calculus in physical world qu (1 Viewer)

[sikeveo]

I haven't touchecd this in 1 yr, so can someone explain how i do this?

dy/dx = 2y and y =5 when x=0, express y as a function of x

I need the working, so i can understand it again

 

[acmilan]

dy/dx = 2y

dy/y = 2dx

integrate both sides:

ln y = 2x + c

and go from there

 

[sikeveo]

I dont get it, how did u get dy/y = 2dx ?

 

[acmilan]

Split the variable and differential. Divide by y, multiply by dx. It's not really taught in 2 unit, but i think 3 unit students learn it.

 

[sikeveo]

sorry yeh thats obvious =/

 

[MarsBarz]

Uh that's a strange method there. I would have just done it like this:

dy/dx = 2y
dx/dy = 1/2y
x = 1/2(lny) + C
when x=0, y=5
0 = 1/2(ln5)+C
C = -1/2(ln5)
x = 1/2(lny)-1/2(ln5)
2x = lny-ln5
lny = 2x+ln5
y = e^(2x+ln5)

Although your method looks awfully neat.

 

[sikeveo]

you forgot to simplify


its y=5e^(2x)

 

[sikeveo]

Uh that's a strange method there. I would have just done it like this:

dy/dx = 2y
dx/dy = 1/2y
x = 1/2(lny) + C
when x=0, y=5
0 = 1/2(ln5)+C
C = -1/2(ln5)
x = 1/2(lny)-1/2(ln5)
2x = lny-ln5
lny = 2x+ln5
y = e^(2x+ln5)

Although your method looks awfully neat.

Thanks, i understand your method better

 

[MarsBarz]

Actually I understand Acmilan's method now. I redid his method but put in all the steps to make it clearer.

dy/dx = 2y
dy = 2y.dx
dy/y = 2.dx
1/y.dy = 2.dx
Integrating both sides
lny + K= 2x + C
lny = 2x + C - K
(remember K and C are constants)
when x=0, y=5
ln5 = 0 + C - K
C - K = ln5
ln y = 2x + ln5
y = e^(2x+ln5)
y = e^(2x) . e^(ln5)
y = e^(2x) . 5
.:. y = 5e^(2x)
Yay.

 

[sikeveo]

Yeh but its longer =/

 

[MarsBarz]

Yeah lol, I rekon the way I first did it was much simpler.

 

[acmilan]

Yeah lol, I rekon the way I first did it was much simpler.

Its exactly the same method

The only difference is when you got up to:

dx/dy = 1/(2y)

You integrated straight away, when deep down you (not you personally but I mean everyone in general) are skipping the step dx = 1/(2y)dy, which is something that almost everyone naturally does as its obvious.

Edit: also when you integrate you do: lny + K= 2x + C

This isnt necessary, the constant only has to appear on one side, ie write it as ln y = 2x + C. Reason:

If lny + K = 2x + C, then lny = 2x + (C - K), and C - K is a constant, which we'll call D. Thus lny = 2x + D, which is what you couldve started with the begin with.

Technically, when you integrated dx/dy = 1/(2y), if you were to follow the same thing as above, you'd get x + K = 1/2 ln(y) + C, which again, is not necessary.

 

[MarsBarz]

Ah yes. I realised that since it's just a constant we could just have it on one side. I guess it saves a bit of time, I was just trying to make it clear .

 

[sikeveo]

another quick qu

if y = Ae^(-kt) and y =1000 when t =0 and y = 368 when t=2, find the values of A and k.

I have A, but not K. Some help would be appreciated.

 

[acmilan]

Are you sure that N = 368 isnt y = 368?

 

[sikeveo]

yeah sorry, post edited accordingly.

 

[acmilan]

y = Ae^kt

e^kt = y/A
kt = ln(y/A)
k = ln(y/A)/t


you know what y (368), A and t (2) are, so sub them in to get k.

 

[Trev]

y=Ae^-kt
[t=0; y=1000]
∴368=A
y=1000e^-kt
[t=2; y=368]
368=1000e^-2k
e^-2k=46/125
-2k=ln(46/125)
k =0.4998blah which is approx .5