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Old 10-17-2012, 12:53 AM   #8961
RedDeath9
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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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Old 10-17-2012, 01:06 AM   #8962
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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!

Last edited by CodChick : 10-17-2012 at 01:15 AM.
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Old 10-17-2012, 01:52 AM   #8963
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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?
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Old 10-17-2012, 02:07 AM   #8964
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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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Old 10-17-2012, 10:55 AM   #8965
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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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Old 10-17-2012, 11:26 AM   #8966
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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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Old 10-17-2012, 10:06 PM   #8967
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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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Old 10-17-2012, 10:30 PM   #8968
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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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Old 10-17-2012, 10:43 PM   #8969
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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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Old 10-17-2012, 10:53 PM   #8970
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Look at my edit. My mind's still a bit foggy on this stuff.
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Old 10-17-2012, 11:22 PM   #8971
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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.
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Old 10-17-2012, 11:31 PM   #8972
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1+2=?
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Old 10-17-2012, 11:33 PM   #8973
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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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Old 10-17-2012, 11:39 PM   #8974
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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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Old 10-17-2012, 11:40 PM   #8975
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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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Old 10-17-2012, 11:42 PM   #8976
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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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Old 10-17-2012, 11:52 PM   #8977
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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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Old 10-19-2012, 12:40 AM   #8978
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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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Old 10-19-2012, 12:46 AM   #8979
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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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Old 10-19-2012, 12:49 AM   #8980
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Using a calculator is unnecessary. Minimum value of the area is given when A'(w) = 0 and A''(w) > 0.
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