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10172012, 12:53 AM  #8961  
UG's Nu/Shyguy
Join Date: Aug 2006

Quote:
How could you do it without logs?... 1. Take the natural logarithm of both sides of the equation. 2. Use logs rules to simplify. 3. Differentiate both sides. (Hint: If you simplified the right side correctly, you'll have to use the product rule when you differentiate) EDIT: The tricky part is actually with the left side. If you can't figure it out, come back and we'll give you some help. Basically, though, y is a function within itself, and when you differentiate it, you have to treat it like you would a function within a function, i.e., use the chain rule. Last edited by RedDeath9 : 10172012 at 12:57 AM. 

10172012, 01:06 AM  #8962  
UG's Cute Canadian :3
Join Date: Jun 2009
Location: Under your bed... waiting for you to fall asleep. >_>

Quote:
Okay, this is what I got so far: lny = ln x^(5x) lny = 5xlnx lny = 5lnx + 5x(1/x) lny = 5(lnx +1) So, for the left side would I have to use implicit differentiation to get y'? Edit: Nevermind! I figured it out. Thanks for the help! Last edited by CodChick : 10172012 at 01:15 AM. 

10172012, 01:52 AM  #8963  
PonyFan #376121
Join Date: Jul 2006

Quote:
But P varies. Are you allowed to do that even though P is not constant?
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10172012, 02:07 AM  #8964  
UG's Nu/Shyguy
Join Date: Aug 2006

Quote:
Hm... Well... You replace P with (nRT)/V, and then integrate from Vi to Vf, which signifies that V, in (nRT)/V, is not constant. And if that V isn't constant, then P must not be constant either (since that's what you substituted for). The integration with respect to V takes into account that P is changing, essentially. Does that make sense? The final formula won't have pressure in it. Again, I don't know if what I'm saying is right. Found a wiki page on isothermal processes: http://en.wikipedia.org/wiki/Isothermal_process The equation that I you're gonna use is under the "Calculation of Work" section. They've got the integration on there too. Quote:
No problem Last edited by RedDeath9 : 10172012 at 02:08 AM. 

10172012, 10:55 AM  #8965  
Makes a fool of you.
Join Date: Jun 2004
Location: Quarkville, Thermodynamicus

Quote:
An isothermal, quasistatic/reversible process yields... w = nRT*ln(Vb/Va) = nRT*ln(p_init/p_final) Vb is the final volume, Va the initial... p_init is the initial pressure, p_final is the final pressure... For an ideal gas both equations are equivalent. You solve the integral substituting p for nRT/V, but can switch back and just use the pressures given by substituting Vb for nRT/p_final and Va for nRT/p_init. The nRT terms cancel, leaving the second form of the solution as shown above.
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i*[∂/∂t]*Ψt = [∇^2]/2*(Ψt) (unitless form)
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Last edited by hethamulburton : 10172012 at 11:07 AM. 

10172012, 11:26 AM  #8966 
UG Bard King
Join Date: Apr 2008
Location: SW England

When applying the equation RC*ln(2)=t to a asymmetric multivibrator what is t measured in?
The time period for one flash (from on to off)?
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10172012, 10:06 PM  #8967  
Look mum, no brakes!
Join Date: Feb 2005
Location: at home

This is a dumb question but it's really bothering me.
Electrical power P = IV From Ohm's law, we can determine that P = I^2 R and also P = V^2 / R In one case, P is directly proportional to R. In the other, it's inversely proportional. They are both derived from the same equation. How can this be true? I'm probably misinterpreting the equation, but it's confusing me. Quote:
A multivibrator usually consists of two transistors, each with its own period T, defined by the R and C values. T is measured in seconds. If you want to find the period of a complete cycle (on and off), then you have to find T for each transistors and add them together. 

10172012, 10:30 PM  #8968  
UG's Nu/Shyguy
Join Date: Aug 2006

Quote:
Yeah, but logically speaking, how can resistance apparently be both directly and inversely proportional to power? I'm trying to figure out a decent way to express it. EDIT: We've got two equations, P = I^2 R, and P = V^2/R. These two equations aren't true at the same time  they're individualized cases, one where current is being held constant and one where voltage is being held constant. In P = I^2 R, you're assuming current is being held constant. This essentially gives you the simplified relationship, P = cR, where c is a constant. You can definitely tell now that P is directly proportional to R. Increasing the resistance when current is constant will cause a larger voltage drop (V=IR), and thus more power will be consumed. In P = V^2/R, you're assuming voltage is held constant. This gives you P = c/R, where c is a constant. You can tell that P is inversely proportional to R. Increasing the resistance here when voltage is constant, will decrease current flow (I=V/R), and thus power consumption will be reduced. So, the equations are only really representative of the two cases. If you allow current and voltage to change as resistance changes, you can't say anything determinate about the proportionality of the variables to power. I think. Because P = c*d, where c and d are both variables and may be dependent on each other, you can't determine how c and d are proportional to P. Or something. Last edited by RedDeath9 : 10172012 at 10:58 PM. 

