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May 10, 2025, 02:12:14 pm

Author Topic: Bucket's Questions  (Read 47789 times)  Share 

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Mao

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Re: Bucket's Questions
« Reply #210 on: July 24, 2008, 08:20:19 pm »
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the equation of a tangent is given by the generic equation


at



since it passes through the origin, (0,0), substituting:






antidifferentiating gives

« Last Edit: July 24, 2008, 08:22:08 pm by Mao »
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Re: Bucket's Questions
« Reply #211 on: July 24, 2008, 08:23:23 pm »
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The equation you have there is the gradient equation for a particular curve, y. With this particular curve, we are told that at , its tangent will intercept the origin. So:




Now, since we know that



...

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Re: Bucket's Questions
« Reply #212 on: July 24, 2008, 08:36:56 pm »
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thankyou my darlings.
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Re: Bucket's Questions
« Reply #213 on: July 24, 2008, 09:35:28 pm »
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help :P
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Re: Bucket's Questions
« Reply #214 on: July 24, 2008, 09:46:00 pm »
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« Last Edit: July 24, 2008, 09:56:36 pm by Toothpick »

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Re: Bucket's Questions
« Reply #215 on: July 24, 2008, 09:50:16 pm »
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thanksssssss
« Last Edit: July 24, 2008, 11:35:08 pm by bucket »
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Re: Bucket's Questions
« Reply #216 on: July 24, 2008, 09:51:27 pm »
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You can anti-differentiate term by term (like differentiation)

How do you integrate ? Or for that matter? Then, .

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Re: Bucket's Questions
« Reply #217 on: July 24, 2008, 09:57:17 pm »
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continuee....i got THAT far :P
eulers constant confuses me >.<
done :)

Mao

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Re: Bucket's Questions
« Reply #218 on: July 25, 2008, 04:15:51 pm »
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help :P




=
=
=
=
=
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=


you could've done it using hyperbolic functions with MUEP/UMEP knowledge :P
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Re: Bucket's Questions
« Reply #219 on: July 27, 2008, 03:28:30 pm »
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 :buck2:
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Re: Bucket's Questions
« Reply #220 on: July 27, 2008, 03:41:31 pm »
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« Last Edit: July 27, 2008, 03:43:57 pm by Toothpick »

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Re: Bucket's Questions
« Reply #221 on: July 27, 2008, 04:11:49 pm »
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ahh i see, so the only way to solve this is to expand the polynomial?
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Mao

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Re: Bucket's Questions
« Reply #222 on: July 28, 2008, 07:05:16 pm »
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ahh i see, so the only way to solve this is to expand the polynomial?

you *could* use linear substitution, but that's beyond the scope of the methods course, and that use of this method for this particularly simple case would be superfluous

the easiest way is to expand the polynomial.
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Re: Bucket's Questions
« Reply #223 on: July 28, 2008, 07:12:48 pm »
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That's one way to do it, but there are surely others, a relatively fast way I thought of for the curious:



Make the substitution,




Where B(m,n) is the Beta function
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Re: Bucket's Questions
« Reply #224 on: July 28, 2008, 08:15:31 pm »
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That's one way to do it, but there are surely others, a relatively fast way I thought of for the curious:



Make the substitution,




Where B(m,n) is the Beta function

Most wouldn't consider it necessary to use the Beta function to integrate a polynomial :P  But then again you are Ahmad :P