Learning calculus

[rockytriton]

Hi everyone, I'm trying to teach myself calculus here and I'm having some algebra problems, I hope you can help me out here.

 

 

I'm having trouble seeing how I get from

 

[−1/(1 + h)^2] − [−1/1^2]

----------------------------

h

 

to

 

−1 − [−(1 + h)^2]

--------------------

h(1 + h)^2

 

 

I know that I should know this stuff before tackling calc but I remember things things all the time. Also I see that it goes from the above form to:

 

2h + h^2

------------------

h + 2h^2 + h^3

 

The bottom part I can understand as first doing (1 + h)^2 as 1 + 2h + h^2 and then multiplying each term by h, but I'm not sure on the top part what went on

 

Thanks in advance!

 

:)

[maddog]

Hi everyone, I'm trying to teach myself calculus here and I'm having some algebra problems, I hope you can help me out here.

 

 

I'm having trouble seeing how I get from

 

[−1/(1 + h)^2] − [−1/1^2]

----------------------------

h

 

to

 

−1 − [−(1 + h)^2]

--------------------

h(1 + h)^2

 

 

I know that I should know this stuff before tackling calc but I remember things things all the time. Also I see that it goes from the above form to:

 

2h + h^2

------------------

h + 2h^2 + h^3

 

The bottom part I can understand as first doing (1 + h)^2 as 1 + 2h + h^2 and then multiplying each term by h, but I'm not sure on the top part what went on

 

Thanks in advance!

Rocky,

 

Your first formula

 

(-1/(1 + h)^2 - (-1/1^2)) / h first simplifies rewritting it a bit (by using distributive property)

 

-1/ h(1 + h)^2 + 1 / h Now make denominator look same by multiplying right term

 

by (1 + h)^2/(1+h)^2

 

(-1 + (1 + h)^2 ) / h(1 +h)^2 <-- to here by gathering terms in numerator

 

This gives the first interim solution multiplying out as

 

(-1 + 1 + 2h + h^2 ) / h(1 + 2h + h^2) =>

 

(2h + h^2) / (h + 2h^2 + h^3) <--- to get your answer.

 

However you can factor more to

 

(h^2 + 2) / (h + 1)^2 <-- to here.

 

Hope this helps. :)

 

Maddog

[rockytriton]

Ok, I think this is the part that I'm having problems with:

 

going from

 

-1/(1 + h)^2 - -1/1^2

----------------------------

h

 

to

 

-1

--------- + 1

h(1 + h)^2

---------------------

h

 

 

I understand how you get "- -1/1^2" to be + 1, but I don't understand how (1 + h)^2 changes to h(1 + h)^2.

 

Sorry, I guess I need a good refresher course on division with fractions and such... I really appreciate your help here.

[maddog]

 

[?1/(1 + h)^2] ? [?1/1^2]

----------------------------

h

 

to

 

?1 ? [?(1 + h)^2]

--------------------

h(1 + h)^2

 

Rocky,

 

Initially --> {[-1 / (1 + h)^2] - [ - 1/1^2] } / h <-- first I'll split in two for you

 

[ -1 / (1 + h)^2] / h - [ -1/1^2] / h <--- simplifying

 

-1 / h(1 + h)^2 + 1/h <-- now to get common terms I need to multiply second by 1

 

1 = (1 + h)^2 / (1 + h)^2 <-- so this comes out to

 

( -1 + (1 + h)^2 ) / h(1 + h)^2 = (-1 - ( - (1 + h)^2) ) / h(1 + h)^2

 

which is your second formula above.

 

I would recommend you go buy a Schaum Outline series on Algebra (NOT Abstract)

(you will not understand that). There are a lot of tricks in there you can use. ;D

 

maddog

[TINNY]

can anyone help me obtain y in terms of x here:

 

dy/dx = (y^2 + y)/(sin x)

 

I couldn't manage to use separable differentiation because i couldn't integrate 1/sin x

[rockytriton]

Rocky,

 

Initially --> {[-1 / (1 + h)^2] - [ - 1/1^2] } / h <-- first I'll split in two for you

 

[ -1 / (1 + h)^2] / h - [ -1/1^2] / h <--- simplifying

 

-1 / h(1 + h)^2 + 1/h <-- now to get common terms I need to multiply second by 1

 

1 = (1 + h)^2 / (1 + h)^2 <-- so this comes out to

 

( -1 + (1 + h)^2 ) / h(1 + h)^2 = (-1 - ( - (1 + h)^2) ) / h(1 + h)^2

 

which is your second formula above.

