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The Math Help Thread

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Beaten to the draw, took longer to type and upload. Here are more steps:
EXAMPLE.png

fixed
 
SniperHunter said:
Ok, anyone what to have a go at this?

tanѲ+cscѲcosѲ = 1/(sinѲcosѲ)
Click to expand...

write left side in sin & cos form and put them over the same numerator and you'll get an easy trig property in the deniominator which will reduce to what you have on right side. JustProgress gives away the surprise. >;O
 
Partial Gamification said:
Beaten to the draw, took longer to type and upload. Here are more steps:
Trig_Example.png

fixed image...hmmm
Click to expand...

wow that great, thanks a lot!! Don't get yourself banned cause I see you helping a lot of people here. You are the best junior member ever!

I need to get a lot of practice with trig id's again, the next pre calculus chapter I have to get through is about trig ids and radians :P

Again, thanks for posting the informative links.

JustProgress said:
(Sin x/cos x) + (cos x/sin x)

(Sin^2 x + cos^2 x)/(sin xcos x)

1/(sin xcos x)

If you need more steps lemme know
Click to expand...

Thanks!
 
Zoibie said:
Just for curiosity's sake, there's a simpler way of doing this isn't there? At the second step, multiply both sides by sinѲcosѲ and you end up with a trig identity.
Click to expand...

Not just any trig identity, the trig identity: the Pythagorean Theorem:
Sine-squared plus cosine-squared equals one!

For your approach to work, it just depends on what the question is requesting. This simplified version is fine if performing operations on both sides of the equation is allowed, and it results is a well-known theorem. If one is showing the right-hand-side is equal to the left-hand-side, operations are typically only performed on the right-hand-side, or vice-versa. The simplest reason why is the trivial case of multiplying both sides by zero, zero equals zero results and doesn't really reveal anything; but again, what you said ought to be a valid approach, in this case.
 
Ok, anyone what to have a go at this?

tanѲ+cscѲcosѲ = 1/(sinѲcosѲ)
Click to expand...

Partial Gamification said:
Beaten to the draw, took longer to type and upload. Here are more steps:
LOTv1.png

fixed
Click to expand...

Just tried it...

You made this so simple!! It's an easy question, I can't believe I got it wrong. I guess my brain wasn't working while I was writing the exam (I still managed to get 88% though).

Anyway, nice to review this since I will be doing trig id's with radians soon. :)
 
SniperHunter said:
Ok, learning about trig ratios and finding exact values. Can someone tell me how they are getting the second step? Book says the ratios have to be written using the special angles from the 30, 60, 90 triangle or 45, 45, 90 triangle (now in radians of course)
rest of the process is simple but I am so confused about the first step. Please help!
Click to expand...
Uh...what? This is basic fractions 101.

3/12 + 4/12 = 7/12.

They're just splitting the fraction. That's all.
 
Seems like "basic splitting" of the values to me. So you get the numbers into a form you can transform.

7/12 = 4/12 + 3/12 and then you can cancel 4 and 3 with 12 to make your equation easier. Sames goes for the other values. This is basicly something you have to "see".

For c.)

13pi/12 is the same as pi*(3/12 + 10/12) = pi * (1/4 + 5/6) = ...

There are of course ways and formulas to reduce a fraction mathematically but you are faster if you do such easy calculations by hand.

EDIT: beaten and changed pie to pi all those numbers have made me hungry :)
 
Feep said:
Uh...what? This is basic fractions 101.

3/12 + 4/12 = 7/12.

They're just splitting the fraction. That's all.
Click to expand...

systemfehler said:
Seems like "basic splitting" of the values to me. So you get the numbers into a form you can transform.

7/12 = 4/12 + 3/12 and then you can cancel 4 and 3 with 12 to make your equation easier. Sames goes for the other values. This is basicly something you have to "see".

For c.)

13pi/12 is the same as pi*(3/12 + 10/12) = pi * (1/4 + 5/6) = ...