10172012, 10:43 PM  #8969  
Look mum, no brakes!
Join Date: Feb 2005
Location: at home

Quote:
I can see how the equations are derived, but I can't really make sense of them. I was thinking about transformers and transmission lines. A transformer is used to step up the voltage, thereby stepping down the current to reduce I^2 R losses. However, now that the voltage is higher, wouldn't the V^2 / R losses be higher? I dun geddit 

10172012, 10:53 PM  #8970 
UG's Nu/Shyguy
Join Date: Aug 2006

Look at my edit. My mind's still a bit foggy on this stuff.

10172012, 11:22 PM  #8971 
Registered User
Join Date: Dec 2007

You guys might be the wrong crowd to ask, but I don't know where else on this forum I'd ask. In Java, can you define a method within another method?
For example, Code:
If you can define a method within a method, what would be the proper syntax for the inner method, which returns void? And yes I tried google but didn't find anything because it's such a specific question (maybe I just suck at google). I'm still a noob at Java, although fairly experienced with coding in general.
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Last edited by guitarplaya322 : 10172012 at 11:32 PM. 
10172012, 11:31 PM  #8972  
Princess Pineapple
Join Date: Apr 2012

1+2=?
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10172012, 11:33 PM  #8973  
1984
Join Date: Aug 2005
Location: Adelaide, Australia

Quote:
may I ask why the hell you'd want to do that? and I'm 99.9% sure that no you can't, but I'm not entirely certain because I've never thought or heard of anyone ever wanting to do it. EDIT: but seriously, no, of course you can't
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10172012, 11:39 PM  #8974  
UG's Nu/Shyguy
Join Date: Aug 2006

Quote:
Well, you can break it down into steps as such: 1+1+1 = ? First do 1+1... One numerical unit added to 1 equals the number one unit directly after 1; this is two. Therefore, 1+1 = 2. You've still got another 1 to add, however. Repeat the process! (1+1)+1 = 2 + 1 = 3! Hope this helped. Also, followed you on tumblr. 

10172012, 11:40 PM  #8975  
Registered User
Join Date: Dec 2007

Quote:
Alright alright thanks for the answer. The reason is that I want to use private variables from the first function one in the nested one. It is, in fact, possible to do in some languages, so don't act like it's such a ridiculous notion.
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10172012, 11:42 PM  #8976  
Princess Pineapple
Join Date: Apr 2012

Quote:
This made me lol. That pretty much summed it up. And thanks I'll follow you back
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10172012, 11:52 PM  #8977  
Registered User
Join Date: Dec 2007

Quote:
One more java question for you man, this is gonna seem pretty dumb, but I'm a beginner to java and recursive functions in general. How do I write a recursive function where the return type isn't void. Everytime I've tried to write a recursive function with a nonvoid return type, it doesn't work? For example, this function isn't working and I can't figure out why. Code:
Eclipse just tells me, 'This method must return a result of type Node.'
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10192012, 12:40 AM  #8978 
Just Bro'in out
Join Date: Jul 2008
Location: Good Ol' Texas

Hey guys, got a question on my homework that I really have no idea how to start to even do. Any help?
Thanks in advance!
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Where's Waldo? yay 
10192012, 12:46 AM  #8979  
Fighter of the Nightman
Join Date: Jun 2007
Location: College Station, TX

Quote:
Just because you're a fellow Texan... A(w) = .5*Xo*Yo where Xo and Yo are the X and Y intercepts of the tangent line respectively Equation of a line: y=mx+b in terms of w, y = 6w^2 x = w m = 2w so 6w^2 = 2w(w) + b solve for b b = 6 + w^2 To find Xo, set y = 0 0 = 2w(Xo) + 6 + w^2 Xo = (6+w^2)/2w To find Yo, set X = 0 Yo = 6 + w^2 Plug those back into the A(w) equation and plot it on a calculator
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10192012, 12:49 AM  #8980 
yoloswag420
Join Date: Nov 2007
Location: Vancouver, BC

Using a calculator is unnecessary. Minimum value of the area is given when A'(w) = 0 and A''(w) > 0.

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