 

I would recommend you go buy a Schaum Outline series on Algebra (NOT Abstract)

(you will not understand that). There are a lot of tricks in there you can use. ;D

 

maddog

 

Thanks, I have a much better understanding now. I do have the Schuam Outline on College Algebra, maybe I should actually open it up. I bought it a few years ago while taking college algebra.

[maddog]

can anyone help me obtain y in terms of x here:

 

dy/dx = (y^2 + y)/(sin x)

 

I couldn't manage to use separable differentiation because i couldn't integrate 1/sin x

To solve for (y^2 + y)/sin x ? Hmm, ok.

 

int(dy) = int((y^2 + y)/sin x * dx)

 

You will need a substitution where y = f(x) or x = g(y) to proceed... hmmm

 

int((1/(y^2 + y)dy) = int((1/sin x)dx) <-- no I need no substitution as before. Instead I need to

simply by using substitution.

 

The right side can be found in a book on calculus (I'm a bit rusty). The left you can do by simple

substitution so try it out.

 

maddog

[Bo]

in my integral books there is no exact olution for 1/sin(x) (only a recursive solution)...

I also cant find a better solution, so what do you need this for tinny? Things would get much simpeler if one can take sin(x)~x for example

 

Bo

[Qfwfq]

Things would get much simpeler if one can take sin(x)~x for example

But that's cheating!!!!!!!! :)

[Bo]

No that's called 'making life easier'. :) Knowing what you can approximate with something easy is one of the most important things you do in physics.

 

Bo

[Qfwfq]

Yeah, that means that, if you equate the other primitive to ln x you get a set of solutions which are tangent to the "exact" ones where they go through the y axis, but wouldn't they be singular there anyway? At the most this boils down to having calculated the tan of an angle for each y value. Considering also the additive constant, I'd say you would hardly have any meat left to sell.

 

Only thinking in my head, no scribbling on paper. :)

[cindy 2005]

can anyone help me obtain y in terms of x here:

 

dy/dx = (y^2 + y)/(sin x)

 

I couldn't manage to use separable differentiation because i couldn't integrate 1/sin x

 

 

Take integral of 1/sinx we need to do as follows:

 

Int 1/sinx dx = Int [sinx/(sinx)^2 ]dx = Int [sinx/(1-(cosx)^2 ]dx

Now

setting u = cosx ==> du = -sinxdx; therefore

Int -du/(1-u^2). You can get this result from Integral table.

[TINNY]

Take integral of 1/sinx we need to do as follows:

 

Int 1/sinx dx = Int [sinx/(sinx)^2 ]dx = Int [sinx/(1-(cosx)^2 ]dx

Now

setting u = cosx ==> du = -sinxdx; therefore

Int -du/(1-u^2). You can get this result from Integral table.

great! but what is Integral table? how do you integrate 1/(1-u^2)? it isn't ln|1-u^2| right?

[Qfwfq]

great! but what is Integral table? how do you integrate 1/(1-u^2)? it isn't ln|1-u^2| right?

It's right if you can neglect d(u^2)/du but this only holds around x = 0 and around any other x = k*pi. Try "Schaum's Mathematical Handbook of Formulas

and Tables" for the integral.

 

Good idea, Cindy.

[cindy 2005]

It's right if you can neglect d(u^2)/du but this only holds around x = 0 and around any other x = k*pi. Try "Schaum's Mathematical Handbook of Formulas

and Tables" for the integral.

 

Good idea, Cindy.

 

If you don't use integral table and continue working on the one that I mentioned by factoring the expression

Int (-du/(1-u^2)) = - Int [(1/(2(u+1)) - 1/(2(u-1))]du

Therefore, applying Int (1/(x+a))dx = ln(x+a), you obtain

Int (-du/(1-u^2)) = -[1/2 ln(u+1) - 1/2ln(u-1)]=-1/2 ln(u+1)/(u-1) = 1/2 ln(u-1)/(u+1)

Finally, by substituting u = cosx, you get

1/2 ln(cosx-1)/(cosx+1)

you can put it in term of csc and cot functions

[Qfwfq]

Clever stuff, Cindy!

[C1ay]

Has anyone here tried running any of these through the Integrator? I've found it handy from time to time.