There are of course ways and formulas to reduce a fraction mathematically but you are faster if you do such easy calculations by hand.

EDIT: beaten and changed pie to pi all those numbers have made me hungry :)
Click to expand...

thanks for the help!
 
GAF, I need a hand. Could someone show the steps for the following:

g(x) = sin(x^2)/x
g'(x) = ?


I know and feel it's real easy but I can't wrap my mind around it at the moment.
 
Danne-Danger said:
GAF, I need a hand. Could someone show the steps for the following:

g(x) = sin(x^2)/x
g'(x) = ?


I know and feel it's real easy but I can't wrap my mind around it at the moment.
Click to expand...

g'(x) = {[sin(x^2)]'*x - [x]'*sinx(^2)}/x^2

<=>

g'(x) = [cos(x^2)*[x^2]'*x - sin(x^2)]/x^2

<=>

g'(x) = [cos(x^2)*2x*x - sin(x^2)]/x^2

<=>

g'(x) = [2*cos(x^2)*x^2 - sin(x^2)]/x^2

:)

--EDIT:

wolframalpha gives the simplified form:

g'(x) = 2*cos(x^2) - sin(x^2)/x^2
 
Danne-Danger said:
GAF, I need a hand. Could someone show the steps for the following:

g(x) = sin(x^2)/x
g'(x) = ?


I know and feel it's real easy but I can't wrap my mind around it at the moment.
Click to expand...

Quotient rule states that:
g(x)/h(x) where:
g'(x)(h(x)- h'(x)(g(x) / h(x)^2
sin(x^2) will be our g(x). Let x be our h(x).
So knowing this, let's take the derivatives of the individual functions:
sin(x^2)-> 2x(cos(x^2)
x-> 1
So let's simply plug our values in:
[2x(cos(x^2)*x - sin(x^2)] / x^2 ->[2x^2(cos(x^2) - sin(x^2)] / x^2
And we're done! We can simply it to:
2cos(x^2)- sin(x^2)/x^2
Edit: wait, I made a mistake somewhere. This what happens when I do it in my head...one second.
Edit2: Now I see it. I forgot the x. lol Either way, I was beaten.
 
mohsinkhan said:
For quotient rule, I usually take the denominator to the numerator and turn it into a product rule problem.
Click to expand...

g''(x)=-7(-4(x-2)^-3))

g''(x)=28/((x-2)^3)

I think, sorry just got up, but remember g(x)h(x)= g'(x)h(x)+g(x)h'(x), derivatives of constants are zero. Hope that helps.
 
mohsinkhan said:
g''(x)=-7(-4(x-2)^-3))

g''(x)=28/((x-2)^3)

I think, sorry just got up, but remember g(x)h(x)= g'(x)h(x)+g(x)h'(x), derivatives of constants are zero. Hope that helps.
Click to expand...

it's actually 7/(x-2)^3 by the chain rule ( the derivative of (-7/2) (x-2)^(-2) is just (-2)(-7/2) (x-2)^(-3) = 7/(x-2)^3) ) unless i'm interpreting the problem incorrectly
 
Goya said:
it's actually 7/(x-2)^3 by the chain rule ( the derivative of (-7/2) (x-2)^(-2) is just (-2)(-7/2) (x-2)^(-3) = 7/(x-2)^3) ) unless i'm interpreting the problem incorrectly
Click to expand...

so only the chain rule is used, interesting. I will do the problem again.

thanks for the help
 
SniperHunter said:
so only the chain rule is used, interesting. I will do the problem again.
Click to expand...

Well if you look at where the exponent is, only the (x-2) comes to the numerator, if you still want to use product rule:

-7(x-2)^-2/2

g(x)=-7/2
h(x)=(x-2)^-2

following product rule, you get = 7/((x-2)^3)
 
Can someone please help me understand how these derivatives were achieved?

1.
vOuwU.jpg


2.
eC94N.jpg


learning optimization problems at the moment, if you get the derivative wrong, the entire answer will be wrong :/ I need to get a good grasp on these.

