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March 29, 2024, 04:05:38 am

Author Topic: Mathematics Question Thread  (Read 1296801 times)  Share 

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szetotess

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Re: Mathematics Question Thread
« Reply #4080 on: March 21, 2019, 06:19:09 pm »
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Hello  :)

I was wondering why Simpson's Rule is more accurate in the equation -
f(x)= 0.00446x^4 - 0.01508x^3 - 0.17816x^2 + 0.73131x +1

(i know its a stupid equation but its my given one...)

Ive written a page on why trapezoidal rule introduces errors with calculations to the original graph but Simpson's rule is very hard...

Thanks

jamonwindeyer

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Re: Mathematics Question Thread
« Reply #4081 on: March 21, 2019, 11:46:18 pm »
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Hello  :)

I was wondering why Simpson's Rule is more accurate in the equation -
f(x)= 0.00446x^4 - 0.01508x^3 - 0.17816x^2 + 0.73131x +1

(i know its a stupid equation but its my given one...)

Ive written a page on why trapezoidal rule introduces errors with calculations to the original graph but Simpson's rule is very hard...

Thanks

Hey there! The answer to this one is actually fairly simple - Simpson's Rule uses parabolas to estimate the area. Thus, Simpson's Rule will be more accurate for polynomials with curvature similar to that of a parabola. That's the case for the curve you've presented (graph it on Desmos to see if you like!

emmajb37

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Re: Mathematics Question Thread
« Reply #4082 on: March 23, 2019, 02:26:28 pm »
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Hey, I can't get this question out please help!!
Find the exact gradient of the normal to the curve y = x-e-x  at the point where x = 2.
I am getting some really yucky numbers and don't know why  :-\
Thanks!

AlphaZero

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Re: Mathematics Question Thread
« Reply #4083 on: March 23, 2019, 02:52:50 pm »
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Hey, I can't get this question out please help!!
Find the exact gradient of the normal to the curve y = x-e-x  at the point where x = 2.
I am getting some really yucky numbers and don't know why  :-\
Thanks!

The numbers are a little 'yucky'. \[y=x-e^{-x}\implies\frac{dy}{dx}=1+e^{-x},\] and so at \(x=2\), \[\text{gradient of normal}=\frac{-1}{1+e^{-2}}\]
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emmajb37

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Re: Mathematics Question Thread
« Reply #4084 on: March 23, 2019, 03:10:03 pm »
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Please help!

Use simpson’s rule with 3 function values to find the area bounded by the curve y = ln x , the x-axis and the lines x = 2 and x = 4 .

jamonwindeyer

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Re: Mathematics Question Thread
« Reply #4085 on: March 23, 2019, 03:19:23 pm »
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Please help!

Use simpson’s rule with 3 function values to find the area bounded by the curve y = ln x , the x-axis and the lines x = 2 and x = 4 .

Hey! So the rule is on your reference sheet, should look like this:



It's just finding the three function values (at x=2,3,4) and subbing them in to the formula! ;D

not a mystery mark

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Re: Mathematics Question Thread
« Reply #4086 on: March 24, 2019, 04:27:17 pm »
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Hey, my friend sent me this question and I'm kinda stumped.

Find the area bounded by the curve y=ln(x), the x-axis, and the lines x=2 and x=4.

Thank you!

EDIT: Funnily enough, I just noticed the question above is apart of the same set of questions this is from.
« Last Edit: March 24, 2019, 04:41:03 pm by not a mystery mark »
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fun_jirachi

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Re: Mathematics Question Thread
« Reply #4087 on: March 24, 2019, 04:43:32 pm »
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Hey there!

I know that the integral of ln x isn't on the reference sheet, but for this, consider the derivative of xlnx.



Have a go, it should be a lot easier to integrate ln x now you have this information. (Hint: manipulate the integral so it appears in the form 1 + lnx, then work from there.)

Hope this helps :)
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not a mystery mark

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Re: Mathematics Question Thread
« Reply #4088 on: March 24, 2019, 05:20:04 pm »
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Hope this helps :)

Hey!! Thanks heaps. Just curious if this is a 2U method? Our teacher is so strict on only using advanced methods.
Thanks man.
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jamonwindeyer

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Re: Mathematics Question Thread
« Reply #4089 on: March 24, 2019, 05:39:27 pm »
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Hey!! Thanks heaps. Just curious if this is a 2U method? Our teacher is so strict on only using advanced methods.
Thanks man.

The more "2U" way could be to find the area with respect to the y-axis in that range, then cleverly use it to find the x-referenced area. That is, do:



That will give you an area with respect to the y-axis. If you draw a diagram, you might be able to see how that might help you - You'll need the area of two other rectangles to make it work! ;D

RuiAce

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Re: Mathematics Question Thread
« Reply #4090 on: March 24, 2019, 05:44:30 pm »
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It's actually more complicated than that, I found.



The bounding by \(x=2\) wrecks things up a bit. I had to do the usual method twice to get the answer.

Edit: Actually I think I was tired. It could be done more quickly with \( \int_{\ln 2}^{\ln 4}e^y\,dy\). Just need to consider the \(2\ln 2\) rectangle differently.

I think it's something like \( 2\ln 2 + \int_{\ln 2}^{\ln 4}e^y\,dy = 4\ln 4\)?
« Last Edit: March 24, 2019, 05:47:22 pm by RuiAce »

jamonwindeyer

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Re: Mathematics Question Thread
« Reply #4091 on: March 24, 2019, 05:54:17 pm »
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Yep, so if the integral I referred to above is \(I\), the area we want is:


not a mystery mark

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Re: Mathematics Question Thread
« Reply #4092 on: March 24, 2019, 06:29:39 pm »
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Yep, so if the integral I referred to above is \(I\), the area we want is:




Ahhh, sneaky. Haha. Thanks man, I'll pass the working on :)
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Re: Mathematics Question Thread
« Reply #4093 on: March 24, 2019, 07:31:11 pm »
+1
Hey!! Thanks heaps. Just curious if this is a 2U method? Our teacher is so strict on only using advanced methods.
Thanks man.

Probably not a 2U method. If your teacher is like that, definitely use Jamon/Rui's method. However, this method is still pretty useful for finding unfamiliar integrals because it's based off the intuition that every derivative gives us another integral ie. you find something with a derivative pretty close to the thing in the integral, then use that information to find the integral. Basically it's still pretty useful for scoping out some integrals, but otherwise stick to some more familiar methods :)
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spnmox

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Re: Mathematics Question Thread
« Reply #4094 on: March 25, 2019, 08:06:34 pm »
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I need some help!

The curve y=2^x is rotated about the x-axis between x=1 and x=2. Use Simpson's value with 3 function values to find an approximation of the volume of the solid formed to 2 s.f.

So I calculated the area and squared it and multiplied by pi to get 26.2, but answers say 27.2 Typo or ?

and: Find the exact area bounded by the parabola y=x^2 and the line y=4-x.

I found the x-intercepts to be (-1+-sqr17)/2. So when I sub into the integral, I have to expand those big fractions?? Am I doing this right/is there an easier method