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April 20, 2019, 10:37:30 am

Author Topic: 4U Maths Question Thread  (Read 229228 times)  Share 

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goodluck

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Re: 4U Maths Question Thread
« Reply #2250 on: April 08, 2019, 02:50:59 pm »
0
hey how do you do this one:

h/(x+h) < ln(x+h) - ln(x) <h/x . (it's meant to be greater than and equal to btw)

RuiAce

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Re: 4U Maths Question Thread
« Reply #2251 on: April 08, 2019, 03:13:57 pm »
+1
hey how do you do this one:

h/(x+h) < ln(x+h) - ln(x) <h/x . (it's meant to be greater than and equal to btw)
Hint: You can use a sketch to verify the following inequality.
\[ h \times \frac{1}{x+h} < \int_x^{x+h} \frac{1}{t}dt < h\times \frac{1}{x} \]

david.wang28

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Re: 4U Maths Question Thread
« Reply #2252 on: April 13, 2019, 10:01:55 pm »
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Hello,
I am stuck on Example 7 part a) in the link below. I have also attached my working out. Can anyone please help me out? Thanks :)
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RuiAce

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Re: 4U Maths Question Thread
« Reply #2253 on: April 13, 2019, 10:40:29 pm »
+2
Hello,
I am stuck on Example 7 part a) in the link below. I have also attached my working out. Can anyone please help me out? Thanks :)
Two things:
\[ \text{Firstly, you've correctly noted that }A = \frac12 x^2 \tan \alpha.\\ \text{So your volume of a cross section should just be }\boxed{\delta V = \frac12 x^2 \tan \alpha\, \delta y}\\ \text{i.e. }\boxed{\delta V = \frac12 (r^2-y^2)\tan \alpha\, \delta y} \]
Not sure why you multiplied \(x^2\) to another \(r^2-y^2\) term in your expression. By doing so, because of the fact that \(x^2+y^2=r^2\), you're basically trying to deal with \(x^4\), or equivalently \((r^2-y^2)^2\)
\[ \text{Secondly, what is }a?\\ \text{The boundaries of integration (and also the sum) are just }-r\text{ and }r \]

david.wang28

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Re: 4U Maths Question Thread
« Reply #2254 on: April 15, 2019, 03:25:52 pm »
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Two things:
\[ \text{Firstly, you've correctly noted that }A = \frac12 x^2 \tan \alpha.\\ \text{So your volume of a cross section should just be }\boxed{\delta V = \frac12 x^2 \tan \alpha\, \delta y}\\ \text{i.e. }\boxed{\delta V = \frac12 (r^2-y^2)\tan \alpha\, \delta y} \]
Not sure why you multiplied \(x^2\) to another \(r^2-y^2\) term in your expression. By doing so, because of the fact that \(x^2+y^2=r^2\), you're basically trying to deal with \(x^4\), or equivalently \((r^2-y^2)^2\)
\[ \text{Secondly, what is }a?\\ \text{The boundaries of integration (and also the sum) are just }-r\text{ and }r \]
I realised where I went wrong; thanks :)
HSC 2019: English Advanced(77) (Forgot everything), Chemistry, Physics, Maths Extension 1(35) (gonna repeat), Maths Extension 2, Business Studies(80) (screw this)

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