How did you guys become expert at deriving derivatives?
 
I can't see the pictures right now, but I looked over them and again they look like applications of the chain rule. The first one had something like (x^2 + 10000)^1/2 + (300-x), so in the first term there was a derivative of the outer function that brought down a factor of 1/2 and reduced the exponent to -1/2. The second term added on should just give you a constant after differentiating.

The second one is a similar thing. They bring the denominator up and wrote it as (something + (.6)^t)^-1. Take the derivative of the outer function which is like X^-1 and you get the factor of -1 out front and an exponent of -2. They then used the rule for derivatives of numbers other than e being raised to the power of x, which is where the factor of ln(.6) came from.
 
krameriffic said:
I can't see the pictures right now, but I looked over them and again they look like applications of the chain rule. The first one had something like (x^2 + 10000)^1/2 + (300-x), so in the first term there was a derivative of the outer function that brought down a factor of 1/2 and reduced the exponent to -1/2. The second term added on should just give you a constant after differentiating.

The second one is a similar thing. They bring the denominator up and wrote it as (something + (.6)^t)^-1. Take the derivative of the outer function which is like X^-1 and you get the factor of -1 out front and an exponent of -2. They then used the rule for derivatives of numbers other than e being raised to the power of x, which is where the factor of ln(.6) came from.
Click to expand...

Thanks for answering so promptly. BTW how did you get so good at solving derivatives? My book is really skimpy on the explanation and examples. I tried searching online but the examples were too easy or not what I was looking for :/
 
SniperHunter said:
any other tips for derivatives?

btw how do I draw a graph of y=xsinx?
Click to expand...

To draw any type of (function)*trig, like xsinx, trace out the positive, an negative versions of the function, and draw sin x inside it. The points where the graph hit the x axis are the same, so use that for reference.

7dde19f4a20b7dad087960f05fd319f0_1482555.jpg


Hey, look, I'm now a member. When did that happen?
 
SniperHunter said:
BTW how did you get so good at solving derivatives?
Click to expand...
By taking a lot of them >> There are only so many rules and it is usually easy to identify when each is relevant once you've had some practice. If you don't know how to take the derivative of a function but you think you should be able to, try rearranging it to a form you can recognize.

This site can be very helpful for all sorts of calculus operations. If you want it to take a derivative, you can just type d/dx(function to derive). If there are multiple steps, you can click "Show steps" to see them. You can make it graph things with Plot[function].
 
I'm going to be taking Physical Chemistry so i need to practice differential equations. It's been a long time since i've taken that class. Does anyone know of a good site to practice them?

Thank you.
 
salva said:
I'm going to be taking Physical Chemistry so i need to practice differential equations. It's been a long time since i've taken that class. Does anyone know of a good site to practice them?

Thank you.
Click to expand...
The same one that I posted just now will solve them for you, if that helps. Just input dy/dx = 1/x, for example, and it will recognize it as a diffeq and spit out a general solution. It will probably do most of what you need for a Pchem class, but the steps it takes for higher-order and otherwise complicated equations can sometimes be hard to follow/different from what you'll probably have learned.

My ODE class was taught out of a book written by the professor, so I don't really know of any good sites for practice problems, if that's what you're looking for. Maybe get your hands on a Pchem book and see what comes up? I don't know what sorts of equations you'd usually see in that application..
 
SniperHunter said:
That's amazing! Good luck with your studies.
Click to expand...

thanks! :)


Shimesaba said:
The same one that I posted just now will solve them for you, if that helps. Just input dy/dx = 1/x, for example, and it will recognize it as a diffeq and spit out a general solution. It will probably do most of what you need for a Pchem class, but the steps it takes for higher-order and otherwise complicated equations can sometimes be hard to follow/different from what you'll probably have learned.

My ODE class was taught out of a book written by the professor, so I don't really know of any good sites for practice problems, if that's what you're looking for. Maybe get your hands on a Pchem book and see what comes up? I don't know what sorts of equations you'd usually see in that application..
Click to expand...

