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        riddles >> putnam exam (pure math) >> Calc 3 help
        
(Message started by: Wardub on Jan 8^th, 2008, 11:27am)

Title: Calc 3 help
Post by Wardub on Jan 8^th, 2008, 11:27am

Hi guys it's me again. I'd like to thank everyone that helped me in Discrete Math. Now I'm taking Calc 3 or multi variable Calc. And i will probably need help in the future. So I'll have problems. My main problem now is I haven't taken Calc in like 8 months and I'm a bit rusty. I remember all the basic derivatives stuff I think :-[ Anyways i need a refresher and I have a few things to ask. Hopefully my parents can find my notes and mail them to me.

I remember there is something special with the exponential function and I forgot how to take the derivative and anti derivative of that.

So could someone explain both of those for f(x) = e^-3x ?

Then I believe the anti derivative of 1/x is ln(x) correct? Just making sure.

Also I want to say the derivative of sin is cosine but i can't remember if its the other way around.

And lastly I can't remember how to do anti derivatives very well. I can do easy things like 6x^2 = 2x^3 but i Can't do more complicated ones. also I forgot integration by parts. So if any of you could help, i would highly appreciate it. I think after I see a few It will all come back hopefully.

Title: Re: Calc 3 help
Post by towr on Jan 8^th, 2008, 3:11pm

on 01/08/08 at 11:27:52, Wardub wrote:

I remember there is something special with the exponential function and I forgot how to take the derivative and anti derivative of that.

So could someone explain both of those for f(x) = e^-3x ?

Two things to know
d/dx e^x = e^x
d/dx e^f(x) = f'(x) e^f(x)

[The latter is just a case of the rule d/dx g(f(x)) = f'(x)g'(f(x)) ]

Quote:

Then I believe the anti derivative of 1/x is ln(x) correct?

Yes

Quote:

Also I want to say the derivative of sin is cosine but i can't remember if its the other way around.

sin -> cos -> -sin -> -cos -> sin
Remember the graphs, sin starts at zero and goes up, so its derivative must be positive at the start, so it must be cos.

Quote:

And lastly I can't remember how to do anti derivatives very well. I can do easy things like 6x^2 = 2x^3 but i Can't do more complicated ones. also I forgot integration by parts. So if any of you could help, i would highly appreciate it. I think after I see a few It will all come back hopefully.

Do you have a specific example?

Title: Re: Calc 3 help
Post by Hippo on Jan 8^th, 2008, 3:56pm

on 01/08/08 at 11:27:52, Wardub wrote:

I can't do more complicated ones. also I forgot integration by parts. So if any of you could help, i would highly appreciate it. I think after I see a few It will all come back hopefully.

It is not so simple in general ;)

Per partes uses formula for derivative of product:
(f.g)'= f'.g+f.g'.
Uf you are looking for "antiderivatives" of f.g', it is f.g - antiderivative of f'.g.

Title: Re: Calc 3 help
Post by Icarus on Jan 9^th, 2008, 7:43am

The big problem with integration by parts is trying to pick your two functions in such a way that you can integrate the one differentiate the other to produce an easier expression. The rule is

http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/int.gifudv = uv - http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/int.gifvdu

or (in a more directly applicable form):

http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/int.giff(x)g'(x) dx = f(x)g(x) - http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/int.giff'(x)g(x) dx

Sometimes, there are obvious ways of writing the expression you are trying to integrate as a product, which allows you to easily figure out how to do something useful:

http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/int.gifxe^x dx = xe^x - http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/int.gif1e^x dx = xe^x - e^x + C (f(x) = x, g'(x) = e^x, g(x) = e^x)

Often, though, you have to be creative to figure out what division of the integrand would be useful:

http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/int.gifln(x) dx = xln(x) - http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/int.gifx/x dx = xln(x) - x + C. (f(x) = ln(x), g'(x) = 1, g(x) = x)

It is very easy to choose poorly, and therefore get something worse than you started with:

http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/int.gifxcos(x) dx = x²cos(x)/2 + http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/int.gifx²sin(x)/2 dx :( (f(x)=cos(x), g'(x) = x, g(x) = x²/2)

