x^2 - 1
is the same thing as:
F(x)= x^2 - 1

Now, F(t) would be thus:

F(t)= t^2 - 1

To take it a step further, let's try F(3):

F(3)= (3)^2 - 1
F(3)= 9 - 1
F(3)= 8

That should make your questions clear.

So its pretty much a normal funtion? Or is it a function with a different end game as in f(x)=y^2/2 where x=y?

Quote:

---

Originally Posted by Veho Nex

So its pretty much a normal funtion? Or is it a function with a different end game as in f(x)=y^2/2 where x=y?

---

Its just a normal function except by using f(t) or f(x) or whatever, you are denoting that y is in respect to t/x/whatever and is dependent upon it.

If

x = t = a

then

f (x) = f (t) = f (a)

and

y = 2x = f (x) = 2y = f (y) = 2a = f (a)


Shortly put: yep, exactly the same thing. Sometimes it's favourable to use other letters; such as t for time, v for velocity, a for acceleration, s for distance etc.