Yes, i know of WolframAlpha. Awesome site! But i was asking if there's a place with lots of differential equation problems (of all kinds, specially partial integrals and derivatives) to practice with.
 
salva said:
Yes, i know of WolframAlpha. Awesome site! But i was asking if there's a place with lots of differential equation problems (of all kinds, specially partial integrals and derivatives) to practice with.
Click to expand...

yea Wolframalpha is amazing and I too was looking for a lot of derivative problems. I don't like studying near a computer. There's next to nothing in this book.....

edit: nm looks like Partial Gamification posted a good link on top of the page.
 
For calc 1-type differentiation/integration problems, the UC Davis link the guy at the top of the page posted looked really good to me.

For salva, I'd like to be able to help you, but what are you talking about? "Partial integral/derivative" differential equations would be solved using methods from PDEs, but I'll be shocked if you are expected to know that material for an undergrad chemistry class. Are you sure that's what you need to study...?
 
Alright I need help with direction fields. I can do simple ones where the x axis does not change, like

y' = y(y-2)^2

I find where the slopes are zero and then find what the slopes are doing in between out side of said slopes to get an idea of the integral curves.

What do I do when the slopes change over the x axis? The equation is particular is...

ty' - y = t^2e^-t

Doing it in the same fashion sounds like it would take a very long time to create a decent dfield.
 
Matlab help guys:

The application that we are hoping to run through Matlab is called Sequence-based Prediction of Protein Interaction Sites with an Integrative Method by Dr. Chen and Dr. Jeong.
http://www.ittc.ku.edu/~xwchen/bindingsite/prediction.htm
and link to the classifier: http://www.ittc.ku.edu/~xwchen/bindingsite/register.php?file=randomforest.zip
and the original paper: http://bioinformatics.oxfordjournals.org/cgi/content/full/btp039/DC1

I can't seem to figure out on how to get it work. Would any of you guys will kindly be able to help me out. :(

I am a total novice with Matlab and I have no clue on how to use it.

Thank you guys in advance.
 
As you'll be able to tell form my question, I'm far from a math advanced student so please bare with me. Currently I need help with difference quotients. Now mind you I don't just need the answers, I can look in the back of the book for that. I need step by step how the heck you managed to get it done if at all possible. Here's the problem the best I can explain it.

Find and simplify the difference quotient
f ( x + h) - f (x)
------------------
h

f (x) = x^2 -4x +3


What I did (that's incredibly wrong apparently)
(x -4x +3) ^2 = (3x +3)^2 = 9x^2 +9x +9x +9 = 9x^2 +18x +9
9x^2 +18x +9 - (x^2 -4x +3) = 8x^2 +22x +6
-----------------
h

Hopefully that makes sense, if not I can try to explain more fully. Thanks in advance.
 
Taking a mandatory (Psych) Stats class, so this thread might become very useful down the line. I spent all of today relearning the necessary algebra for Stats. And I read a little bit more about the overall concept of Psych Stats.

A little scared, took a lot longer than I expected to finish all of my homework for the first week. But I acquired an average of a 92% for the first batch of assignments, so there is hope that I can get an A.
 
I have a calc problem that I need some help with.

lim (2-2e^(-x)) / ((e^x)-1)
x->0


I know that the answer is 2, but I have no idea why... any help would be greatly appreciated!
 
Ya no said:
I have a calc problem that I need some help with.

lim (2-2e^(-x)) / ((e^x)-1)
x->0


I know that the answer is 2, but I have no idea why... any help would be greatly appreciated!
Click to expand...

The easiest way to do this looks like L'Hopital's rule. If you plug in the number and you get 0/0 or Infinity/Infinity, you take the derivative of the top and the bottom and then take the limit. So you get

lim (2e^(-x))/(e^x)
x --> 0

you can plug that in and you get 2

http://en.wikipedia.org/wiki/L'Hospital_rule
 
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