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10-17-2012, 12:53 AM   #8961
RedDeath9
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Join Date: Aug 2006
Location: Vancouver, British Columbia
Quote:
 Originally Posted by CodChick Use logarithmic differentiation to find the derivative of the function. y = x^(5x) I can figure it out without logs, but I'm unsure how with logs.

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.
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Last edited by RedDeath9 : 10-17-2012 at 12:57 AM.

10-17-2012, 01:06 AM   #8962
CodChick

Join Date: Jun 2009
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Quote:
 Originally Posted by RedDeath9 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.

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!
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Last edited by CodChick : 10-17-2012 at 01:15 AM.

10-17-2012, 01:52 AM   #8963
metal4eva_22
UG Senior Member

Join Date: Jul 2006
Location: Too Far Gone
Quote:
 Originally Posted by RedDeath9 But doesn't the spring absorb some of the impact/force? I'm sort of clueless about this stuff too. Wait, it doesn't give you a spring constant, so... Not sure why that's there. I also don't know what center of mass means. EDIT: ^Hm, since it's isothermal, change in temperature equals zero. Thus, you also know U = CvdT = w - q = 0. Or in other words, q = w. Uh... Not sure if that's getting us anywhere. You have the work formula for an isothermal process, right? w = integral of PdV, from Va to Vb? I think you might need to replace P with (nRT)/V and then integrate. EDIT2: Aaanddd, you might need to figure out what volumes 100bar and 10bar correspond to. Once again, I might be completely wrong.

But P varies. Are you allowed to do that even though P is not constant?

10-17-2012, 02:07 AM   #8964
RedDeath9
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Join Date: Aug 2006
Location: Vancouver, British Columbia
Quote:
 Originally Posted by metal4eva_22 But P varies. Are you allowed to do that even though P is not constant?

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:
 Originally Posted by CodChick 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!

No problem
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Last edited by RedDeath9 : 10-17-2012 at 02:08 AM.

10-17-2012, 10:55 AM   #8965
hethamulburton
Makes a fool of you.

Join Date: Jun 2004
Location: Quarkville, Thermodynamicus
Quote:
 Originally Posted by metal4eva_22 But P varies. Are you allowed to do that even though P is not constant?

An isothermal, quasi-static/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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Last edited by hethamulburton : 10-17-2012 at 11:07 AM.

 10-17-2012, 11:26 AM #8966 IYanoplathizoI 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)? __________________ ❀ ✿ ❁ ✿ Time on earth is like butterscotch; you really want more, even though it will probably just make you ill. ❀ ✿ ❁ ✿ Certified lurker
10-17-2012, 10:06 PM   #8967
sashki
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:
 Originally Posted by IYanoplathizoI 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)?

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.
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10-17-2012, 10:30 PM   #8968
RedDeath9
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Join Date: Aug 2006
Location: Vancouver, British Columbia
Quote:
 Originally Posted by magnus_maximus P = I V. V = I R, therefore P either equals I^2 R, V^2/R It's just maths brah.

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.
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Last edited by RedDeath9 : 10-17-2012 at 10:58 PM.

10-17-2012, 10:43 PM   #8969
sashki
Look mum, no brakes!

Join Date: Feb 2005
Location: at home
Quote:
 Originally Posted by RedDeath9 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.

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
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 10-17-2012, 10:53 PM #8970 RedDeath9 UG's Nu/Shyguy     Join Date: Aug 2006 Location: Vancouver, British Columbia Look at my edit. My mind's still a bit foggy on this stuff. __________________ My RPG/Soundtrack Music This creature sleeps beyond the flow of time
 10-17-2012, 11:22 PM #8971 guitarplaya322 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: ``` public class Poop { ///variables and constructor and shit public int feces(int k) { //heres where I want to define another method, which returns void } }``` 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. __________________ "I wanna see movies of my dreams" Last edited by guitarplaya322 : 10-17-2012 at 11:32 PM.
10-17-2012, 11:31 PM   #8972
kimberlydawn
Princess Pineapple

Join Date: Apr 2012
1+2=?
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10-17-2012, 11:33 PM   #8973
Diamond Dave
1984

Join Date: Aug 2005
Quote:
 Originally Posted by guitarplaya322 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: ``` public class Poop { ///variables and constructor and shit public int feces(int k) { //heres where I want to define another method, which returns void } }``` 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.

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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10-17-2012, 11:39 PM   #8974
RedDeath9
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Join Date: Aug 2006
Location: Vancouver, British Columbia
Quote:
 Originally Posted by kimberlydawn 1+2=?

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.
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10-17-2012, 11:40 PM   #8975
guitarplaya322
Registered User

Join Date: Dec 2007
Quote:
 Originally Posted by Diamond Dave 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

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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10-17-2012, 11:42 PM   #8976
kimberlydawn
Princess Pineapple

Join Date: Apr 2012
Quote:
 Originally Posted by RedDeath9 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.

This made me lol. That pretty much summed it up. And thanks I'll follow you back
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Quote:
 Originally Posted by Todd Hart So 'crunk and 'gandhi are already pussy-whipped, impressive.

Quote:
 Originally Posted by Burgery you just think they're being mean to you because you have fragile girl feelings

10-17-2012, 11:52 PM   #8977
guitarplaya322
Registered User

Join Date: Dec 2007
Quote:
 Originally Posted by Diamond Dave 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

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 non-void return type, it doesn't work?

For example, this function isn't working and I can't figure out why.
Code:
```
public Node find(int key, Node current) {
if(key == current.keyValue) {
return current;
}
else if(key < current.keyValue) {
if(current.leftChild == null) {
return null;
}
else {
return find(key, current.leftChild);
}
}
else if(key > current.keyValue) {
if(current.rightChild == null) {
return null;
}
else {
return find(key, current.leftChild);
}
}

}```

Eclipse just tells me, 'This method must return a result of type Node.'
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 10-19-2012, 12:40 AM #8978 SLCdragons102 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! __________________ Where's Waldo? yay
10-19-2012, 12:46 AM   #8979
sacamano79
Fighter of the Nightman

Join Date: Jun 2007
Location: College Station, TX
Quote:
 Originally Posted by SLCdragons102 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!

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 = 6-w^2
x = w
m = -2w

so

6-w^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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 10-19-2012, 12:49 AM #8980 Avedas 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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