Sometimes, you can seem to get nowhere, but in fact, if you do additional integrations by parts, the answer slips out in an unexpected direction:

http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/int.gife^xcos(x) dx = e^xcos(x) + http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/int.gife^xsin(x) dx = e^xcos(x) + e^xsin(x) - http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/int.gife^xcos(x) dx. Solving for the integral gives http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/int.gife^xcos(x) dx = e^x(cos(x) + sin(x))/2 + C.
(first int by parts: f(x)=cos(x), g(x)=g'(x)=e^x; 2nd int by parts: f(x)=sin(x), g(x)=g'(x)=e^x)

Title: Re: Calc 3 help
Post by Wardub on Jan 9^th, 2008, 8:32am

Thanks everyone I am starting to remember :). Ok I knew there was something special about e^x so when taking the anti derivative is it e^(w/e)/ derivative of (w/e) for example e^2x = e^2x/2? Im pretty sure thats how you take them with the expo function.

And I get the cycle for sin to cos to -sin ... so thanks for that.

Lastly Icarus thanks for trying to help me understand integration by parts i still am not fully sure of how to do it. I remember it can often take doing the integration by parts multiple times. Anyway how i was taught was with the U V DU and DV but i cant remember exactly how to do it. For your example of xe^x which is the U? and which is the DV. I remember there was like a rule to which was which. Hmm I'm not explaining myself well. Could you for your example explain which is the U and why? also for the other V and DU/DV Because i think that would help me get it. Thanks again everyone. I'm starting my first chapter of calc 3 tonight, So i might post question but hopefully i get it :) Good thing is homework isn't graded and we have quizzes that are extracredit so I think I should do pretty well I just need to make sure I understand it for the tests. Thanks again everyone.

Title: Re: Calc 3 help
Post by Grimbal on Jan 9^th, 2008, 9:02am

on 01/09/08 at 08:32:12, Wardub wrote:

... so when taking the anti derivative is it e^(w/e)/ derivative of (w/e) for example e^2x = e^2x/2? Im pretty sure thats how you take them with the expo function.

No. Not if "w/e" means "whatever".
(e^f(x))' = e^f(x)·f'(x)
doesn't imply
http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/int.gif(e^f(x)) = e^f(x)/f'(x)

It works in your example because f'(x) is a constant, i.e. f(x) = a·x.

Title: Re: Calc 3 help
Post by pex on Jan 9^th, 2008, 9:06am

on 01/09/08 at 08:32:12, Wardub wrote:

This only works if (w/e) is linear - indeed, an antiderivative of e^ax+b is (1/a)e^ax+b, if a is nonzero.

However, it does not work for nonlinear exponents. Clearly, e^{ln x} = x, so an antiderivative is (1/2)x². Your "rule" would give an antiderivative e^{ln x} / (1/x) = x², which is incorrect.

Taking it even further, consider e^{x^2}. Taking antiderivatives as you propose leads to e^{x^2} / (2x), but it has been proven that no "elementary" antiderivative of e^{x^2} exists at all!

Edit: I'm slow...

Title: Re: Calc 3 help
Post by Icarus on Jan 10^th, 2008, 11:29am

http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/int.gifxe^x dx = xe^x - http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/int.gif1e^x dx = xe^x - e^x + C

u = x; dv = e^xdx (so v = e^x).

http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/int.gifln(x) dx = xln(x) - http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/int.gifx/x dx = xln(x) - x + C.

u = ln(x); dv = dx (so v = x)

http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/int.gifxcos(x) dx = x²cos(x)/2 + http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/int.gifx²sin(x)/2 dx :(

u = cos(x); dv = xdx (so v = x²/2). A much better choice would have been to make u = x; dv = cos(x)dx, but that would spoil my example!

http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/int.gife^xcos(x) dx = e^xcos(x) + http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/int.gife^xsin(x) dx = e^xcos(x) + e^xsin(x) - http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/int.gife^xcos(x) dx. Solving for the integral gives http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/int.gife^xcos(x) dx = e^x(cos(x) + sin(x))/2 + C.

First integration by parts: u = cos(x); dv = e^xdx (so v = e^x)
second integration by parts: u = sin(x); dv = e^xdx (so v = e^x)